**Objectives**

After completing this chapter, you should be able to

**1**Determine sample spaces and find the probability of an event, using classical probability or empirical probability.

**2**Find the probability of compound events, using the addition rules.

**3**Find the probability of compound events, using the multiplication rules.

**4**Find the conditional probability of an event.

**5**Find the total number of outcomes in a sequence of events, using the fundamental counting rule.

**6**Find the number of ways that *r* objects can be selected from *n* objects, using the permutation rule.

**7**Find the number of ways that *r* objects can be selected from *n* objects without regard to order, using the combination rule.

**8**Find the probability of an event, using the counting rules.

**Outline**

**Introduction**

**4–1Sample Spaces and Probability**

**4–2The Addition Rules for Probability**

**4–3The Multiplication Rules and Conditional Probability**

**4–4Counting Rules**

**4–5Probability and Counting Rules**

**Summary**

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**Statistics Today**

**Would You Bet Your Life?**

Humans not only bet money when they gamble, but also bet their lives by engaging in unhealthy activities such as smoking, drinking, using drugs, and exceeding the speed limit when driving. Many people don’t care about the risks involved in these activities since they do not understand the concepts of probability. On the other hand, people may fear activities that involve little risk to health or life because these activities have been sensationalized by the press and media.

In his book *Probabilities in Everyday Life* (Ivy Books, p. 191), John D. McGervey states

*When people have been asked to estimate the frequency of death from various causes, the most overestimated categories are those involving pregnancy, tornadoes, floods, fire, and homicide. The most underestimated categories include deaths from diseases such as diabetes, strokes, tuberculosis, asthma, and stomach cancer (although cancer in general is overestimated)*.

The question then is, Would you feel safer if you flew across the United States on a commercial airline or if you drove? How much greater is the risk of one way to travel over the other? See Statistics Today—Revisited at the end of the chapter for the answer.

In this chapter, you will learn about probability—its meaning, how it is computed, and how to evaluate it in terms of the likelihood of an event actually happening.

**Introduction**

A cynical person once said, “The only two sure things are death and taxes.” This philosophy no doubt arose because so much in people’s lives is affected by chance. From the time you awake until you go to bed, you make decisions regarding the possible events that are governed at least in part by chance. For example, should you carry an umbrella to work today? Will your car battery last until spring? Should you accept that new job?

**Probability** as a general concept can be defined as the chance of an event occurring. Many people are familiar with probability from observing or playing games of chance, such as card games, slot machines, or lotteries. In addition to being used in games of chance, probability theory is used in the fields of insurance, investments, and weather forecasting and in various other areas. Finally, as stated in Chapter 1, probability is the basis of inferential statistics. For example, predictions are based on probability, and hypotheses are tested by using probability.

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The basic concepts of probability are explained in this chapter. These concepts include *probability experiments, sample spaces*, the *addition* and *multiplication rules*, and the *probabilities of complementary events*. Also in this chapter, you will learn the rule for counting, the differences between permutations and combinations, and how to figure out how many different combinations for specific situations exist. Finally, Section 4–5 explains how the counting rules and the probability rules can be used together to solve a wide variety of problems.

Objective 1

Determine sample spaces and find the probability of an event, using classical probability or empirical probability.

**4–1Sample Spaces and Probability**

The theory of probability grew out of the study of various games of chance using coins, dice, and cards. Since these devices lend themselves well to the application of concepts of probability, they will be used in this chapter as examples. This section begins by explaining some basic concepts of probability. Then the types of probability and probability rules are discussed.

**Basic Concepts**

Processes such as flipping a coin, rolling a die, or drawing a card from a deck are called *probability experiments*.

A **probability experiment** is a chance process that leads to well-defined results called outcomes.

An **outcome** is the result of a single trial of a probability experiment.

A trial means flipping a coin once, rolling one die once, or the like. When a coin is tossed, there are two possible outcomes: head or tail. (*Note:* We exclude the possibility of a coin landing on its edge.) In the roll of a single die, there are six possible outcomes: 1, 2, 3, 4, 5, or 6. In any experiment, the set of all possible outcomes is called the *sample space*.

A **sample space** is the set of all possible outcomes of a probability experiment.

Some sample spaces for various probability experiments are shown here.

Experiment | Sample space |

Toss one coin | Head, tail |

Roll a die | 1, 2, 3, 4, 5, 6 |

Answer a true/false question | True, false |

Toss two coins | Head-head, tail-tail, head-tail, tail-head |

It is important to realize that when two coins are tossed, there are *four* possible outcomes, as shown in the fourth experiment above. Both coins could fall heads up. Both coins could fall tails up. Coin 1 could fall heads up and coin 2 tails up. Or coin 1 could fall tails up and coin 2 heads up. Heads and tails will be abbreviated as H and T throughout this chapter.

Example 4–1

Rolling Dice

Find the sample space for rolling two dice.

Solution

Since each die can land in six different ways, and two dice are rolled, the sample space can be presented by a rectangular array, as shown in Figure 4–1. The sample space is the list of pairs of numbers in the chart.

Figure 4–1

Sample Space for Rolling Two Dice (Example 4–1)

Example 4–2

Drawing Cards

Find the sample space for drawing one card from an ordinary deck of cards.

Solution

Since there are 4 suits (hearts, clubs, diamonds, and spades) and 13 cards for each suit (ace through king), there are 52 outcomes in the sample space. See Figure 4–2.

Figure 4–2

Sample Space for Drawing a Card (Example 4–2)

Example 4–3

Gender of Children

Find the sample space for the gender of the children if a family has three children. Use B for boy and G for girl.

Solution

There are two genders, male and female, and each child could be either gender. Hence, there are eight possibilities, as shown here.

BBB BBG BGB GBB GGG GGB GBG BGG

In Examples 4–1 through 4–3, the sample spaces were found by observation and reasoning; however, another way to find all possible outcomes of a probability experiment is to use a *tree diagram*.

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A **tree diagram** is a device consisting of line segments emanating from a starting point and also from the outcome point. It is used to determine all possible outcomes of a probability experiment.

Example 4–4

Gender of Children

Use a tree diagram to find the sample space for the gender of three children in a family, as in Example 4–3.

Solution

Since there are two possibilities (boy or girl) for the first child, draw two branches from a starting point and label one B and the other G. Then if the first child is a boy, there are two possibilities for the second child (boy or girl), so draw two branches from B and label one B and the other G. Do the same if the first child is a girl. Follow the same procedure for the third child. The completed tree diagram is shown in Figure 4–3. To find the outcomes for the sample space, trace through all the possible branches, beginning at the starting point for each one.

Figure 4–3

Tree Diagram for Example 4–4

Historical Note

The famous Italian astronomer Galileo (1564–1642) found that a sum of 10 occurs more often than any other sum when three dice are tossed. Previously, it was thought that a sum of 9 occurred more often than any other sum.

An outcome was defined previously as the result of a single trial of a probability experiment. In many problems, one must find the probability of two or more outcomes. For this reason, it is necessary to distinguish between an outcome and an event.

Historical Note

A mathematician named Jerome Cardan (1501–1576) used his talents in mathematics and probability theory to make his living as a gambler. He is thought to be the first person to formulate the definition of classical probability.

An **event** consists of a set of outcomes of a probability experiment.

An event can be one outcome or more than one outcome. For example, if a die is rolled and a 6 shows, this result is called an *outcome*, since it is a result of a single trial. An event with one outcome is called a **simple event.** The event of getting an odd number when a die is rolled is called a **compound event,** since it consists of three outcomes or three simple events. In general, a compound event consists of two or more outcomes or simple events.

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There are three basic interpretations of probability:

**1.**Classical probability

**2.**Empirical or relative frequency probability

**3.**Subjective probability

Historical Note

During the mid-1600s, a professional gambler named Chevalier de Méré made a considerable amount of money on a gambling game. He would bet unsuspecting patrons that in four rolls of a die, he could get at least one 6. He was so successful at the game that some people refused to play. He decided that a new game was necessary to continue his winnings. By reasoning, he figured he could roll at least one double 6 in 24 rolls of two dice, but his reasoning was incorrect and he lost systematically. Unable to figure out why, he contacted a mathematician named Blaise Pascal (1623–1662) to find out why.

Pascal became interested and began studying probability theory. He corresponded with a French government official, Pierre de Fermat (1601–1665), whose hobby was mathematics. Together the two formulated the beginnings of probability theory.

**Classical Probability**

**Classical probability** uses sample spaces to determine the numerical probability that an event will happen. You do not actually have to perform the experiment to determine that probability. Classical probability is so named because it was the first type of probability studied formally by mathematicians in the 17th and 18th centuries.

*Classical probability assumes that all outcomes in the sample space are equally likely to occur*. For example, when a single die is rolled, each outcome has the same probability of occurring. Since there are six outcomes, each outcome has a probability of . When a card is selected from an ordinary deck of 52 cards, you assume that the deck has been shuffled, and each card has the same probability of being selected. In this case, it is .

**Equally likely events** are events that have the same probability of occurring.

**Formula for Classical Probability**

The probability of any event *E* is

This probability is denoted by

This probability is called *classical probability*, and it uses the sample space *S*.

Probabilities can be expressed as fractions, decimals, or—where appropriate—percentages. If you ask, “What is the probability of getting a head when a coin is tossed?” typical responses can be any of the following three.

“One-half.”

“Point five.”

“Fifty percent.”^{1}

These answers are all equivalent. In most cases, the answers to examples and exercises given in this chapter are expressed as fractions or decimals, but percentages are used where appropriate.

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**Rounding Rule for Probabilities** Probabilities should be expressed as reduced fractions or rounded to two or three decimal places. When the probability of an event is an extremely small decimal, it is permissible to round the decimal to the first nonzero digit after the point. For example, 0.0000587 would be 0.00006. When obtaining probabilities from one of the tables in Appendix C, use the number of decimal places given in the table. If decimals are converted to percentages to express probabilities, move the decimal point two places to the right and add a percent sign.

Example 4–5

Drawing Cards

Find the probability of getting a red ace when a card is drawn at random from an ordinary deck of cards.

Solution

Since there are 52 cards and there are 2 red aces, namely, the ace of hearts and the ace of diamonds, *P*(red ace) = = .

Example 4–6

Gender of Children

If a family has three children, find the probability that two of the three children are girls.

Solution

The sample space for the gender of the children for a family that has three children has eight outcomes, that is, BBB, BBG, BGB, GBB, GGG, GGB, GBG, and BGG. (See Examples 4–3 and 4–4.) Since there are three ways to have two girls, namely, GGB, GBG, and BGG, *P*(two girls) = .

In probability theory, it is important to understand the meaning of the words *and* and *or*. For example, if you were asked to find the probability of getting a queen *and* a heart when you were drawing a single card from a deck, you would be looking for the queen of hearts. Here the word *and* means “at the same time.” The word *or* has two meanings. For example, if you were asked to find the probability of selecting a queen *or* a heart when one card is selected from a deck, you would be looking for one of the 4 queens or one of the 13 hearts. In this case, the queen of hearts would be included in both cases and counted twice. So there would be 4 + 13 – 1 = 16 possibilities.

On the other hand, if you were asked to find the probability of getting a queen *or* a king, you would be looking for one of the 4 queens or one of the 4 kings. In this case, there would be 4 + 4 = 8 possibilities. In the first case, both events can occur at the same time; we say that this is an example of the *inclusive or*. In the second case, both events cannot occur at the same time, and we say that this is an example of the *exclusive or*.

Historical Note

Ancient Greeks and Romans made crude dice from animal bones, various stones, minerals, and ivory. When the dice were tested mathematically, some were found to be quite accurate.

Example 4–7

Drawing Cards

A card is drawn from an ordinary deck. Find these probabilities.

*a*.Of getting a jack

*b*.Of getting the 6 of clubs (i.e., a 6 and a club)

*c*.Of getting a 3 or a diamond

*d*.Of getting a 3 or a 6

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Solution

*a*.Refer to the sample space in Figure 4–2. There are 4 jacks so there are 4 outcomes in event *E* and 52 possible outcomes in the sample space. Hence,

*b*.Since there is only one 6 of clubs in event E, the probability of getting a 6 of clubs is

*c*.There are four 3s and 13 diamonds, but the 3 of diamonds is counted twice in this listing. Hence, there are 16 possibilities of drawing a 3 or a diamond, so

This is an example of the inclusive or.

*d*.Since there are four 3s and four 6s,

This is an example of the exclusive or.

There are four basic probability rules. These rules are helpful in solving probability problems, in understanding the nature of probability, and in deciding if your answers to the problems are correct.

Historical Note

Paintings in tombs excavated in Egypt show that the Egyptians played games of chance. One game called *Hounds and Jackals* played in 1800 B.C. is similar to the present-day game of *Snakes and Ladders*.

**Probability Rule 1**

The probability of any event *E* is a number (either a fraction or decimal) between and including 0 and 1. This is denoted by 0 ≤ *P*(*E*) ≤ 1.

Rule 1 states that probabilities cannot be negative or greater than 1.

**Probability Rule 2**

If an event *E* cannot occur (i.e., the event contains no members in the sample space), its probability is 0.

Example 4–8

Rolling a Die

When a single die is rolled, find the probability of getting a 9.

Solution

Since the sample space is 1, 2, 3, 4, 5, and 6, it is impossible to get a 9. Hence, the probability is *P*(9) = = 0.

**Probability Rule 3**

If an event *E* is certain, then the probability of *E* is 1.

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In other words, if *P*(*E*) = 1, then the event *E* is certain to occur. This rule is illustrated in Example 4–9.

Example 4–9

Rolling a Die

When a single die is rolled, what is the probability of getting a number less than 7?

Solution

Since all outcomes—1, 2, 3, 4, 5, and 6—are less than 7, the probability is

The event of getting a number less than 7 is certain.

In other words, probability values range from 0 to 1. When the probability of an event is close to 0, its occurrence is highly unlikely. When the probability of an event is near 0.5, there is about a 50-50 chance that the event will occur; and when the probability of an event is close to 1, the event is highly likely to occur.

**Probability Rule 4**

The sum of the probabilities of all the outcomes in the sample space is 1.

For example, in the roll of a fair die, each outcome in the sample space has a probability of . Hence, the sum of the probabilities of the outcomes is as shown.

**Complementary Events**

Another important concept in probability theory is that of *complementary events*. When a die is rolled, for instance, the sample space consists of the outcomes 1, 2, 3, 4, 5, and 6. The event *E* of getting odd numbers consists of the outcomes 1, 3, and 5. The event of not getting an odd number is called the *complement* of event *E*, and it consists of the outcomes 2, 4, and 6.

The **complement of an event***E* is the set of outcomes in the sample space that are not included in the outcomes of event *E*. The complement of *E* is denoted by (read “*E* bar”).

Example 4–10 further illustrates the concept of complementary events.

Example 4–10

Finding Complements

Find the complement of each event.

*a*.Rolling a die and getting a 4

*b*.Selecting a letter of the alphabet and getting a vowel

*c*.Selecting a month and getting a month that begins with a J

*d*.Selecting a day of the week and getting a weekday

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Solution

*a*.Getting a 1, 2, 3, 5, or 6

*b*.Getting a consonant (assume *y* is a consonant)

*c*.Getting February, March, April, May, August, September, October, November, or December

*d*.Getting Saturday or Sunday

The outcomes of an event and the outcomes of the complement make up the entire sample space. For example, if two coins are tossed, the sample space is HH, HT, TH, and TT The complement of “getting all heads” is not “getting all tails,” since the event “all heads” is HH, and the complement of HH is HT, TH, and TT. Hence, the complement of the event “all heads” is the event “getting at least one tail.”

Since the event and its complement make up the entire sample space, it follows that the sum of the probability of the event and the probability of its complement will equal 1. That is, . In Example 4–10, let *E* = all heads, or HH, and let = at least one tail, or HT, TH, TT. Then and ; hence, .

The rule for complementary events can be stated algebraically in three ways.

**Rule for Complementary Events**

Stated in words, the rule is: *If the probability of an event or the probability of its complement is known*, *then the other can be found by subtracting the probability from 1*. This rule is important in probability theory because at times the best solution to a problem is to find the probability of the complement of an event and then subtract from 1 to get the probability of the event itself.

Example 4–11

Residence of People

If the probability that a person lives in an industrialized country of the world is , find the probability that a person does not live in an industrialized country.

Source: *Harper’s Index*.

Solution

*P*(not living in an industrialized country) = 1 – *P*(living in an industrialized country)

Probabilities can be represented pictorially by **Venn diagrams.**Figure 4–4(a) shows the probability of a simple event *E*. The area inside the circle represents the probability of event *E*, that is, P(*E*). The area inside the rectangle represents the probability of all the events in the sample space *P*(*S*).

Figure 4–4

Venn Diagram for the Probability and Complement

The Venn diagram that represents the probability of the complement of an event is shown in Figure 4–4(b). In this case, , which is the area inside the rectangle but outside the circle representing P(*E*). Recall that *P(S)* = 1 and *P(E) =* 1 – *P*(). The reasoning is that *P*(*E*) is represented by the area of the circle and *P*() is the probability of the events that are outside the circle.

**Empirical Probability**

The difference between classical and **empirical probability** is that classical probability assumes that certain outcomes are equally likely (such as the outcomes when a die is rolled), while empirical probability relies on actual experience to determine the likelihood of outcomes. In empirical probability, one might actually roll a given die 6000 times, observe the various frequencies, and use these frequencies to determine the probability of an outcome.

Suppose, for example, that a researcher for the American Automobile Association (AAA) asked 50 people who plan to travel over the Thanksgiving holiday how they will get to their destination. The results can be categorized in a frequency distribution as shown.

Method | Frequency |

Drive | 41 |

Fly | 6 |

Train or bus | 3 |

50 |

Now probabilities can be computed for various categories. For example, the probability of selecting a person who is driving is , since 41 out of the 50 people said that they were driving.

**Formula for Empirical Probability**

Given a frequency distribution, the probability of an event being in a given class is

This probability is called *empirical probability* and is based on observation.

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Example 4–12

Travel Survey

In the travel survey just described, find the probability that a person will travel by airplane over the Thanksgiving holiday.

Solution

*Note:* These figures are based on a AAA survey.

Example 4–13

Distribution of Blood Types

In a sample of 50 people, 21 had type O blood, 22 had type A blood, 5 had type B blood, and 2 had type AB blood. Set up a frequency distribution and find the following probabilities.

*a*.A person has type O blood.

*b*.A person has type A or type B blood.

*c*.A person has neither type A nor type O blood.

*d*.A person does not have type AB blood.

Source: The American Red Cross.

Solution

Type | Frequency |

A | 22 |

B | 5 |

AB | 2 |

O | 21 Total 50 |

*a*.

*b*.

(Add the frequencies of the two classes.)

*c*.

(Neither A nor O means that a person has either type B or type AB blood.)

*d*.

(Find the probability of not AB by subtracting the probability of type AB from 1.)

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Example 4–14

Hospital Stays for Maternity Patients

Hospital records indicated that maternity patients stayed in the hospital for the number of days shown in the distribution.

Number of days stayed | Frequency |

3 | 15 |

4 | 32 |

5 | 56 |

6 | 19 |

7 | 5 |

127 |

Find these probabilities.

*a*.A patient stayed exactly 5 days.

*b*.A patient stayed less than 6 days.

*c*.A patient stayed at most 4 days.

*d*.A patient stayed at least 5 days.

Solution

*a*.*b*.

(Fewer than 6 days means 3, 4, or 5 days.)

*c*.

(At most 4 days means 3 or 4 days.)

*d*.

(At least 5 days means 5, 6, or 7 days.)

Empirical probabilities can also be found by using a relative frequency distribution, as shown in Section 2–2. For example, the relative frequency distribution of the travel survey shown previously is

Method | Frequency | Relative frequency |

Drive | 41 | 0.82 |

Fly | 6 | 0.12 |

Train or bus | 3 | 0.06 |

50 | 1.00 |

These frequencies are the same as the relative frequencies explained in Chapter 2.

**Law of Large Numbers**

When a coin is tossed one time, it is common knowledge that the probability of getting a head is . But what happens when the coin is tossed 50 times? Will it come up heads 25 times? Not all the time. You should expect about 25 heads if the coin is fair. But due to chance variation, 25 heads will not occur most of the time.

If the empirical probability of getting a head is computed by using a small number of trials, it is usually not exactly . However, as the number of trials increases, the empirical probability of getting a head will approach the theoretical probability of , if in fact the coin is fair (i.e., balanced). This phenomenon is an example of the **law of large numbers.**

You should be careful to not think that the number of heads and number of tails tend to “even out.” As the number of trials increases, the proportion of heads to the total number of trials will approach . This law holds for any type of gambling game—tossing dice, playing roulette, and so on.

It should be pointed out that the probabilities that the proportions steadily approach may or may not agree with those theorized in the classical model. If not, it can have important implications, such as “the die is not fair.” Pit bosses in Las Vegas watch for empirical trends that do not agree with classical theories, and they will sometimes take a set of dice out of play if observed frequencies are too far out of line with classical expected frequencies.

**Subjective Probability**

The third type of probability is called *subjective probability*. **Subjective probability** uses a probability value based on an educated guess or estimate, employing opinions and inexact information.

In subjective probability, a person or group makes an educated guess at the chance that an event will occur. This guess is based on the person’s experience and evaluation of a solution. For example, a sportswriter may say that there is a 70% probability that the Pirates will win the pennant next year. A physician might say that, on the basis of her diagnosis, there is a 30% chance the patient will need an operation. A seismologist might say there is an 80% probability that an earthquake will occur in a certain area. These are only a few examples of how subjective probability is used in everyday life.

All three types of probability (classical, empirical, and subjective) are used to solve a variety of problems in business, engineering, and other fields.

**Probability and Risk Taking**

An area in which people fail to understand probability is risk taking. Actually, people fear situations or events that have a relatively small probability of happening rather than those events that have a greater likelihood of occurring. For example, many people think that the crime rate is increasing every year. However, in his book entitled *How Risk Affects Your Everyday Life*, author James Walsh states: “Despite widespread concern about the number of crimes committed in the United States, FBI and Justice Department statistics show that the national crime rate has remained fairly level for 20 years. It even dropped slightly in the early 1990s.”

He further states, “Today most media coverage of risk to health and well-being focuses on shock and outrage.” Shock and outrage make good stories and can scare us about the wrong dangers. For example, the author states that if a person is 20% overweight, the loss of life expectancy is 900 days (about 3 years), but loss of life expectancy from exposure to radiation emitted by nuclear power plants is 0.02 day. As you can see, being overweight is much more of a threat than being exposed to radioactive emission.

Many people gamble daily with their lives, for example, by using tobacco, drinking and driving, and riding motorcycles. When people are asked to estimate the probabilities or frequencies of death from various causes, they tend to overestimate causes such as accidents, fires, and floods and to underestimate the probabilities of death from diseases (other than cancer), strokes, etc. For example, most people think that their chances of dying of a heart attack are 1 in 20, when in fact they are almost 1 in 3; the chances of dying by pesticide poisoning are 1 in 200,000 *(True Odds* by James Walsh). The reason people think this way is that the news media sensationalize deaths resulting from catastrophic events and rarely mention deaths from disease.

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When you are dealing with life-threatening catastrophes such as hurricanes, floods, automobile accidents, or smoking, it is important to get the facts. That is, get the actual numbers from accredited statistical agencies or reliable statistical studies, and then compute the probabilities and make decisions based on your knowledge of probability and statistics.

In summary, then, when you make a decision or plan a course of action based on probability, make sure that you understand the true probability of the event occurring. Also, find out how the information was obtained (i.e., from a reliable source). Weigh the cost of the action and decide if it is worth it. Finally, look for other alternatives or courses of action with less risk involved.

*Applying the Concepts* 4–1

**Tossing a Coin**

Assume you are at a carnival and decide to play one of the games. You spot a table where a person is flipping a coin, and since you have an understanding of basic probability, you believe that the odds of winning are in your favor. When you get to the table, you find out that all you have to do is to guess which side of the coin will be facing up after it is tossed. You are assured that the coin is fair, meaning that each of the two sides has an equally likely chance of occurring. You think back about what you learned in your statistics class about probability before you decide what to bet on. Answer the following questions about the coin-tossing game.

1.What is the sample space?

2.What are the possible outcomes?

3.What does the classical approach to probability say about computing probabilities for this type of problem?

You decide to bet on heads, believing that it has a 50% chance of coming up. A friend of yours, who had been playing the game for awhile before you got there, tells you that heads has come up the last 9 times in a row. You remember the law of large numbers.

4.What is the law of large numbers, and does it change your thoughts about what will occur on the next toss?

5.What does the empirical approach to probability say about this problem, and could you use it to solve this problem?

6.Can subjective probabilities be used to help solve this problem? Explain.

7.Assume you could win $1 million if you could guess what the results of the next toss will be. What would you bet on? Why?

See page 257 for the answers.

Exercises 4–1 |

**1.**What is a probability experiment?

**2.**Define *sample space*.

**3.**What is the difference between an outcome and an event?

**4.**What are equally likely events?

**5.**What is the range of the values of the probability of an event?

**6.**When an event is certain to occur, what is its probability?

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**7.**If an event cannot happen, what value is assigned to its probability?

**8.**What is the sum of the probabilities of all the outcomes in a sample space?

**9.**If the probability that it will rain tomorrow is 0.20, what is the probability that it won’t rain tomorrow? Would you recommend taking an umbrella?

**10.**A probability experiment is conducted. Which of these cannot be considered a probability of an outcome?

*a*.

*b*.

*c*.0.80

*d*.–0.59

*e*.0

*f*.1.45

*g*.1

*h*.33%

*i*.112%

**11.**Classify each statement as an example of classical probability, empirical probability, or subjective probability.

*a*.The probability that a person will watch the 6 o’clock evening news is 0.15.

*b*.The probability of winning at a Chuck-a-Luck game is .

*c*.The probability that a bus will be in an accident on a specific run is about 6%.

*d*.The probability of getting a royal flush when five cards are selected at random is .

*e*.The probability that a student will get a C or better in a statistics course is about 70%.

*f*.The probability that a new fast-food restaurant will be a success in Chicago is 35%.

*g*.The probability that interest rates will rise in the next 6 months is 0.50.

**12.(ans) Rolling a Die** If a die is rolled one time, find these probabilities.

*a*.Of getting a 4

*b*.Of getting an even number

*c*.Of getting a number greater than 4

*d*.Of getting a number less than 7

*e*.Of getting a number greater than 0

*f*.Of getting a number greater than 3 or an odd number

*g*.Of getting a number greater than 3 and an odd number

**13.Rolling Two Dice** If two dice are rolled one time, find the probability of getting these results.

*a*.A sum of 6

*b*.Doubles

*c*.A sum of 7 or 11

*d*.A sum greater than 9

*e*.A sum less than or equal to 4

**14.(ans) Drawing a Card** If one card is drawn from a deck, find the probability of getting these results.

*a*.An ace

*b*.A diamond

*c*.An ace of diamonds

*d*.A 4 or a 6

*e*.A 4 or a club

*f*.A 6 or a spade

*g*.A heart or a club

*h*.A red queen

*i*.A red card or a 7

*j*.A black card and a 10

**15.Shopping Mall Promotion** A shopping mall has set up a promotion as follows. With any mall purchase of $50 or more, the customer gets to spin the wheel shown here. If a number 1 comes up, the customer wins $10. If the number 2 comes up, the customer wins $5; and if the number 3 or 4 comes up, the customer wins a discount coupon. Find the following probabilities.

*a*.The customer wins $10.

*b*.The customer wins money.

*c*.The customer wins a coupon.

**16.Selecting a State** Choose one of the 50 states at random.

*a*.What is the probability that it begins with M?

*b*.What is the probability that it doesn’t begin with a vowel?

**17.Human Blood Types** Human blood is grouped into four types. The percentages of Americans with each type are listed below.

O 43% A 40% B 12% AB 5%

Choose one American at random. Find the probability that this person

*a*.Has type O blood

*b*.Has type A or B

*c*.Does not have type O or A

Source: www.infoplease.com

**18.Gender of College Students** In 2004, 57.2% of all enrolled college students were female. Choose one enrolled student at random. What is the probability that the student was male?

Source: www.nces.ed.gov

**19.Prime Numbers** A prime number is a number that is evenly divisible only by 1 and itself. The prime numbers less than 100 are listed below.

Choose one of these numbers at random. Find the probability that

*a*.The number is even

*b*.The sum of the number’s digits is even

*c*.The number is greater than 50

**20.Rural Speed Limits** Rural speed limits for all 50 states are indicated below.

Choose one state at random. Find the probability that its speed limit is

*a*.60 or 70 miles per hour

*b*.Greater than 65 miles per hour

*c*.70 miles per hour or less

Source: *World Almanac*.

**21.Gender of Children** A couple has three children. Find each probability.

*a*.All boys

*b*.All girls or all boys

*c*.Exactly two boys or two girls

*d*.At least one child of each gender

**22.Craps Game** In the game of craps using two dice, a person wins on the first roll if a 7 or an 11 is rolled. Find the probability of winning on the first roll.

**23.Craps Game** In a game of craps, a player loses on the roll if a 2, 3, or 12 is tossed on the first roll. Find the probability of losing on the first roll.

**24.Computers in Elementary Schools** Elementary and secondary schools were classified by the number of computers they had. Choose one of these schools at random.

Choose one school at random. Find the probability that it has

*a*.50 or fewer computers

*b*.More than 100 computers

*c*.No more than 20 computers

Source: *World Almanac*.

**25.Roulette** A roulette wheel has 38 spaces numbered 1 through 36, 0, and 00. Find the probability of getting these results.

*a*.An odd number (Do not count 0 or 00.)

*b*.A number greater than 27

*c*.A number that contains the digit 0

*d*.Based on the answers to parts *a, b*, and *c*, which is most likely to occur? Explain why.

**26.Gasoline Mileage for Autos and Trucks** Of the top 10 cars and trucks based on gas mileage, 4 are Hondas, 3 are Toyotas, and 3 are Volkswagens. Choose one at random. Find the probability that it is

*a*.Japanese

*b*.Japanese or German

*c*.Not foreign

Source: www.autobytel.com

**27.Large Monetary Bills in Circulation** There are 1,765,000 five thousand dollar bills in circulation and 3,460,000 ten thousand dollar bills in circulation. Choose one bill at random (wouldn’t that be nice!). What is the probability that it is a ten thousand dollar bill?

Source: *World Almanac*.

**28.Sources of Energy Uses in the United States** A breakdown of the sources of energy used in the United States is shown below. Choose one energy source at random. Find the probability that it is

*a*.Not oil

*b*.Natural gas or oil

*c*.Nuclear

Oil 39% Natural gas 24% Coal 23%

Nuclear 8% Hydropower 3% Other 3%

Source: www.infoplease.com

**29.Rolling Dice** Roll two dice and multiply the numbers.

*a*.Write out the sample space.

*b*.What is the probability that the product is a multiple of 6?

*c*.What is the probability that the product is less than 10?

**30.Federal Government Revenue** The source of federal government revenue for a specific year is

50% from individual income taxes

32% from social insurance payroll taxes

10% from corporate income taxes

3% from excise taxes

5% other

If a revenue source is selected at random, what is the probability that it comes from individual or corporate income taxes?

Source: *New York Times Almanac*.

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**31.Selecting a Bill** A box contains a $1 bill, a $5 bill, a $10 bill, and a $20 bill. A bill is selected at random, and it is not replaced; then a second bill is selected at random. Draw a tree diagram and determine the sample space.

**32.Tossing Coins** Draw a tree diagram and determine the sample space for tossing four coins.

**33.Selecting Numbered Balls** Four balls numbered 1 through 4 are placed in a box. A ball is selected at random, and its number is noted; then it is replaced. A second ball is selected at random, and its number is noted. Draw a tree diagram and determine the sample space.

**34.Family Dinner Combinations** A family special at a neighborhood restaurant offers dinner for four for $19.99. There are 3 appetizers available, 4 entrees, and 3 desserts from which to choose. The special includes one of each. Represent the possible dinner combinations with a tree diagram.

**35.Required First-Year College Courses** First-year students at a particular college must take one English class, one class in mathematics, a first-year seminar, and an elective. There are 2 English classes to choose from, 3 mathematics classes, 5 electives, and everyone takes the same first-year seminar. Represent the possible schedules using a tree diagram.

**36.Tossing a Coin and Rolling a Die** A coin is tossed; if it falls heads up, it is tossed again. If it falls tails up, a die is rolled. Draw a tree diagram and determine the outcomes.

**Extending the Concepts**

**37.Distribution of CEO Ages** The distribution of ages of CEOs is as follows:

Age | Frequency |

21–30 | 1 |

31–40 | 8 |

41–50 | 27 |

51–60 | 29 |

61–70 | 24 |

71-up | 11 |

Source: Information based on *USA TODAY* Snapshot.

If a CEO is selected at random, find the probability that his or her age is

*a*.Between 31 and 40

*b*.Under 31

*c*.Over 30 and under 51

*d*.Under 31 or over 60

**38.Tossing a Coin** A person flipped a coin 100 times and obtained 73 heads. Can the person conclude that the coin was unbalanced?

**39.Medical Treatment** A medical doctor stated that with a certain treatment, a patient has a 50% chance of recovering without surgery. That is, “Either he will get well or he won’t get well.” Comment on this statement.

**40.Wheel Spinner** The wheel spinner shown here is spun twice. Find the sample space, and then determine the probability of the following events.

*a*.An odd number on the first spin and an even number on the second spin *(Note:* 0 is considered even.)

*b*.A sum greater than 4

*c*.Even numbers on both spins

*d*.A sum that is odd

*e*.The same number on both spins

**41.Tossing Coins** Toss three coins 128 times and record the number of heads (0, 1, 2, or 3); then record your results with the theoretical probabilities. Compute the empirical probabilities of each.

**42.Tossing Coins** Toss two coins 100 times and record the number of heads (0, 1, 2). Compute the probabilities of each outcome, and compare these probabilities with the theoretical results.

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**43.Odds** Odds are used in gambling games to make them fair. For example, if you rolled a die and won every time you rolled a 6, then you would win on average once every 6 times. So that the game is fair, the odds of 5 to 1 are given. This means that if you bet $1 and won, you could win $5. On average, you would win $5 once in 6 rolls and lose $1 on the other 5 rolls—hence the term *fair game*.

In most gambling games, the odds given are not fair. For example, if the odds of winning are really 20 to 1, the house might offer 15 to 1 in order to make a profit.

Odds can be expressed as a fraction or as a ratio, such as f, , or 5 to 1. Odds are computed in favor of the event or against the event. The formulas for odds are

In the die example,

Find the odds in favor of and against each event.

*a*.Rolling a die and getting a 2

*b*.Rolling a die and getting an even number

*c*.Drawing a card from a deck and getting a spade

*d*.Drawing a card and getting a red card

*e*.Drawing a card and getting a queen

*f*.Tossing two coins and getting two tails

*g*.Tossing two coins and getting one tail

Objective 2

Find the probability of compound events, using the addition rules.

**4–2The Addition Rules for Probability**

Many problems involve finding the probability of two or more events. For example, at a large political gathering, you might wish to know, for a person selected at random, the probability that the person is a female or is a Republican. In this case, there are three possibilities to consider:

**1.**The person is a female.

**2.**The person is a Republican.

**3.**The person is both a female and a Republican.

Consider another example. At the same gathering there are Republicans, Democrats, and Independents. If a person is selected at random, what is the probability that the person is a Democrat or an Independent? In this case, there are only two possibilities:

**1.**The person is a Democrat.

**2.**The person is an Independent.

The difference between the two examples is that in the first case, the person selected can be a female and a Republican at the same time. In the second case, the person selected cannot be both a Democrat and an Independent at the same time. In the second case, the two events are said to be *mutually exclusive;* in the first case, they are not mutually exclusive.

Historical Note

The first book on probability, *The Book of Chance and Games*, was written by Jerome Cardan (1501–1576). Cardan was an astrologer, philosopher, physician, mathematician, and gambler. This book contained techniques on how to cheat and how to catch others at cheating.

Two events are **mutually exclusive events** if they cannot occur at the same time (i.e., they have no outcomes in common).

In another situation, the events of getting a 4 and getting a 6 when a single card is drawn from a deck are mutually exclusive events, since a single card cannot be both a 4 and a 6. On the other hand, the events of getting a 4 and getting a heart on a single draw are not mutually exclusive, since you can select the 4 of hearts when drawing a single card from an ordinary deck.

Example 4–15

Rolling a Die

Determine which events are mutually exclusive and which are not, when a single die is rolled.

*a*.Getting an odd number and getting an even number

*b*.Getting a 3 and getting an odd number

*c*.Getting an odd number and getting a number less than 4

*d*.Getting a number greater than 4 and getting a number less than 4

Solution

*a*.The events are mutually exclusive, since the first event can be 1, 3, or 5 and the second event can be 2, 4, or 6.

*b*.The events are not mutually exclusive, since the first event is a 3 and the second can be 1, 3, or 5. Hence, 3 is contained in both events.

*c*.The events are not mutually exclusive, since the first event can be 1, 3, or 5 and the second can be 1, 2, or 3. Hence, 1 and 3 are contained in both events.

*d*.The events are mutually exclusive, since the first event can be 5 or 6 and the second event can be 1, 2, or 3.

Example 4–16

Drawing a Card

Determine which events are mutually exclusive and which are not, when a single card is drawn from a deck.

*a*.Getting a 7 and getting a jack

*b*.Getting a club and getting a king

*c*.Getting a face card and getting an ace

*d*.Getting a face card and getting a spade

Solution

Only the events in parts *a* and *c* are mutually exclusive.

The probability of two or more events can be determined by the *addition rules*. The first addition rule is used when the events are mutually exclusive.

**Addition Rule 1**

When two events *A* and *B* are mutually exclusive, the probability that *A* or *B* will occur is

*P*(*A* or *B*) = *P*(*A*) + *P*(*B*)

Example 4–17

Selecting a Doughnut

A box contains 3 glazed doughnuts, 4 jelly doughnuts, and 5 chocolate doughnuts. If a person selects a doughnut at random, find the probability that it is either a glazed doughnut or a chocolate doughnut.

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Solution

Since the box contains 3 glazed doughnuts, 5 chocolate doughnuts, and a total of 12 doughnuts, *P*(glazed or chocolate) = *P*(glazed) + *P*(chocolate) = . The events are mutually exclusive.

Example 4–18

Political Affiliation at a Rally

At a political rally, there are 20 Republicans, 13 Democrats, and 6 Independents. If a person is selected at random, find the probability that he or she is either a Democrat or an Independent.

Solution

Example 4–19

Selecting a Day of the Week

A day of the week is selected at random. Find the probability that it is a weekend day.

Solution

When two events are not mutually exclusive, we must subtract one of the two probabilities of the outcomes that are common to both events, since they have been counted twice. This technique is illustrated in Example 4–20.

Example 4–20

Drawing a Card

A single card is drawn at random from an ordinary deck of cards. Find the probability that it is either an ace or a black card.

Solution

Since there are 4 aces and 26 black cards (13 spades and 13 clubs), 2 of the aces are black cards, namely, the ace of spades and the ace of clubs. Hence the probabilities of the two outcomes must be subtracted since they have been counted twice.

When events are not mutually exclusive, addition rule 2 can be used to find the probability of the events.

**Addition Rule 2**

If *A* and *B* are *not* mutually exclusive, then

*P*(*A* or *B*) = *P*(*A*) + *P*(*B*) – *P*(*A* and *B*)

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*Note:* This rule can also be used when the events are mutually exclusive, since *P(A* and *B)* will always equal 0. However, it is important to make a distinction between the two situations.

Example 4–21

Selecting a Medical Staff Person

In a hospital unit there are 8 nurses and 5 physicians; 7 nurses and 3 physicians are females. If a staff person is selected, find the probability that the subject is a nurse or a male.

Solution

The sample space is shown here.

The probability is

Example 4–22

Driving While Intoxicated

On New Year’s Eve, the probability of a person driving while intoxicated is 0.32, the probability of a person having a driving accident is 0.09, and the probability of a person having a driving accident while intoxicated is 0.06. What is the probability of a person driving while intoxicated or having a driving accident?

Solution

*P*(intoxicated or accident) = *P*(intoxicated) + *P*(accident) – *P*(intoxicated and accident)

= 0.32 + 0.09 – 0.06 = 0.35

In summary, then, when the two events are mutually exclusive, use addition rule 1. When the events are not mutually exclusive, use addition rule 2.

The probability rules can be extended to three or more events. For three mutually exclusive events *A, B*, and *C*,

*P*(*A* or *B* or *C*) = *P*(*A*) + *P*(*B*) + *P*(*C*)

For three events that are *not* mutually exclusive,

*P*(*A* or *B* or *C*) = *P*(*A*) + *P*(*B*) + *P*(*C*) – *P*(*A* and *B*) – *P*(*A* and *C*) – *P*(*B* and *C*) + *P*(*A* and *B* and *C*)

See Exercises 23 and 24 in this section.

Figure 4–5(a) shows a Venn diagram that represents two mutually exclusive events *A* and *B*. In this case, *P*(*A* or *B*) = *P*(*A*) + *P*(*B*), since these events are mutually exclusive and do not overlap. In other words, the probability of occurrence of event *A* or event *B* is the sum of the areas of the two circles.

Venn Diagrams for the Addition Rules

Figure 4–5(b) represents the probability of two events that are *not* mutually exclusive. In this case, *P*(*A* or *B*) = *P*(*A*) + *P*(*B*) – *P*(*A* and *B*). The area in the intersection or overlapping part of both circles corresponds to *P*(*A* and *B*); and when the area of circle *A* is added to the area of circle *B*, the overlapping part is counted twice. It must therefore be subtracted once to get the correct area or probability.

*Note:* Venn diagrams were developed by mathematician John Venn (1834–1923) and are used in set theory and symbolic logic. They have been adapted to probability theory also. In set theory, the symbol represents the *union* of two sets, and *A**B* corresponds to *A* or *B*. The symbol represents the *intersection* of two sets, and *A**B* corresponds to *A* and *B*. Venn diagrams show only a general picture of the probability rules and do not portray all situations, such as *P*(*A*) = 0, accurately.

*Applying the Concepts* 4–2

**Which Pain Reliever Is Best?**

Assume that following an injury you received from playing your favorite sport, you obtain and read information on new pain medications. In that information you read of a study that was conducted to test the side effects of two new pain medications. Use the following table to answer the questions and decide which, if any, of the two new pain medications you will use.

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1.How many subjects were in the study?

2.How long was the study?

3.What were the variables under study?

4.What type of variables are they, and what level of measurement are they on?

5.Are the numbers in the table exact figures?

6.What is the probability that a randomly selected person was receiving a placebo?

7.What is the probability that a person was receiving a placebo or drug A? Are these mutually exclusive events? What is the complement to this event?

8.What is the probability that a randomly selected person was receiving a placebo or experienced a neurological headache?

9.What is the probability that a randomly selected person was not receiving a placebo or experienced a sinus headache?

See page 257 for the answers.

Exercises 4–2 |

**1.**Define mutually exclusive events, and give an example of two events that are mutually exclusive and two events that are not mutually exclusive.

**2**.Determine whether these events are mutually exclusive.

*a*.Roll a die: Get an even number, and get a number less than 3.

*b*.Roll a die: Get a prime number (2, 3, 5), and get an odd number.

*c*.Roll a die: Get a number greater than 3, and get a number less than 3.

*d*.Select a student in your class: The student has blond hair, and the student has blue eyes.

*e*.Select a student in your college: The student is a sophomore, and the student is a business major.

*f*.Select any course: It is a calculus course, and it is an English course.

*g*.Select a registered voter: The voter is a Republican, and the voter is a Democrat.

**3**.**College Degrees Awarded** The table below represents the college degrees awarded in a recent academic year by gender.

Choose a degree at random. Find the probability that it is

*a*.A bachelor’s degree

*b*.A doctorate or a degree awarded to a woman

*c*.A doctorate awarded to a woman

*d*.Not a master’s degree

Source: www.nces.ed.gov

**4**.**Selecting a Staff Person** At a community swimming pool there are 2 managers, 8 lifeguards, 3 concession stand clerks, and 2 maintenance people. If a person is selected at random, find the probability that the person is either a lifeguard or a manager.

**5**.**Selecting an Instructor** At a convention there are 7 mathematics instructors, 5 computer science instructors, 3 statistics instructors, and 4 science instructors. If an instructor is selected, find the probability of getting a science instructor or a math instructor.

**6**.**Selecting a Movie** A media rental store rented the following number of movie titles in each of these categories: 170 horror, 230 drama, 120 mystery, 310 romance, and 150 comedy. If a person selects a movie to rent, find the probability that it is a romance or a comedy. Is this event likely or unlikely to occur? Explain your answer.

**7**.**Selecting a Nurse** A recent study of 200 nurses found that of 125 female nurses, 56 had bachelor’s degrees; and of 75 male nurses, 34 had bachelor’s degrees. If a nurse is selected at random, find the probability that the nurse is

*a*.A female nurse with a bachelor’s degree

*b*.A male nurse

*c*.A male nurse with a bachelor’s degree

*d*.Based on your answers to parts *a, b*, and *c*, explain which is most likely to occur. Explain why.

**8**.**Tourist Destinations** The probability that a given tourist goes to the amusement park is 0.47, and the probability that she goes to the water park is 0.58. If the probability that she goes to either the water park or the amusement park is 0.95, what is the probability that she visits both of the parks on vacation?

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**9.Sports Participation** At a particular school with 200 male students, 58 play football, 40 play basketball, and 8 play both. What is the probability that a randomly selected male student plays neither sport?

**10**.**Selecting a Card** A single card is drawn from a deck. Find the probability of selecting the following.

*a*.A 4 or a diamond

*b*.A club or a diamond

*c*.A jack or a black card

**11**.**Selecting a Student** In a statistics class there are 18 juniors and 10 seniors; 6 of the seniors are females, and 12 of the juniors are males. If a student is selected at random, find the probability of selecting the following.

*a*.A junior or a female

*b*.A senior or a female

*c*.A junior or a senior

**12**.**Selecting a Book** At a used-book sale, 100 books are adult books and 160 are children’s books. Of the adult books, 70 are nonfiction while 60 of the children’s books are nonfiction. If a book is selected at random, find the probability that it is

*a*.Fiction

*b*.Not a children’s nonfiction book

*c*.An adult book or a children’s nonfiction book

**13**.**Young Adult Residences** According to the Bureau of the Census, the following statistics describe the number (in thousands) of young adults living at home or in a dormitory in the year 2004.

Ages 18–24 | Ages 25–34 | |

Male | 7922 | 2534 |

Female | 5779 | 995 |

Source: *World Almanac*.

Choose one student at random. Find the probability that the student is

*a*.A female student aged 25–34

*b*.Male or aged 18–24

*c*.Under 25 years of age and not male

**14**.**Endangered Species** The chart below shows the numbers of endangered and threatened species both here in the United States and abroad.

Source: www.infoplease.com

Choose one species at random. Find the probability that it is

*a*.Threatened and in the United States

*b*.An endangered foreign bird

*c*.A mammal or a threatened foreign species

**15**.**Multiple Births** The number of multiple births in the United States for a recent year indicated that there were 128,665 sets of twins, 7110 sets of triplets, 468 sets of quadruplets, and 85 sets of quintuplets. Choose one set of siblings at random. Find the probability that it

*a*.Represented more than two babies

*b*.Represented quads or quints

*c*.Now choose one baby from these multiple births. What is the probability that the baby was a triplet?

**16**.**Licensed Drivers in the United States** In a recent year there were the following numbers (in thousands) of licensed drivers in the United States.

Male | Female | |

Age 19 and under | 4746 | 4517 |

Age 20 | 1625 | 1553 |

Age 21 | 1679 | 1627 |

Source: *World Almanac*.

Choose one driver at random. Find the probability that the driver is

*a*.Male and 19 or under

*b*.Age 20 or female

*c*.At least 20 years old

**17**.**Cable Channel Programming** Three cable channels (6, 8, and 10) have quiz shows, comedies, and dramas. The number of each is shown here.

If a show is selected at random, find these probabilities.

*a*.The show is a quiz show, or it is shown on channel 8.

*b*.The show is a drama or a comedy.

*c*.The show is shown on channel 10, or it is a drama.

**18**.**Mail Delivery** A local postal carrier distributes first-class letters, advertisements, and magazines. For a certain day, she distributed the following numbers of each type of item.

If an item of mail is selected at random, find these probabilities.

*a*.The item went to a home.

*b*.The item was an ad, or it went to a business.

*c*.The item was a first-class letter, or it went to a home.

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**19**.**Medical Tests on Emergency Patients** The frequency distribution shown here illustrates the number of medical tests conducted on 30 randomly selected emergency patients.

Number of tests performed | Number of patients |

0 | 12 |

1 | 8 |

2 | 2 |

3 | 3 |

4 or more | 5 |

If a patient is selected at random, find these probabilities.

*a*.The patient has had exactly 2 tests done.

*b*.The patient has had at least 2 tests done.

*c*.The patient has had at most 3 tests done.

*d*.The patient has had 3 or fewer tests done.

*e*.The patient has had 1 or 2 tests done.

**20**.A social organization of 32 members sold college sweatshirts as a fundraiser. The results of their sale are shown below.

No. of sweatshirts | No. of students |

0 | 2 |

1–5 | 13 |

6–10 | 8 |

11–15 | 4 |

16–20 | 4 |

20+ | 1 |

Choose one student at random. Find the probability that the student sold

*a*.More than 10 sweatshirts

*b*.At least one sweatshirt

*c*.1–5 or more than 15 sweatshirts

**21.Door-to-Door Sales** A sales representative who visits customers at home finds she sells 0, 1, 2, 3, or 4 items according to the following frequency distribution.

Items sold | Frequency |

0 | 8 |

1 | 10 |

2 | 3 |

3 | 2 |

4 | 1 |

Find the probability that she sells the following.

*a*.Exactly 1 item

*b*.More than 2 items

*c*.At least 1 item

*d*.At most 3 items

**22.Medical Patients** A recent study of 300 patients found that of 100 alcoholic patients, 87 had elevated cholesterol levels, and of 200 nonalcoholic patients, 43 had elevated cholesterol levels. If a patient is selected at random, find the probability that the patient is the following.

*a*.An alcoholic with elevated cholesterol level

*b*.A nonalcoholic

*c*.A nonalcoholic with nonelevated cholesterol level

**23.Selecting a Card** If one card is drawn from an ordinary deck of cards, find the probability of getting the following.

*a*.A king or a queen or a jack

*b*.A club or a heart or a spade

*c*.A king or a queen or a diamond

*d*.An ace or a diamond or a heart

*e*.A 9 or a 10 or a spade or a club

**24.Rolling a Die** Two dice are rolled. Find the probability of getting

*a*.A sum of 5, 6, or 7

*b*.Doubles or a sum of 6 or 8

*c*.A sum greater than 8 or less than 3

*d*.Based on the answers to parts *a, b*, and *c*, which is least likely to occur? Explain why.

**25**.**Corn Products** U.S. growers harvested 11 billion bushels of corn in 2005. About 1.9 billion bushels were exported, and 1.6 billion bushels were used for ethanol. Choose one bushel of corn at random. What is the probability that it was used either for export or for ethanol?

Source: www.census.gov

**26**.**Rolling Dice** Three dice are rolled. Find the probability of getting

*a*.Triples

*b*.A sum of 5

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**Extending the Concepts**

**27.Purchasing a Pizza** The probability that a customer selects a pizza with mushrooms or pepperoni is 0.43, and the probability that the customer selects only mushrooms is 0.32. If the probability that he or she selects only pepperoni is 0.17, find the probability of the customer selecting both items.

**28.Building a New Home** In building new homes, a contractor finds that the probability of a home buyer selecting a two-car garage is 0.70 and of selecting a one-car garage is 0.20. Find the probability that the buyer will select no garage. The builder does not build houses with three-car or more garages.

**29.**In Exercise 28, find the probability that the buyer will not want a two-car garage.

**30.**Suppose that *P*(*A*) = 0.42, *P(B) =* 0.38, and *P*(*A**B*) = 0.70. Are *A* and *B* mutually exclusive? Explain.

“I know you haven’t had an accident in thirteen years. We’re raising your rates because you’re about due one.”

Source: © King Features Syndicate.

**Technology Step by Step**

**MINITAB**

Step by Step

**Calculate Relative Frequency Probabilities**

The random variable X represents the number of days patients stayed in the hospital from Example 4–14.

**1.**In C1 of a worksheet, type in the values of X. Name the column **X.**

**2.**In C2 enter the frequencies. Name the column **f.**

**3.**To calculate the relative frequencies and store them in a new column named Px:

a)Select **Calc>Calculator.**

b)Type **Px** in the box for Store result in variable:.

c)Click in the Expression box, then double-click C2 f.

d)Type or click the division operator.

e)Scroll down the function list to Sum, then click [Select].

f)Double-click C2 f to select it.

g)Click [OK].

The dialog box and completed worksheet are shown.

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If the original data, rather than the table, are in a worksheet, use **Stat>Tables>Tally** to make the tables with percents (Section 2–1).

MINITAB can also make a two-way classification table.

**Construct a Contingency Table**

**1.**Select **File>Open Worksheet** to open the Databank.mtw file.

**2.**Select **Stat>Tables>Crosstabulation …**

a)Double-click C4 SMOKING STATUS to select it For rows:.

b)Select C11 GENDER for the For Columns: Field.

c)Click on option for Counts and then [OK].

The session window and completed dialog box are shown.

In this sample of 100 there are 25 females who do not smoke compared to 22 men. Sixteen individuals smoke 1 pack or more per day.

**TI–83 Plus or TI–84 Plus**

**Step by Step**

To construct a relative frequency table:

**1.**Enter the data values in L_{1} and the frequencies in L_{2}.

**2.**Move the cursor to the top of the L_{3} column so that L_{3} is highlighted.

**3.**Type **L _{2}** divided by the sample size, then press

**ENTER.**

Use the data from Example 4–14.

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**Excel**

**Step by Step**

**Constructing a Relative Frequency Distribution**

Use the data from Example 4–14.

**1.**In a new worksheet, type the label **DAYS** in cell A1. Beginning in cell A2, type in the data for the variable representing the number of days maternity patients stayed in the hospital.

**2.**In cell B1, type the label for the frequency, **COUNT.** Beginning in cell B2, type in the frequencies.

**3.**In cell B7, compute the total frequency by selecting the sum icon from the toolbar and press **Enter.**

**4.**In cell C1, type a label for the relative frequencies, **Rf.** In cell C2, type = (**B2**)/(**B7**) and **Enter.** In cell C2, type =(**B3**)/(**B7**) and **Enter.** Repeat this for each of the remaining frequencies.

**5.**To find the total relative frequency, select the sum icon from the toolbar and **Enter.** This sum should be 1.

**Constructing a Contingency Table**

**Example XL4–1**

For this example, you will need to have the MegaStat Add-In installed on Excel (refer to Chapter 1, Excel Step by Step instructions for instructions on installing MegaStat).

**1.**Open the Databank.xls file from the CD-ROM that came with your text. To do this:

Double-click My Computer on the Desktop.

Double-click the Bluman CD-ROM icon in the CD drive holding the disk.

Double-click the datasets folder. Then double-click the all_data-sets folder.

Double-click the bluman_es_data-sets_excel-windows folder. In this folder double-click the Databank.xls file. The Excel program will open automatically once you open this file.

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**2.**Highlight the column labeled SMOKING STATUS to copy these data onto a new Excel worksheet.

**3.**Click the Microsoft Office Button , select New Blank Workbook, then Create.

**4.**With cell A1 selected, click the Paste icon on the toolbar to paste the data into the new workbook.

**5.**Return to the Databank.xls file. Highlight the column labeled Gender. Copy and paste these data into column B of the worksheet containing the SMOKING STATUS data.

**6.**Type in the categories for SMOKING STATUS, **0, 1,** and 2 into cells C2–C4. In cell D2, type M for male and in cell D3, type F for female.

**7.**On the toolbar, select Add-Ins. Then select MegaStat. *Note:* You may need to open MegaStat from the file MegaStat.xls saved on your computer’s hard drive.

**8.**Select **Chi-Square/Crosstab>Crosstabulation.**

**9.**In the Row variable Data range box, type A1:A101. In the Row variable Specification range box, type C2:C4. In the Column variable Data range box, type B1:B101. In the Column variable Specification range box, type D2:D3. Remove any checks from the Output Options. Then click [OK].

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Objective 3

Find the probability of compound events, using the multiplication rules.

**4–3The Multiplication Rules and Conditional Probability**

Section 4–2 showed that the addition rules are used to compute probabilities for mutually exclusive and non-mutually exclusive events. This section introduces the multiplication rules.

**The Multiplication Rules**

The *multiplication rules* can be used to find the probability of two or more events that occur in sequence. For example, if you toss a coin and then roll a die, you can find the probability of getting a head on the coin *and* a 4 on the die. These two events are said to be *independent* since the outcome of the first event (tossing a coin) does not affect the probability outcome of the second event (rolling a die).

Two events *A* and *B* are **independent events** if the fact that *A* occurs does not affect the probability of *B* occurring.

Here are other examples of independent events:

Rolling a die and getting a 6, and then rolling a second die and getting a 3.

Drawing a card from a deck and getting a queen, replacing it, and drawing a second card and getting a queen.

To find the probability of two independent events that occur in sequence, you must find the probability of each event occurring separately and then multiply the answers. For example, if a coin is tossed twice, the probability of getting two heads is . This result can be verified by looking at the sample space HH, HT, TH, TT. Then *P*(HH) = .

**Multiplication Rule 1**

When two events are independent, the probability of both occurring is

*P*(*A* and *B*) = *P*(*A*) · *P*(*B*)

Example 4–23

Tossing a Coin

A coin is flipped and a die is rolled. Find the probability of getting a head on the coin and a 4 on the die.

Solution

Note that the sample space for the coin is H, T; and for the die it is 1, 2, 3, 4, 5, 6.

The problem in Example 4–23 can also be solved by using the sample space

H1 H2 H3 H4 H5 H6 T1 T2 T3 T4 T5 T6

The solution is , since there is only one way to get the head-4 outcome.

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Example 4–24

**Drawing a Card**

A card is drawn from a deck and replaced; then a second card is drawn. Find the probability of getting a queen and then an ace.

Solution

The probability of getting a queen is , and since the card is replaced, the probability of getting an ace is . Hence, the probability of getting a queen and an ace is

Example 4–25

Selecting a Colored Ball

An urn contains 3 red balls, 2 blue balls, and 5 white balls. A ball is selected and its color noted. Then it is replaced. A second ball is selected and its color noted. Find the probability of each of these.

*a*.Selecting 2 blue balls

*b*.Selecting 1 blue ball and then 1 white ball

*c*.Selecting 1 red ball and then 1 blue ball

Solution

*a*.*P*(blue and blue) =*P*(blue) · P(blue) =*b*.*P*(blue and white) =*P*(blue) · P(white) =*c*.*P*(red and blue) =*P*(red) · P(blue) =

Multiplication rule 1 can be extended to three or more independent events by using the formula

P(*A* and *B* and *C* and … and *K*) = *P*(*A*) · *P*(*B*) · *P*(*C*) … *P*(*K*)

When a small sample is selected from a large population and the subjects are not replaced, the probability of the event occurring changes so slightly that for the most part, it is considered to remain the same. Examples 4–26 and 4–27 illustrate this concept.

Example 4–26

Survey on Stress

A Harris poll found that 46% of Americans say they suffer great stress at least once a week. If three people are selected at random, find the probability that all three will say that they suffer great stress at least once a week.

Source: *100% American*.

Solution

Let *S* denote stress. Then

P(*S* and *S* and *S*) = P(*S*) · P(*S*) · *P(S)*

= (0.46)(0.46)(0.46) ≈ 0.097

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Example 4–27

Male Color Blindness

Approximately 9% of men have a type of color blindness that prevents them from distinguishing between red and green. If 3 men are selected at random, find the probability that all of them will have this type of red-green color blindness.

Source: *USA TODAY*.

Solution

Let *C* denote red-green color blindness. Then

*P(C* and *C* and *C*) = *P(C*) · *P(C*) · *P(C*)

= (0.09)(0.09)(0.09)

= 0.000729

Hence, the rounded probability is 0.0007.

In Examples 4–23 through 4–27, the events were independent of one another, since the occurrence of the first event in no way affected the outcome of the second event. On the other hand, when the occurrence of the first event changes the probability of the occurrence of the second event, the two events are said to be *dependent*. For example, suppose a card is drawn from a deck and *not* replaced, and then a second card is drawn. What is the probability of selecting an ace on the first card and a king on the second card?

Before an answer to the question can be given, you must realize that the events are dependent. The probability of selecting an ace on the first draw is . If that card is *not* replaced, the probability of selecting a king on the second card is , since there are 4 kings and 51 cards remaining. The outcome of the first draw has affected the outcome of the second draw.

Dependent events are formally defined now.

When the outcome or occurrence of the first event affects the outcome or occurrence of the second event in such a way that the probability is changed, the events are said to be **dependent events.**

Here are some examples of dependent events:

Drawing a card from a deck, not replacing it, and then drawing a second card.

Selecting a ball from an urn, not replacing it, and then selecting a second ball.

Being a lifeguard and getting a suntan.

Having high grades and getting a scholarship.

Parking in a no-parking zone and getting a parking ticket.

To find probabilities when events are dependent, use the multiplication rule with a modification in notation. For the problem just discussed, the probability of getting an ace on the first draw is , and the probability of getting a king on the second draw is . By the multiplication rule, the probability of both events occurring is

The event of getting a king on the second draw *given* that an ace was drawn the first time is called a *conditional probability*.

The **conditional probability** of an event *B* in relationship to an event *A* is the probability that event *B* occurs after event *A* has already occurred. The notation for conditional probability is *P*(*B*|*A*). This notation does not mean that *B* is divided by *A;* rather, it means the probability that event *B* occurs given that event *A* has already occurred. In the card example, *P(B|A)* is the probability that the second card is a king given that the first card is an ace, and it is equal to since the first card was *not* replaced.

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**Multiplication Rule 2**

When two events are dependent, the probability of both occurring is

*P*(*A* and *B*) = *P*(*A*) · *P(B|A)*

Example 4–28

University Crime

At a university in western Pennsylvania, there were 5 burglaries reported in 2003, 16 in 2004, and 32 in 2005. If a researcher wishes to select at random two burglaries to further investigate, find the probability that both will have occurred in 2004.

Source: IUP Police Department.

Solution

In this case, the events are dependent since the researcher wishes to investigate two distinct cases. Hence the first case is selected and not replaced.

Example 4–29

Homeowner’s and Automobile Insurance

World Wide Insurance Company found that 53% of the residents of a city had homeowner’s insurance (H) with the company. Of these clients, 27% also had automobile insurance (A) with the company. If a resident is selected at random, find the probability that the resident has both homeowner’s and automobile insurance with World Wide Insurance Company.

Solution

*P*(H and A) = *P*(H) · *P*(A|H) = (0.53)(0.27) = 0.1431

This multiplication rule can be extended to three or more events, as shown in Example 4–30.

Example 4–30

**Drawing Cards**

Three cards are drawn from an ordinary deck and not replaced. Find the probability of these events.

*a*.Getting 3 jacks

*b*.Getting an ace, a king, and a queen in order

*c*.Getting a club, a spade, and a heart in order

*d*.Getting 3 clubs

Solution

*a*.P(3 jacks) =

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*b*.P(ace and king and queen) =

*c*.P(club and spade and heart) =

*d*.P(3 clubs) =

Tree diagrams can be used as an aid to finding the solution to probability problems when the events are sequential. Example 4–31 illustrates the use of tree diagrams.

Example 4–31

Selecting Colored Balls

Box 1 contains 2 red balls and 1 blue ball. Box 2 contains 3 blue balls and 1 red ball. A coin is tossed. If it falls heads up, box 1 is selected and a ball is drawn. If it falls tails up, box 2 is selected and a ball is drawn. Find the probability of selecting a red ball.

Solution

The first two branches designate the selection of either box 1 or box 2. Then from box 1, either a red ball or a blue ball can be selected. Likewise, a red ball or blue ball can be selected from box 2. Hence a tree diagram of the example is shown in Figure 4–6.

Next determine the probabilities for each branch. Since a coin is being tossed for the box selection, each branch has a probability of , that is, heads for box 1 or tails for box 2. The probabilities for the second branches are found by using the basic probability rule. For example, if box 1 is selected and there are 2 red balls and 1 blue ball, the probability of selecting a red ball is and the probability of selecting a blue ball is . If box 2 is selected and it contains 3 blue balls and 1 red ball, then the probability of selecting a red ball is and the probability of selecting a blue ball is .

Next multiply the probability for each outcome, using the rule *P*(*A* and *B*) = *P(A)·P(B|A)*. For example, the probability of selecting box 1 and selecting a red ball is . The probability of selecting box 1 and a blue ball is . The probability of selecting box 2 and selecting a red ball is . The probability of selecting box 2 and a blue ball is . (Note that the sum of these probabilities is 1.)

Finally a red ball can be selected from either box 1 or box 2 so .

Figure 4–6

Tree Diagram for Example 4–31

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Tree diagrams can be used when the events are independent or dependent, and they can also be used for sequences of three or more events.

Objective 4

Find the conditional probability of an event.

**Conditional Probability**

The conditional probability of an event *B* in relationship to an event *A* was defined as the probability that event *B* occurs after event *A* has already occurred.

The conditional probability of an event can be found by dividing both sides of the equation for multiplication rule 2 by *P(A)*, as shown:

**Formula for Conditional Probability**

The probability that the second event *B* occurs given that the first event *A* has occurred can be found by dividing the probability that both events occurred by the probability that the first event has occurred. The formula is

Examples 4–32, 4–33, and 4–34 illustrate the use of this rule.

Example 4–32

Selecting Colored Chips

A box contains black chips and white chips. A person selects two chips without replacement. If the probability of selecting a black chip *and* a white chip is , and the probability of selecting a black chip on the first draw is , find the probability of selecting the white chip on the second draw, *given* that the first chip selected was a black chip.

Solution

Let

*B =* selecting a black chip

*W =* selecting a white chip

Then

Hence, the probability of selecting a white chip on the second draw given that the first chip selected was black is .

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Example 4–33

Parking Tickets

The probability that Sam parks in a no-parking zone *and* gets a parking ticket is 0.06, and the probability that Sam cannot find a legal parking space and has to park in the noparking zone is 0.20. On Tuesday, Sam arrives at school and has to park in a no-parking zone. Find the probability that he will get a parking ticket.

Solution

Let

*N =* parking in a no-parking zone

*T =* getting a ticket

Then

Hence, Sam has a 0.30 probability of getting a parking ticket, given that he parked in a no-parking zone.

The conditional probability of events occurring can also be computed when the data are given in table form, as shown in Example 4–34.

Example 4–34

Survey on Women in the Military

A recent survey asked 100 people if they thought women in the armed forces should be permitted to participate in combat. The results of the survey are shown.

Find these probabilities.

*a*.The respondent answered yes, given that the respondent was a female.

*b*.The respondent was a male, given that the respondent answered no.

Solution

Let

*M =* respondent was a male

*Y =* respondent answered yes

*F =* respondent was a female

*N =* respondent answered no

*a*.The problem is to find *P(Y|F)*. The rule states

The probability *P*(*F* and *Y*) is the number of females who responded yes, divided by the total number of respondents:

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The probability *P(F*) is the probability of selecting a female:

Then

*b*.The problem is to find *P(M|N*).

The Venn diagram for conditional probability is shown in Figure 4–7. In this case,

which is represented by the area in the intersection or overlapping part of the circles *A* and *B*, divided by the area of circle *A*. The reasoning here is that if you assume *A* has occurred, then *A*becomes the sample space for the next calculation and is the denominator of the probability fraction . The numerator *P*(*A* and *B*) represents the probability of the part of *B* that is contained in *A*. Hence, *P*(*A* and *B*) becomes the numerator of the probability fraction . Imposing a condition reduces the sample space.

**Probabilities for “At Least”**

The multiplication rules can be used with the complementary event rule (Section 4–1) to simplify solving probability problems involving “at least.” Examples 4–35, 4–36, and 4–37 illustrate how this is done.

Figure 4–7

Venn Diagram for Conditional Probability

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Example 4–35

**Drawing Cards**

A game is played by drawing 4 cards from an ordinary deck and replacing each card after it is drawn. Find the probability that at least 1 ace is drawn.

Solution

It is much easier to find the probability that no aces are drawn (i.e., losing) and then subtract that value from 1 than to find the solution directly, because that would involve finding the probability of getting 1 ace, 2 aces, 3 aces, and 4 aces and then adding the results.

Let *E =* at least 1 ace is drawn and = no aces drawn. Then

Hence,

or a hand with at least 1 ace will win about 27% of the time.

Example 4–36

Tossing Coins

A coin is tossed 5 times. Find the probability of getting at least 1 tail.

Solution

It is easier to find the probability of the complement of the event, which is “all heads,” and then subtract the probability from 1 to get the probability of at least 1 tail.

Hence,

Example 4–37

The Neckware Association of America reported that 3% of ties sold in the United States are bow ties. If 4 customers who purchased a tie are randomly selected, find the probability that at least 1 purchased a bow tie.

Solution

Let *E =* at least 1 bow tie is purchased and = no bow ties are purchased. Then

*P(E*) = 0.03 and *P*() = 1 – 0.03 = 0.97

*P*(no bow ties are purchased) = (0.97)(0.97)(0.97)(0.97) ≈ 0.885; hence,

*P*(at least one bow tie is purchased) = 1 – 0.885 = 0.115.

Similar methods can be used for problems involving “at most.”

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*Applying the Concepts* 4–3

**Guilty or Innocent?**

In July 1964, an elderly woman was mugged in Costa Mesa, California. In the vicinity of the crime a tall, bearded man sat waiting in a yellow car. Shortly after the crime was committed, a young, tall woman, wearing her blond hair in a ponytail, was seen running from the scene of the crime and getting into the car, which sped off. The police broadcast a description of the suspected muggers. Soon afterward, a couple fitting the description was arrested and convicted of the crime. Although the evidence in the case was largely circumstantial, the two people arrested were nonetheless convicted of the crime. The prosecutor based his entire case on basic probability theory, showing the unlikeness of another couple being in that area while having all the same characteristics that the elderly woman described. The following probabilities were used.

Characteristic | Assumed probability |

Drives yellow car | 1 out of 12 |

Man over 6 feet tall | 1 out of 10 |

Man wearing tennis shoes | 1 out of 4 |

Man with beard | 1 out of 11 |

Woman with blond hair | 1 out of 3 |

Woman with hair in a ponytail | 1 out of 13 |

Woman over 6 feet tall | 1 out of 100 |

1.Compute the probability of another couple being in that area with the same characteristics.

2.Would you use the addition or multiplication rule? Why?

3.Are the characteristics independent or dependent?

4.How are the computations affected by the assumption of independence or dependence?

5.Should any court case be based solely on probabilities?

6.Would you convict the couple who was arrested even if there were no eyewitnesses?

7.Comment on why in today’s justice system no person can be convicted solely on the results of probabilities.

8.In actuality, aren’t most court cases based on uncalculated probabilities?

See pages 257 and 258 for the answers.

Exercises 4–3 |

**1.**State which events are independent and which are dependent.

*a*.Tossing a coin and drawing a card from a deck

*b*.Drawing a ball from an urn, not replacing it, and then drawing a second ball

*c*.Getting a raise in salary and purchasing a new car

*d*.Driving on ice and having an accident

*e*.Having a large shoe size and having a high IQ

*f*.A father being left-handed and a daughter being left-handed

*g*.Smoking excessively and having lung cancer

*h*.Eating an excessive amount of ice cream and smoking an excessive amount of cigarettes

**2.Exercise** If 37% of high school students said that they exercise regularly, find the probability that 5 randomly selected high school students will say that they exercise regularly. Would you consider this event likely or unlikely to occur? Explain your answer.

**3.Video and Computer Games** Sixty-nine percent of U.S. heads of households play video or computer games. Choose 4 heads of households at random. Find the probability that

*a*.None play video or computer games

*b*.All four do

Source: www.theesa.com

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**4.Seat Belt Use** The Gallup Poll reported that 52% of Americans used a seat belt the last time they got into a car. If 4 people are selected at random, find the probability that they all used a seat belt the last time they got into a car.

Source: *100% American*.

**5.Medical Degrees** If 28% of U.S. medical degrees are conferred to women, find the probability that 3 randomly selected medical school graduates are men. Would you consider this event likely or unlikely to occur? Explain your answer.

**6.Prison Populations** If 25% of U.S. federal prison inmates are not U.S. citizens, find the probability that 2 randomly selected federal prison inmates will not be U.S. citizens.

Source: *Harper’s Index*.

**7.Computer Ownership** At a local university 54.3% of incoming first-year students have computers. If 3 students are selected at random, find the following probabilities.

*a*.None have computers.

*b*.At least one has a computer.

*c*.All have computers.

**8.Cards** If 2 cards are selected from a standard deck of 52 cards without replacement, find these probabilities.

*a*.Both are spades.

*b*.Both are the same suit.

*c*.Both are kings.

**9.MLS Players** Of the 216 players on major league soccer rosters, 80.1% are U.S. citizens. If 3 players are selected at random for an exhibition, what is the probability that all are U.S. citizens?

Source: *USA TODAY*.

**10.Cable Television** In 2006, 86% of U.S. households had cable TV. Choose 3 households at random. Find the probability that

*a*.None of the 3 households had cable TV

*b*.All 3 households had cable TV

*c*.At least 1 of the 3 households had cable TV

Source: www.infoplease.com

**11.Working Women and Computer Use** It is reported that 72% of working women use computers at work. Choose 5 working women at random. Find

*a*.The probability that at least 1 doesn’t use a computer at work

*b*.The probability that all 5 use a computer in their jobs

Source: www.infoplease.com

**12.Flashlight Batteries** A flashlight has 6 batteries, 2 of which are defective. If 2 are selected at random without replacement, find the probability that both are defective.

**13.Leisure Time Exercise** Only 27% of U.S. adults get enough leisure time exercise to achieve cardiovascular fitness. Choose 3 adults at random. Find the probability that

*a*.All 3 get enough daily exercise

*b*.At least 1 of the 3 gets enough exercise

Source: www.infoplease.com

**14.Text Messages via Cell Phones** Thirty-five percent of people who own cell phones use their phones to send and receive text messages. Choose 4 cell phone owners at random. What is the probability that none use their phones for texting?

**15.Customer Purchases** In a department store there are 120 customers, 90 of whom will buy at least 1 item. If 5 customers are selected at random, one by one, find the probability that all will buy at least 1 item.

**16.Drawing Cards** Three cards are drawn from a deck *without* replacement. Find these probabilities.

*a*.All are jacks.

*b*.All are clubs.

*c*.All are red cards.

**17.Scientific Study** In a scientific study there are 8 guinea pigs, 5 of which are pregnant. If 3 are selected at random without replacement, find the probability that all are pregnant.

**18.**In Exercise 17, find the probability that none are pregnant.

**19.Membership in a Civic Organization** In a civic organization, there are 38 members; 15 are men and 23 are women. If 3 members are selected to plan the July 4th parade, find the probability that all 3 are women. Would you consider this event likely or unlikely to occur? Explain your answer.

**20.**In Exercise 19, find the probability that all 3 members are men.

**21.Sales** A manufacturer makes two models of an item: model I, which accounts for 80% of unit sales, and model II, which accounts for 20% of unit sales. Because of defects, the manufacturer has to replace (or exchange) 10% of its model I and 18% of its model II. If a model is selected at random, find the probability that it will be defective.

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**22.Student Financial Aid** In a recent year 8,073,000 male students and 10,980,000 female students were enrolled as undergraduates. Receiving aid were 60.6% of the male students and 65.2% of the female students. Of those receiving aid, 44.8% of the males got federal aid and 50.4% of the females got federal aid. Choose 1 student at random. *(Hint:* Make a tree diagram.) Find the probability that the student is

*a*.A male student without aid

*b*.A male student, given that the student has aid

*c*.A female student or a student who receives federal aid

Source: www.nces.gov

**23.Automobile Insurance** An insurance company classifies drivers as low-risk, medium-risk, and high-risk. Of those insured, 60% are low-risk, 30% are medium-risk, and 10% are high-risk. After a study, the company finds that during a 1-year period, 1% of the low-risk drivers had an accident, 5% of the medium-risk drivers had an accident, and 9% of the high-risk drivers had an accident. If a driver is selected at random, find the probability that the driver will have had an accident during the year.

**24.Defective Items** A production process produces an item. On average, 15% of all items produced are defective. Each item is inspected before being shipped, and the inspector misclassifies an item 10% of the time. What proportion of the items will be “classified as good”? What is the probability that an item is defective given that it was classified as good?

**25.Selecting Colored Balls** Urn 1 contains 5 red balls and 3 black balls. Urn 2 contains 3 red balls and 1 black ball. Urn 3 contains 4 red balls and 2 black balls. If an urn is selected at random and a ball is drawn, find the probability it will be red.

**26.Prison Populations** For a recent year, 0.99 of the incarcerated population is adults and 0.07 is female. If an incarcerated person is selected at random, find the probability that the person is a female given that the person is an adult.

Source: Bureau of Justice.

**27.Rolling Dice** Roll two standard dice and add the numbers. What is the probability of getting a number larger than 9 for the first time on the third roll?

**28.Model Railroad Circuit** A circuit to run a model railroad has 8 switches. Two are defective. If you select 2 switches at random and test them, find the probability that the second one is defective, given that the first one is defective.

**29.Country Club Activities** At the Avonlea Country Club, 73% of the members play bridge and swim, and 82% play bridge. If a member is selected at random, find the probability that the member swims, given that the member plays bridge.

**30.College Courses** At a large university, the probability that a student takes calculus and is on the dean’s list is 0.042. The probability that a student is on the dean’s list is 0.21. Find the probability that the student is taking calculus, given that he or she is on the dean’s list.

**31.House Types** In Rolling Acres Housing Plan, 42% of the houses have a deck and a garage; 60% have a deck. Find the probability that a home has a garage, given that it has a deck.

**32.Pizza and Salads** In a pizza restaurant, 95% of the customers order pizza. If 65% of the customers order pizza and a salad, find the probability that a customer who orders pizza will also order a salad.

**33.Gift Baskets** The Gift Basket Store had the following premade gift baskets containing the following combinations in stock.

Choose 1 basket at random. Find the probability that it contains

*a*.Coffee or candy

*b*.Tea given that it contains mugs

*c*.Tea and cookies

Source: www.infoplease.com

**34.Blood Types and Rh Factors** In addition to being grouped into four types, human blood is grouped by its Rhesus (Rh) factor. Consider the figures below which show the distributions of these groups for Americans.

Choose 1 American at random. Find the probability that the person

*a*.Is a universal donor, i.e., has O negative blood

*b*.Has type O blood given that the person is Rh+

*c*.Has A+ or AB– blood

*d*.Has Rh– given that the person has type B

Source: www.infoplease.com

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**35.Doctor Specialties** Below are listed the numbers of doctors in various specialties by gender.

Choose 1 doctor at random.

*a*.Find *P* (Male|pediatrician).

*b*.Find *P* (Pathologist|female).

*c*.Are the characteristics “female” and “pathologist” independent? Explain.

Source: *World Almanac*.

**36.Olympic Medals** The medal distribution from the 2004 Summer Olympic Games for the top 23 countries is shown below.

Choose 1 medal winner at random.

*a*.Find the probability that the winner won the gold medal, given that the winner was from the United States.

*b*.Find the probability that the winner was from the United States, given that she or he won a gold medal.

*c*.Are the events “medal winner is from United States” and “gold medal won” independent? Explain.

Source: *New York Times Almanac*.

**37.Marital Status of Women** According to the *Statistical Abstract of the United States*, 70.3% of females ages 20 to 24 have never been married. Choose 5 young women in this age category at random. Find the probability that

*a*.None have ever been married

*b*.At least 1 has been married

Source: *New York Times Almanac*.

**38.Fatal Accidents** The American Automobile Association (AAA) reports that of the fatal car and truck accidents, 54% are caused by car driver error. If 3 accidents are chosen at random, find the probability that

*a*.All are caused by car driver error

*b*.None are caused by car driver error

*c*.At least 1 is caused by car driver error

Source: AAA quoted on CNN.

**39.On-Time Airplane Arrivals** The greater Cincinnati airport led major U.S. airports in on-time arrivals in the last quarter of 2005 with an 84.3% on-time rate. Choose 5 arrivals at random and find the probability that at least 1 was not on time.

Source: www.census.gov

**40.Online Electronic Games** Fifty-six percent of electronic gamers play games online, and sixty-four percent of those gamers are female. What is the probability that a randomly selected gamer plays games online and is male?

Source: www.tech.msn.com

**41.Reading to Children** Fifty-eight percent of American children (ages 3 to 5) are read to every day by someone at home. Suppose 5 children are randomly selected. What is the probability that at least 1 is read to every day by someone at home?

Source: Federal Interagency Forum on Child and Family Statistics.

**42.Doctoral Assistantships** Of Ph.D. students, 60% have paid assistantships. If 3 students are selected at random, find the probabilities

*a*.All have assistantships

*b*.None have assistantships

*c*.At least 1 has an assistantship

Source: U.S. Department of Education/*Chronicle of Higher Education*.

**43.Selecting Cards If** 4 cards are drawn from a deck of 52 and not replaced, find the probability of getting at least 1 club.

**44.Full-Time College Enrollment** The majority (69%) of undergraduate students were enrolled in a 4-year college in a recent year. Eighty-one percent of those enrolled attended full-time. Choose 1 enrolled undergraduate student at random. What is the probability that she or he is a part-time student at a 4-year college?

Source: www.census.gov

**45.Family and Children’s Computer Games** It was reported that 19.8% of computer games sold in 2005 were classified as “family and children’s.” Choose 5 purchased computer games at random. Find the probability that

*a*.None of the 5 were family and children’s

*b*.At least 1 of the 5 was family and children’s

Source: www.theesa.com

**46.Medication Effectiveness** A medication is 75% effective against a bacterial infection. Find the probability that if 12 people take the medication, at least 1 person’s infection will not improve.

**47.Tossing a Coin** A coin is tossed 5 times; find the probability of getting at least 1 tail. Would you consider this event likely to happen? Explain your answer.

**48.Selecting a Letter of the Alphabet** If 3 letters of the alphabet are selected at random, find the probability of getting at least 1 letter x. Letters can be used more than once. Would you consider this event likely to happen? Explain your answer.

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**49.Rolling a Die** A die is rolled 7 times. Find the probability of getting at least one 3. Would you consider this event likely to occur? Explain your answer.

**50.High School Grades of First-Year College Students** Forty-seven percent of first-year college students enrolled in 2005 had an average grade of A in high school compared to 20 percent of first-year college students in 1970. Choose 6 first-year college students at random enrolled in 2005. Find the probability that

*a*.All had an A average in high school

*b*.None had an A average in high school

*c*.At least 1 had an A average in high school

Source: www.census.gov

**51.Rolling a Die** If a die is rolled 3 times, find the probability of getting at least 1 even number.

**52.Selecting a Flower** In a large vase, there are 8 roses, 5 daisies, 12 lilies, and 9 orchids. If 4 flowers are selected at random, find the probability that at least 1 of the flowers is a rose. Would you consider this event likely to occur? Explain your answer.

**Extending the Concepts**

**53.**Let *A* and *B* be two mutually exclusive events. Are *A* and *B* independent events? Explain your answer.

**54.Types of Vehicles** The Bargain Auto Mall has the following cars in stock.

Are the events “compact” and “domestic” independent? Explain.

**55.College Enrollment** An admissions director knows that the probability a student will enroll after a campus visit is 0.55, or *P(E*) = 0.55. While students are on campus visits, interviews with professors are arranged. The admissions director computes these conditional probabilities for students enrolling after visiting three professors, DW, LP, and MH.

*P(E*|DW) = 0.95 *P(E*|LP) = 0.55 *P(E*|MH) = 0.15

Is there something wrong with the numbers? Explain.

**56.Commercials** Event *A* is the event that a person remembers a certain product commercial. Event *B* is the event that a person buys the product. If *P*(*B*) = 0.35, comment on each of these conditional probabilities if you were vice president for sales.

*a*.*P(B|A)* = 0.20

*b*.*P(B|A)* = 0.35

*c*.*P(B|A)* = 0.55

Objective 5

Find the total number of outcomes in a sequence of events, using the fundamental counting rule.

**4–4Counting Rules**

Many times a person must know the number of all possible outcomes for a sequence of events. To determine this number, three rules can be used: the *fundamental counting rule*, the *permutation rule*, and the *combination rule*. These rules are explained here, and they will be used in Section 4–5 to find probabilities of events.

The first rule is called the **fundamental counting rule.**

**The Fundamental Counting Rule**

**Fundamental Counting Rule**

In a sequence of *n* events in which the first one has *k _{1}* possibilities and the second event has

*k*and the third has

_{2}*k*, and so forth, the total number of possibilities of the sequence will be

_{3}*k _{1}* ·

*k*·

_{2}*k*…

_{3}*k*

_{n}*Note:* In this case *and* means to multiply.

Examples 4–38 through 4–41 illustrate the fundamental counting rule.

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Example 4–38

Tossing a Coin and Rolling a Die

A coin is tossed and a die is rolled. Find the number of outcomes for the sequence of events.

Figure 4–8

Complete Tree Diagram for Example 4–38

*Interesting Fact*

Possible games of chess: 25 × 10^{115.}

Solution

Since the coin can land either heads up or tails up and since the die can land with any one of six numbers showing face up, there are 2 · 6 = 12 possibilities. A tree diagram can also be drawn for the sequence of events. See Figure 4–8.

Example 4–39

Types of Paint

A paint manufacturer wishes to manufacture several different paints. The categories include

Color | Red, blue, white, black, green, brown, yellow |

Type | Latex, oil |

Texture | Flat, semigloss, high gloss |

Use | Outdoor, indoor |

How many different kinds of paint can be made if you can select one color, one type, one texture, and one use?

Solution

You can choose one color and one type and one texture and one use. Since there are 7 color choices, 2 type choices, 3 texture choices, and 2 use choices, the total number of possible different paints is

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Example 4–40

Distribution of Blood Types

There are four blood types, A, B, AB, and O. Blood can also be Rh+ and Rh–. Finally, a blood donor can be classified as either male or female. How many different ways can a donor have his or her blood labeled?

Figure 4–9

Complete Tree Diagram for Example 4–40

Solution

Since there are 4 possibilities for blood type, 2 possibilities for Rh factor, and 2 possibilities for the gender of the donor, there are 4 · 2 · 2, or 16, different classification categories, as shown.

A tree diagram for the events is shown in Figure 4–9.

When determining the number of different possibilities of a sequence of events, you must know whether repetitions are permissible.

Example 4–41

Identification Cards

The manager of a department store chain wishes to make four-digit identification cards for her employees. How many different cards can be made if she uses the digits 1, 2, 3, 4, 5, and 6 and repetitions are permitted?

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Solution

Since there are 4 spaces to fill on each card and there are 6 choices for each space, the total number of cards that can be made is 6 · 6 · 6 · 6 = 1296.

Now, what if repetitions are not permitted? For Example 4–41, the first digit can be chosen in 6 ways. But the second digit can be chosen in only 5 ways, since there are only five digits left, etc. Thus, the solution is

6 · 5 · 4 · 3 = 360

The same situation occurs when one is drawing balls from an urn or cards from a deck. If the ball or card is replaced before the next one is selected, then repetitions are permitted, since the same one can be selected again. But if the selected ball or card is not replaced, then repetitions are not permitted, since the same ball or card cannot be selected the second time.

These examples illustrate the fundamental counting rule. In summary: *If repetitions are permitted, then the numbers stay the same going from left to right. If repetitions are not permitted, then the numbers decrease by 1 for each place left to right*.

Two other rules that can be used to determine the total number of possibilities of a sequence of events are the permutation rule and the combination rule.

**Factorial Notation**

These rules use *factorial notation*. The factorial notation uses the exclamation point.

5! = 5 · 4 · 3 · 2 · 1

9! = 9 · 8 · 7 · 6 · 5 · 4 · 3 · 2 · 1

To use the formulas in the permutation and combination rules, a special definition of 0! is needed. 0! = 1.

Historical Note

In 1808 Christian Kramp first used the factorial notation.

Factorial Formulas

For any counting *n*

*n*! = *n*(*n* – 1)(n – 2) · · · 1

0! = 1

**Permutations**

A **permutation** is an arrangement of *n* objects in a specific order.

Examples 4–42 and 4–43 illustrate permutations.

Example 4–42

Business Location

Suppose a business owner has a choice of 5 locations in which to establish her business. She decides to rank each location according to certain criteria, such as price of the store and parking facilities. How many different ways can she rank the 5 locations?

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Solution

There are

5! = 5 · 4 · 3 · 2 · 1 = 120

different possible rankings. The reason is that she has 5 choices for the first location, 4 choices for the second location, 3 choices for the third location, etc.

In Example 4–42 all objects were used up. But what happens when not all objects are used up? The answer to this question is given in Example 4–43.

Example 4–43

Business Location

Suppose the business owner in Example 4–42 wishes to rank only the top 3 of the 5 locations. How many different ways can she rank them?

Solution

Using the fundamental counting rule, she can select any one of the 5 for first choice, then any one of the remaining 4 locations for her second choice, and finally, any one of the remaining locations for her third choice, as shown.

The solutions in Examples 4–42 and 4–43 are permutations.

Objective 6

Find the number of ways that *r* objects can be selected from *n* objects, using the permutation rule.

**Permutation Rule**

The arrangement of *n* objects in a specific order using *r* objects at a time is called a *permutation of n objects taking r objects at a time*. It is written as * _{n}P_{r}*, and the formula is

The notation * _{n}P_{r}* is used for permutations.

Although Examples 4–42 and 4–43 were solved by the multiplication rule, they can now be solved by the permutation rule.

In Example 4–42, 5 locations were taken and then arranged in order; hence,

(Recall that 0! = 1.)

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In Example 4–43, 3 locations were selected from 5 locations, so *n* = 5 and *r* = 3; hence

Examples 4–44 and 4–45 illustrate the permutation rule.

Example 4–44

Television News Stories

A television news director wishes to use 3 news stories on an evening show. One story will be the lead story, one will be the second story, and the last will be a closing story. If the director has a total of 8 stories to choose from, how many possible ways can the program be set up?

Solution

Since order is important, the solution is

Hence, there would be 336 ways to set up the program.

Example 4–45

School Musical Plays

A school musical director can select 2 musical plays to present next year. One will be presented in the fall, and one will be presented in the spring. If she has 9 to pick from, how many different possibilities are there?

Solution

Order is important since one play can be presented in the fall and the other play in the spring.

There are 72 different possibilities.

Objective 7

Find the number of ways that *r* objects can be selected from *n* objects without regard to order, using the combination rule.

**Combinations**

Suppose a dress designer wishes to select two colors of material to design a new dress, and she has on hand four colors. How many different possibilities can there be in this situation?

This type of problem differs from previous ones in that the order of selection is not important. That is, if the designer selects yellow and red, this selection is the same as the selection red and yellow. This type of selection is called a *combination*. The difference between a permutation and a combination is that in a combination, the order or arrangement of the objects is not important; by contrast, order *is* important in a permutation. Example 4–46 illustrates this difference.

A selection of distinct objects without regard to order is called a **combination.**

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Example 4–46

Letters

Given the letters A, B, C, and D, list the permutations and combinations for selecting two letters.

Solution

The permutations are

In permutations, AB is different from BA. But in combinations, AB is the same as BA since the order of the objects does not matter in combinations. Therefore, if duplicates are removed from a list of permutations, what is left is a list of combinations, as shown.

Hence the combinations of A, B, C, and D are AB, AC, AD, BC, BD, and CD. (Alternatively, BA could be listed and AB crossed out, etc.) The combinations have been listed alphabetically for convenience, but this is not a requirement.

*Combinations are used when the order or arrangement is not important, as in the selecting process*. Suppose a committee of 5 students is to be selected from 25 students. The five selected students represent a combination, since it does not matter who is selected first, second, etc.

*Interesting Fact*

The total number of hours spent mowing lawns in the United States each year: 2,220,000,000.

**Combination Rule**

The number of combinations of *r* objects selected from *n* objects is denoted by * _{n}C_{r}* and is given by the formula

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Example 4–47

Combinations

How many combinations of 4 objects are there, taken 2 at a time?

Solution

Since this is a combination problem, the answer is

This is the same result shown in Example 4–46.

Notice that the expression for * _{n}C_{r}* is

which is the formula for permutations with r! in the denominator. In other words,

This *r*! divides out the duplicates from the number of permutations, as shown in Example 4–46. For each two letters, there are two permutations but only one combination. Hence, dividing the number of permutations by *r*! eliminates the duplicates. This result can be verified for other values of *n* and *r. Note: _{n}C_{n} =* 1.

Example 4–48

Book Reviews

A newspaper editor has received 8 books to review. He decides that he can use 3 reviews in his newspaper. How many different ways can these 3 reviews be selected?

Solution

There are 56 possibilities.

Example 4–49

Committee Selection

In a club there are 7 women and 5 men. A committee of 3 women and 2 men is to be chosen. How many different possibilities are there?

Solution

Here, you must select 3 women from 7 women, which can be done in _{7}*C*_{3}, or 35, ways. Next, 2 men must be selected from 5 men, which can be done in _{5}*C*_{2}, or 10, ways. Finally, by the fundamental counting rule, the total number of different ways is 35 · 10 = 350, since you are choosing both men and women. Using the formula gives Table 4–1 summarizes the counting rules.

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Table 4–1 | Summary of Counting Rules | |

Rule | Definition | Formula |

Fundamental counting rule | The number of ways a sequence of n events can occur if the first event can occur in k ways, the second event can occur in _{1}k_{2} ways, etc. | k_{1} · k_{2} · k_{3} … k_{n} |

Permutation rule | The number of permutations of n objects taking r objects at a time (order is important) | |

Combination rule | The number of combinations of r objects taken from n objects (order is not important) |

A list of permutations and combinations is found in Table C of Appendix C.

*Applying the Concepts* 4–4

**Garage Door Openers**

Garage door openers originally had a series of four on/off switches so that homeowners could personalize the frequencies that opened their garage doors. If all garage door openers were set at the same frequency, anyone with a garage door opener could open anyone else’s garage door.

1.Use a tree diagram to show how many different positions 4 consecutive on/off switches could be in.

After garage door openers became more popular, another set of 4 on/off switches was added to the systems.

2.Find a pattern of how many different positions are possible with the addition of each on/off switch.

3.How many different positions are possible with 8 consecutive on/off switches?

4.Is it reasonable to assume, if you owned a garage door opener with 8 switches, that someone could use his or her garage door opener to open your garage door by trying all the different possible positions?

In 1989 it was reported that the ignition keys for 1988 Dodge Caravans were made from a single blank that had five cuts on it. Each cut was made at one out of five possible levels. In 1988, assume there were 420,000 Dodge Caravans sold in the United States.

5.How many different possible keys can be made from the same key blank?

6.How many different 1988 Dodge Caravans could any one key start?

Look at the ignition key for your car and count the number of cuts on it. Assume that the cuts are made at one of any of five possible levels. Most car companies use one key blank for all their makes and models of cars.

7.Conjecture how many cars your car company sold over recent years, and then figure out how many other cars your car key could start. What would you do to decrease the odds of someone being able to open another vehicle with his or her key?

See page 258 for the answers.

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Exercises 4–4 |

**1.Zip Codes** How many 5-digit zip codes are possible if digits can be repeated? If there cannot be repetitions?

**2.Batting Order** How many ways can a baseball manager arrange a batting order of 9 players?

**3.Video Games** How many different ways can 7 different video game cartridges be arranged on a shelf?

**4.Seating Arrangements** In how many ways can 5 speakers be seated in a row on a stage?

**5.Shampoo Display** A store manager wishes to display 8 different brands of shampoo in a row. How many ways can this be done?

**6.Show Programs** Three bands and two comics are performing for a student talent show. How many different programs (in terms of order) can be arranged? How many if the comics must perform between bands?

**7.Campus Tours** Student volunteers take visitors on a tour of 7 campus buildings. How many different tours are possible? (Assume order is important.)

**8.Radio Station Call Letters** The call letters of a radio station must have 4 letters. The first letter must be a K or a W. How many different station call letters can be made if repetitions are not allowed? If repetitions are allowed?

**9.Identification Tags** How many different 3-digit identification tags can be made if the digits can be used more than once? If the first digit must be a 5 and repetitions are not permitted?

**10.Book Arrangements** A reference encyclopedia has 12 volumes. Disregarding alphabetical or numerical order, in how many ways can the books be arranged on a shelf?

**11.Selection of Officers** Six students are running for the positions of president and vice-president, and five students are running for secretary and treasurer. If the two highest vote getters in each of the two contests are elected, how many winning combinations can there be?

**12.Automobile Trips** There are 2 major roads from city *X* to city *Y* and 4 major roads from city *Y* to city *Z*. How many different trips can be made from city *X* to city *Z* passing through city *Y*?

**13.**Evaluate each of these.

*a*.8!

*b*.10!

*c*.0!

*d*.1!

*e*._{7}*P*_{5}

*f*._{12}*P*_{4}

*g*._{5}*P*_{3}

*h*._{6}*P*_{0}

*i*._{5}*P*_{5}

*j*._{6}*P*_{2}

**14.County Assessments** The County Assessment Bureau decides to reassess homes in 8 different areas. How many different ways can this be accomplished?

**15.Sports Car Stripes** How many different 4-color code stripes can be made on a sports car if each code consists of the colors green, red, blue, and white? All colors are used only once.

**16.Manufacturing Tests** An inspector must select 3 tests to perform in a certain order on a manufactured part. He has a choice of 7 tests. How many ways can he perform 3 different tests?

**17.Threatened Species of Reptiles** There are 22 threatened species of reptiles in the United States. In how many ways can you choose 4 to write about? (Order is not important.)

Source: www.infoplease.com

**18.Inspecting Restaurants** How many different ways can a city health department inspector visit 5 restaurants in a city with 10 restaurants?

**19.**How many different 4-letter permutations can be formed from the letters in the word *decagon?*

**20.Cell Phone Models** A particular cell phone company offers 4 models of phones, each in 6 different colors and each available with any one of 5 calling plans. How many combinations are possible?

**21.ID Cards** How many different ID cards can be made if there are 6 digits on a card and no digit can be used more than once?

**22.Free-Sample Requests** An online coupon service has 13 offers for free samples. How may different requests are possible if a customer must request exactly 3 free samples? How many are possible if the customer may request up to 3 free samples?

**23.Ticket Selection** How many different ways can 4 tickets be selected from 50 tickets if each ticket wins a different prize?

**24.Movie Selections** The Foreign Language Club is showing a four-movie marathon of subtitled movies. How many ways can they choose 4 from the 11 available?

**25.Task Assignments** How many ways can an adviser choose 4 students from a class of 12 if they are all assigned the same task? How many ways can the students be chosen if they are each given a different task?

**26.Agency Cases** An investigative agency has 7 cases and 5 agents. How many different ways can the cases be assigned if only 1 case is assigned to each agent?

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**27.(ans)** Evaluate each expression.

*a*._{5}*C*_{2}

*b*._{8}*C*_{3}

*c*._{7}*C*_{4}

*d*._{6}*C*_{2}

*e*._{6}*C*_{4}

*f*._{3}*C*_{0}

*g*._{3}*C*_{3}

*h*._{9}*C*_{7}

*i*._{12}*C*_{2}

*j*._{4}*C*_{3}

**28.Selecting Cards** How many ways can 3 cards be selected from a standard deck of 52 cards, disregarding the order of selection?

**29.Selecting Bracelets** How many ways are there to select 3 bracelets from a box of 10 bracelets, disregarding the order of selection?

**30.Selecting Players** How many ways can 4 baseball players and 3 basketball players be selected from 12 baseball players and 9 basketball players?

**31.Selecting a Committee** How many ways can a committee of 4 people be selected from a group of 10 people?

**32.Selecting Christmas Presents** If a person can select 3 presents from 10 presents under a Christmas tree, how many different combinations are there?

**33.Questions for a Test** How many different tests can be made from a test bank of 20 questions if the test consists of 5 questions?

**34.Promotional Program** The general manager of a fast-food restaurant chain must select 6 restaurants from 11 for a promotional program. How many different possible ways can this selection be done?

**35.Music Program Selections** A jazz band has prepared 18 selections for a concert tour. At each stop they will perform 10. How many different programs are possible? How many programs are possible if they always begin with the same song and end with the same song?

**36.Freight Train Cars** In a train yard there are 4 tank cars, 12 boxcars, and 7 flatcars. How many ways can a train be made up consisting of 2 tank cars, 5 boxcars, and 3 flatcars? (In this case, order is not important.)

**37.Selecting a Committee** There are 7 women and 5 men in a department. How many ways can a committee of 4 people be selected? How many ways can this committee be selected if there must be 2 men and 2 women on the committee? How many ways can this committee be selected if there must be at least 2 women on the committee?

**38.Selecting Cereal Boxes** Wake Up cereal comes in 2 types, crispy and crunchy. If a researcher has 10 boxes of each, how many ways can she select 3 boxes of each for a quality control test?

**39.Hawaiian Words** The Hawaiian alphabet consists of 7 consonants and 5 vowels. How many three-letter “words” are possible if there are never two consonants together and if a word must always end in a vowel?

**40.Selecting a Jury** How many ways can a jury of 6 women and 6 men be selected from 10 women and 12 men?

**41.Selecting a Golf Foursome** How many ways can a foursome of 2 men and 2 women be selected from 10 men and 12 women in a golf club?

**42.Investigative Team** The state narcotics bureau must form a 5-member investigative team. If it has 25 agents from which to choose, how many different possible teams can be formed?

**43.Dominoes** A domino is a flat rectangular block the face of which is divided into two square parts, each part showing from zero to six pips (or dots). Playing a game consists of playing dominoes with a matching number of pips. Explain why there are 28 dominoes in a complete set.

**44.Charity Event Participants** There are 16 seniors and 15 juniors in a particular social organization. In how many ways can 4 seniors and 2 juniors be chosen to participate in a charity event?

**45.Selecting Commercials** How many ways can a person select 7 television commercials from 11 television commercials?

**46.DVD Selection** How many ways can a person select 8 DVDs from a display of 13 DVDs?

**47.Candy Bar Selection** How many ways can a person select 6 candy bars from a list of 10 and 6 salty snacks from a list of 12 to put in a vending machine?

**48.Selecting a Location** An advertising manager decides to have an ad campaign in which 8 special calculators will be hidden at various locations in a shopping mall. If he has 17 locations from which to pick, how many different possible combinations can he choose?

**Permutations and Combinations**

**49.Selecting Posters** A buyer decides to stock 8 different posters. How many ways can she select these 8 if there are 20 from which to choose?

**50.Test Marketing Products** Anderson Research Company decides to test-market a product in 6 areas. How many different ways can 3 areas be selected in a certain order for the first test?

**51.Selecting Rats** How many different ways can a researcher select 5 rats from 20 rats and assign each to a different test?

**52.Selecting Musicals** How many different ways can a theatrical group select 2 musicals and 3 dramas from 11 musicals and 8 dramas to be presented during the year?

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**53.Textbook Selections** How many different ways can an instructor select 2 textbooks from a possible 17?

**54.Tape Selections** How many ways can a person select 8 videotapes from 10 videotapes?

**55.Public Service Announcements** How many different ways can 5 public service announcements be run during 1 hour?

**56.Signal Flags** How many different signals can be made by using at least 3 different flags if there are 5 different flags from which to select?

**57.Dinner Selections** How many ways can a dinner patron select 3 appetizers and 2 vegetables if there are 6 appetizers and 5 vegetables on the menu?

**58.Air Pollution** The Environmental Protection Agency must investigate 9 mills for complaints of air pollution. How many different ways can a representative select 5 of these to investigate this week?

**59.Selecting Officers** In a board of directors composed of 8 people, how many ways can one chief executive officer, one director, and one treasurer be selected?

**Extending the Concepts**

**60.Selecting Coins** How many different ways can you select one or more coins if you have 2 nickels, 1 dime, and 1 half-dollar?

**61.People Seated in a Circle** In how many ways can 3 people be seated in a circle? 4? *n? (Hint:* Think of them standing in a line before they sit down and/or draw diagrams.)

**62.Seating in a Movie Theater** How many different ways can 5 people—A, B, C, D, and E—sit in a row at a movie theater if (*a*) A and B must sit together; (*b*) C must sit to the right of, but not necessarily next to, B; (*c*) D and E will not sit next to each other?

**63.Poker Hands** Using combinations, calculate the number of each poker hand in a deck of cards. (A poker hand consists of 5 cards dealt in any order.)

*a*.Royal flush

*b*.Straight flush

*c*.Four of a kind

*a*.Full house

**Technology Step by Step**

**TI-83 Plus or TI-84 Plus**

Step by Step

**Factorials, Permutations, and Combinations**

**Factorials n!**

**1.**Type the value of *n*.

**2.**Press **MATH** and move the cursor to PRB, then press **4** for !.

**3.**Press **ENTER**.

**Permutations _{n}P_{r}**

**1.**Type the value of *n*.

**2.**Press **MATH** and move the cursor to PRB, then press **2** for * _{n}P_{r}*.

**3.**Type the value of *r*.

**4.**Press **ENTER.**

**Combinations**_{n}*C _{r}*

**1.**Type the value of *n*.

**2.**Press **MATH** and move the cursor to PRB, then press **3** for * _{n}C_{r}*.

**3.**Type the value of *r*.

**4.**Press **ENTER**.

Calculate 5!, _{8}*P*_{3}, and _{12}*C*_{5} (Examples 4–42, 4–44, and 4–48 from the text).

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Excel

Step by Step

**Permutations, Combinations, and Factorials**

To find a value of a permutation, for example, _{5}*P*_{3}:

**1.**In an open cell in an Excel worksheet, select the Formulas tab on the toolbar. Then click the Insert function icon .

**2.**Select the Statistical function category, then the PERMUT function, and click [OK].

**3.**Type **5** in the Number box.

**4.**Type **3** in the Number_chosen box and click [OK].

The selected cell will display the answer: 60.

To find a value of a combination, for example, _{5}*C*_{3}:

**1.**In an open cell, select the Formulas tab on the toolbar. Click the Insert function icon.

**2.**Select the All function category, then the COMBIN function, and click [OK].

**3.**Type **5** in the Number box.

**4.**Type **3** in the Number_chosen box and click [OK].

The selected cell will display the answer: 10.

To find a factorial of a number, for example, 7!:

**1.**In an open cell, select the Formulas tab on the toolbar. Click the Insert function icon.

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**2.**Select the Math & Trig function category, then the FACT function, and click [OK].

**3.**Type 7 in the Number box and click [OK].

The selected cell will display the answer: 5040.

Objective 8

Find the probability of an event, using the counting rules.

**4–5Probability and Counting Rules**

The counting rules can be combined with the probability rules in this chapter to solve many types of probability problems. By using the fundamental counting rule, the permutation rules, and the combination rule, you can compute the probability of outcomes of many experiments, such as getting a full house when 5 cards are dealt or selecting a committee of 3 women and 2 men from a club consisting of 10 women and 10 men.

Example 4–50

Four Aces

Find the probability of getting 4 aces when 5 cards are drawn from an ordinary deck of cards.

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Solution

There are _{52}*C*_{5} ways to draw 5 cards from a deck. There is only 1 way to get 4 aces (i.e., _{4}*C*_{4}), but there are 48 possibilities to get the fifth card. Therefore, there are 48 ways to get 4 aces and 1 other card. Hence,

Example 4–51

Defective Transistors

A box contains 24 transistors, 4 of which are defective. If 4 are sold at random, find the following probabilities.

*a*.Exactly 2 are defective.

*b*.None is defective.

*c*.All are defective.

*d*.At least 1 is defective.

Solution

There are _{24}*C*_{4} ways to sell 4 transistors, so the denominator in each case will be 10,626.

*a*.Two defective transistors can be selected as _{4}*C*_{2} and two nondefective ones as _{20}*C*_{2}. Hence,

*b*.The number of ways to choose no defectives is _{20}*C*_{4}. Hence,

*c*.The number of ways to choose 4 defectives from 4 is * _{4}C_{4}*, or 1. Hence,

*d*.To find the probability of at least 1 defective transistor, find the probability that there are no defective transistors, and then subtract that probability from 1.

Example 4–52

Magazines

A store has 6 *TV Graphic* magazines and 8 *Newstime* magazines on the counter. If two customers purchased a magazine, find the probability that one of each magazine was purchased.

Solution

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Example 4–53

**Combination Lock**

A combination lock consists of the 26 letters of the alphabet. If a 3-letter combination is needed, find the probability that the combination will consist of the letters ABC in that order. The same letter can be used more than once. *(Note:* A combination lock is really a permutation lock.)

Solution

Since repetitions are permitted, there are 26 · 26 · 26 = 17,576 different possible combinations. And since there is only one ABC combination, the probability is *P*(ABC) = 1/26^{3} = 1/17,576.

Example 4–54

Tennis Tournament

There are 8 married couples in a tennis club. If 1 man and 1 woman are selected at random to plan the summer tournament, find the probability that they are married to each other.

Solution

Since there are 8 ways to select the man and 8 ways to select the woman, there are 8 · 8, or 64, ways to select 1 man and 1 woman. Since there are 8 married couples, the solution is .

As indicated at the beginning of this section, the counting rules and the probability rules can be used to solve a large variety of probability problems found in business, gambling, economics, biology, and other fields.

*Applying the Concepts* 4–5

**Counting Rules and Probability**

One of the biggest problems for students when doing probability problems is to decide which formula or formulas to use. Another problem is to decide whether two events are independent or dependent. Use the following problem to help develop a better understanding of these concepts.

Assume you are given a 5-question multiple-choice quiz. Each question has 5 possible answers: A, B, C, D, and E.

1.How many events are there?

2.Are the events independent or dependent?

3.If you guess at each question, what is the probability that you get all of them correct?

4.What is the probability that a person would guess answer A for each question?

Assume that you are given a test in which you are to match the correct answers in the right column with the questions in the left column. You can use each answer only once.

5.How many events are there?

6.Are the events independent or dependent?

7.What is the probability of getting them all correct if you are guessing?

8.What is the difference between the two problems?

See page 258 for the answers.

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*Speaking of* Statistics

**The Mathematics of Gambling**

Gambling is big business. There are state lotteries, casinos, sports betting, and church bingos. It seems that today everybody is either watching or playing Texas Hold ’Em Poker.

Using permutations, combinations, and the probability rules, mathematicians can find the probabilities of various gambling games. Here are the probabilities of the various 5-card poker hands.

Hand | Number of ways | Probability |

Straight flush | 40 | 0.000015 |

Four of a kind | 624 | 0.000240 |

Full house | 3,744 | 0.001441 |

Flush | 5,108 | 0.001965 |

Straight | 10,200 | 0.003925 |

Three of a kind | 54,912 | 0.021129 |

Two pairs | 123,552 | 0.047539 |

One pair | 1,098,240 | 0.422569 |

Less than one pair | 1,302,540 | 0.501177 |

Total | 2,598,960 | 1.000000 |

The chance of winning at gambling games can be compared by using what is called the house advantage, house edge, or house percentage. For example, the house advantage for roulette is about 5.26%, which means in the long run, the house wins 5.26 cents on every $1 bet; or you will lose, on average, 5.26 cents on every $1 you bet. The lower the house advantage, the more favorable the game is to you.

For the game of craps, the house advantage is anywhere between 1.4 and 15%, depending on what you bet on. For the game called keno, the house advantage is 29.5%. The house of advantage for Chuck-a-Luck is 7.87%, and for baccarat, it is either 1.36 or 1.17% depending on your bet.

Slot machines have a house advantage anywhere from about 4 to 10% depending on the geographic location, such as Atlantic City, Las Vegas, and Mississippi, and the amount put in the machine, such as 5¢, 25¢, and $1.

Actually, gamblers found winning strategies for the game blackjack or 21 such as card counting. However, the casinos retaliated by using multiple decks and by banning card counters.

Exercises 4–5 |

**1.Selecting Cards** Find the probability of getting 2 face cards (king, queen, or jack) when 2 cards are drawn from a deck without replacement.

**2.Selecting a Committee** A parent-teacher committee consisting of 4 people is to be formed from 20 parents and 5 teachers. Find the probability that the committee will consist of these people. (Assume that the selection will be random.)

*a*.All teachers

*b*.2 teachers and 2 parents

*c*.All parents

*d*.1 teacher and 3 parents

**3.Management Seminar** In a company there are 7 executives: 4 women and 3 men. Three are selected to attend a management seminar. Find these probabilities.

*a*.All 3 selected will be women.

*b*.All 3 selected will be men.

*c*.2 men and 1 woman will be selected.

*d*.1 man and 2 women will be selected.

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**4.Senate Partisanship** The composition of the Senate of the 107th Congress is

49 Republicans 1 Independent 50 Democrats

A new committee is being formed to study ways to benefit the arts in education. If 3 Senators are selected at random to head the committee, what is the probability that they will all be Republicans? What is the probability that they will all be Democrats? What is the probability that there will be 1 from each party, including the Independent?

Source: *New York Times Almanac*.

**5.Congressional Committee Memberships** The composition of the 108th Congress is 51 Republicans, 48 Democrats, and 1 Independent. A committee on aid to higher education is to be formed with 3 Senators to be chosen at random to head the committee. Find the probability that the group of 3 consists of

*a*.All Republicans

*b*.All Democrats

*c*.One Democrat, one Republican, and one Independent

**6.Defective Resistors** A package contains 12 resistors, 3 of which are defective. If 4 are selected, find the probability of getting

*a*.0 defective resistors

*b*.1 defective resistor

*c*.3 defective resistors

**7.Winning Tickets** If 50 tickets are sold and 2 prizes are to be awarded, find the probability that one person will win 2 prizes if that person buys 2 tickets.

**8.Getting a Full House** Find the probability of getting a full house (3 cards of one denomination and 2 of another) when 5 cards are dealt from an ordinary deck.

**9.Flight School Graduation** At a recent graduation at a naval flight school, 18 Marines, 10 members of the Navy, and 3 members of the Coast Guard got their wings. Choose three pilots at random to feature on a training brochure. Find the probability that there will be

*a*.1 of each

*b*.0 members of the Navy

*c*.3 Marines

**10.Selecting Cards** The red face cards and the black cards numbered 2–9 are put into a bag. Four cards are drawn at random without replacement. Find the following probabilities:

*a*.All 4 cards are red.

*b*.2 cards are red and 2 cards are black.

*c*.At least 1 of the cards is red.

*d*.All 4 cards are black.

**11.Socks in a Drawer** A drawer contains 11 identical red socks and 8 identical black socks. Suppose that you choose 2 socks at random in the dark.

*a*.What is the probability that you get a pair of red socks?

*b*.What is the probability that you get a pair of black socks?

*c*.What is the probability that you get 2 unmatched socks?

*d*.Where did the other red sock go?

**12.Selecting Books** Find the probability of selecting 3 science books and 4 math books from 8 science books and 9 math books. The books are selected at random.

**13.Rolling Three Dice** When 3 dice are rolled, find the probability of getting a sum of 7.

**14.Football Team Selection** A football team consists of 20 each freshmen and sophomores, 15 juniors, and 10 seniors. Four players are selected at random to serve as captains. Find the probability that

*a*.All 4 are seniors

*b*.There is 1 each: freshman, sophomore, junior, and senior

*c*.There are 2 sophomores and 2 freshmen

*d*.At least 1 of the students is a senior

**15.Arrangement of Washers** Find the probability that if 5 different-sized washers are arranged in a row, they will be arranged in order of size.

**16.**Using the information in Exercise 63 in Section 4–4, find the probability of each poker hand.

*a*.Royal flush

*b*.Straight flush

*c*.Four of a kind

**17.Plant Selection** All holly plants are dioecious—a male plant must be planted within 30 to 40 feet of the female plants in order to yield berries. A home improvement store has 12 unmarked holly plants for sale, 8 of which are female. If a homeowner buys 3 plants at random, what is the probability that berries will be produced?

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**Summary**

In this chapter, the basic concepts and rules of probability are explained. The three types of probability are classical, empirical, and subjective. Classical probability uses sample spaces. Empirical probability uses frequency distributions and is based on observation. In subjective probability, the researcher makes an educated guess about the chance of an event occurring.

A probability event consists of one or more outcomes of a probability experiment. Two events are said to be mutually exclusive if they cannot occur at the same time. Events can also be classified as independent or dependent. If events are independent, whether or not the first event occurs does not affect the probability of the next event occurring. If the probability of the second event occurring is changed by the occurrence of the first event, then the events are dependent. The complement of an event is the set of outcomes in the sample space that are not included in the outcomes of the event itself. Complementary events are mutually exclusive.

Probability problems can be solved by using the addition rules, the multiplication rules, and the complementary event rules.

Finally, the fundamental counting rule, the permutation rule, and the combination rule can be used to determine the number of outcomes of events; then these numbers can be used to determine the probabilities of events.

**Important Terms**

classical probability

combination

complement of an event

compound event

conditional probability

dependent events

empirical probability

equally likely events

event

fundamental counting rule

independent events

law of large numbers

mutually exclusive events

outcome

permutation

probability

probability experiment

sample space

simple event

subjective probability

tree diagram

Venn diagrams

**Important Formulas**

Formula for classical probability:

Formula for empirical probability:

Addition rule 1, for two mutually exclusive events:

*P***( A or B) = P(A) + P(B)**

Addition rule 2, for events that are not mutually exclusive:

*P***( A or B) = P(A) + P(B)** –

*p*(*A*and*B*)Multiplication rule 1, for independent events:

*P(A*** and B) = P(A)**

**‧**

*P(B)*Multiplication rule 2, for dependent events:

*P(A*** and B) = P(A)**

**‧**

*P(B***|**

*A*)Formula for conditional probability:

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Formula for complementary events:

Fundamental counting rule: In a sequence of *n* events in which the first one has *k _{1}* possibilities, the second event has

*k*possibilities, the third has

_{2}*k*possibilities, etc., the total number possibilities of the sequence will be

_{3}*k*_{1}** · k_{2} · k_{3} … k_{n}**

Permutation rule: The number of permutations of *n* objects taking *r* objects at a time when order is important is

Combination rule: The number of combinations of *r* objects selected from *n* objects when order is not important is

**Review Exercises**

**1**.When a standard die is rolled, find the probability of getting

*a*.*A* 5

*b*.A number larger than 2

*c*.An odd number

**2.Selecting a Card** When a card is selected from a deck, find the probability of getting

*a*.A club

*b*.A face card or a heart

*c*.A 6 and a spade

*d*.A king

*e*.A red card

**3.Software Selection** The top-10 selling computer software titles last year consisted of 3 for doing taxes, 5 antivirus or security programs, and 2 “other.” Choose one title at random.

*a*.What is the probability that it is not used for doing taxes?

*b*.What is the probability that it is used for taxes or is one of the “other” programs?

Source: www.infoplease.com

**4.**A six-sided die is printed with the numbers 1, 2, 3, 5, 8, and 13. Roll the die once—what is the probability of getting an even number? Roll the die twice and add the numbers. What is the probability of getting an odd sum on the dice?

**5.****Cordless Phone Survey** A recent survey indicated that in a town of 1500 households, 850 had cordless telephones. If a household is randomly selected, find the probability that it has a cordless telephone.

**6.Purchasing Sweaters** During a sale at a men’s store, 16 white sweaters, 3 red sweaters, 9 blue sweaters, and 7 yellow sweaters were purchased. If a customer is selected at random, find the probability that he bought

*a*.A blue sweater

*b*.A yellow or a white sweater

*c*.A red, a blue, or a yellow sweater

*d*.A sweater that was not white

**7.****Budget Rental Cars** Cheap Rentals has nothing but budget cars for rental. The probability that a car has air conditioning is 0.5, and the probability that a car has a CD player is 0.37. The probability that a car has both air conditioning and a CD player is 0.06. What is the probability that a randomly selected car has neither air conditioning nor a CD player?

**8.Rolling Two Dice** When two dice are rolled, find the probability of getting

*a*.A sum of 5 or 6

*b*.A sum greater than 9

*c*.A sum less than 4 or greater than 9

*d*.A sum that is divisible by 4

*e*.A sum of 14

*f*.A sum less than 13

**9.****Car and Boat Ownership** The probability that a person owns a car is 0.80, that a person owns a boat is 0.30, and that a person owns both a car and a boat is 0.12. Find the probability that a person owns either a boat or a car.

**10.Car Purchases** There is a 0.39 probability that John will purchase a new car, a 0.73 probability that Mary will purchase a new car, and a 0.36 probability that both will purchase a new car. Find the probability that neither will purchase a new car.

**11.Online Course Selection** Roughly 1 in 6 students enrolled in higher education took at least one online course last fall. Choose 5 enrolled students at random. Find the probability that

*a*.All 5 took online courses

*b*.None of the 5 took a course online

*c*.At least 1 took an online course

Source: www.encarta.msn.com

**12.Borrowing Books** Of Americans using library services, 67% borrow books. If 5 patrons are chosen at random, what is the probability that all borrowed books? That none borrowed books?

Source: American Library Association.

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**13.Drawing Cards** Three cards are drawn from an ordinary deck *without* replacement. Find the probability of getting

*a*.All black cards

*b*.All spades

*c*.All queens

**14.Coin Toss and Card Drawn** A coin is tossed and a card is drawn from a deck. Find the probability of getting

*a*.A head and a 6

*b*.A tail and a red card

*c*.A head and a club

**15.Movie Releases** The top five countries for movie releases so far this year are the United States with 471 releases, United Kingdom with 386, Japan with 79, Germany with 316, and France with 132. Choose one new release at random. Find the probability that it is

*a*.European

*b*.From the United States

*c*.German or French

*d*.German given that it is European

Source: www.showbizdata.com

**16.Factory Output** A manufacturing company has three factories: X, Y, and Z. The daily output of each is shown here.

If one item is selected at random, find these probabilities.

*a*.It was manufactured at factory X or is a stereo.

*b*.It was manufactured at factory Y or factory Z.

*c*.It is a TV or was manufactured at factory Z.

**17.****Effectiveness of Vaccine** A vaccine has a 90% probability of being effective in preventing a certain disease. The probability of getting the disease if a person is not vaccinated is 50%. In a certain geographic region, 25% of the people get vaccinated. If a person is selected at random, find the probability that he or she will contract the disease.

**18.Television Models** A manufacturer makes three models of a television set, models A, B, and C. A store sells 40% of model A sets, 40% of model B sets, and 20% of model C sets. Of model A sets, 3% have stereo sound; of model B sets, 7% have stereo sound; and of model C sets, 9% have stereo sound. If a set is sold at random, find the probability that it has stereo sound.

**19.****Car Purchase** The probability that Sue will live on campus and buy a new car is 0.37. If the probability that she will live on campus is 0.73, find the probability that she will buy a new car, given that she lives on campus.

**20.Applying Shipping Labels** Four unmarked packages have lost their shipping labels, and you must reapply them. What is the probability that you apply the labels and get all four of them correct? Exactly three correct? Exactly two? At least one correct?

**21.****Health Club Membership** Of the members of the Blue River Health Club, 43% have a lifetime membership and exercise regularly (three or more times a week). If 75% of the club members exercise regularly, find the probability that a randomly selected member is a life member, given that he or she exercises regularly.

**22.Bad Weather** The probability that it snows and the bus arrives late is 0.023. José hears the weather forecast, and there is a 40% chance of snow tomorrow. Find the probability that the bus will be late, given that it snows.

**23.Education Level and Smoking** At a large factory, the employees were surveyed and classified according to their level of education and whether they smoked. The data are shown in the table.

If an employee is selected at random, find these probabilities.

*a*.The employee smokes, given that he or she graduated from college.

*b*.Given that the employee did not graduate from high school, he or she is a smoker.

**24.War Veterans** Approximately 11% of the civilian population are veterans. Choose 5 civilians at random. What is the probability that none are veterans? What is the probability that at least 1 is a veteran?

Source: www.factfinder.census.gov

**25.DVD Players** Eighty-one percent of U.S. households have DVD players. Choose 6 households at random. What is the probability that at least one does not have a DVD player?

Source: www.infoplease.com

**26.Chronic Sinusitis** The U.S. Department of Health and Human Services reports that 15% of Americans have chronic sinusitis. If 5 people are selected at random, find the probability that at least 1 has chronic sinusitis.

Source: *100% American*.

**27.****Automobile License Plate** An automobile license plate consists of 3 letters followed by 4 digits. How many different plates can be made if repetitions are allowed? If repetitions are not allowed? If repetitions are allowed in the letters but not in the digits?

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**28.Types of Copy Paper** White copy paper is offered in 5 different strengths and 11 different degrees of brightness, recycled or not, and acid-free or not. How many different types of paper are available for order?

**29.****Baseball Players** How many ways can 3 outfielders and 4 infielders be chosen from 5 outfielders and 7 infielders?

**30.Computer Operators** How many different ways can 8 computer operators be seated in a row?

**31.****Student Representatives** How many ways can a student select 2 electives from a possible choice of 10 electives?

**32.Committee Representation** There are 6 Republican 5 Democrat, and 4 Independent candidates. How many different ways can a committee of 3 Republicans, 2 Democrats, and 1 Independent be selected?

**33.****Song Selections** A promotional MP3 player is available with the capacity to store 100 songs which can be reordered at the push of a button. How many different arrangements of these songs are possible?

(*Note:* Factorials get very big, very fast! How large a factorial will your calculator calculate?)

**34.Employee Health Care Plans** A new employee has a choice of 5 health care plans, 3 retirement plans, and 2 different expense accounts. If a person selects one of each option, how many different options does he or she have?

**35.****Course Enrollmen**t There are 12 students who wish to enroll in a particular course. There are only 4 seats left in the classroom. How many different ways can 4 students be selected to attend the class?

**36.Candy Selection** A candy store allows customers to select 3 different candies to be packaged and mailed. If there are 13 varieties available, how many possible selections can be made?

**37.****Book Selection** If a student can select 5 novels from a reading list of 20 for a course in literature, how many different possible ways can this selection be done?

**38.Course Selection** If a student can select one of 3 language courses, one of 5 mathematics courses, and one of 4 history courses, how many different schedules can be made?

**39.****License Plates** License plates are to be issued with 3 letters followed by 4 single digits. How many such license plates are possible? If the plates are issued at random, what is the probability that the license plate says USA followed by a number that is divisible by 5?

**40.Leisure Activities** A newspaper advertises 5 different movies, 3 plays, and 2 baseball games for the weekend. If a couple selects 3 activities, find the probability that they attend 2 plays and 1 movie.

**41.****Territorial Selection** Several territories and colonies today are still under the jurisdiction of another country. France holds the most with 16 territories, the United Kingdom has 15, the United States has 14, and several other countries have territories as well. Choose 3 territories at random from those held by France, the United Kingdom, and the United States. What is the probability that all 3 belong to the same country?

Source: www.infoplease.com

**42.Yahtzee **Yahtzee is a game played with 5 dice. Players attempt to score points by rolling various combinations. When all 5 dice show the same number, it is called a *Yahtzee* and scores 50 points for the first one and 100 points for each subsequent Yahtzee in the same game. What is the probability that a person throws a Yahtzee on the very first roll? What is the probability that a person throws two Yahtzees on two successive turns?

**43.****Personnel Classification** For a survey, a subject can be classified as follows:

Gender: male or female

Martial status: single, married, widowed, divorced

Occupation: administration, faculty, staff

Draw a tree diagram for the different ways a person can be classified.

**Statistics Today**

**Would You Bet Your Life?—Revisited**

In his book *Probabilities in Everyday Life*, John D. McGervey states that the chance of being killed on any given commercial airline flight is almost 1 in 1 million and that the chance of being killed during a transcontinental auto trip is about 1 in 8000. The corresponding probabilities are 1/1,000,000 = 0.000001 as compared to 1/8000 = 0.000125. Since the second number is 125 times greater than the first number, you have a much higher risk driving than flying across the United States.

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**Chapter Quiz**

**Determine whether each statement is true or false. If the statement is false, explain why.**

**1.**Subjective probability has little use in the real world.

**2.**Classical probability uses a frequency distribution to compute probabilities.

**3.**In classical probability, all outcomes in the sample space are equally likely.

**4.**When two events are not mutually exclusive, *P*(*A* or *B*) = *P*(*A*) + *P*(*B*).

**5.**If two events are dependent, they must have the same probability of occurring.

**6.**An event and its complement can occur at the same time.

**7.**The arrangement ABC is the same as BAC for combinations.

**8.**When objects are arranged in a specific order, the arrangement is called a combination.

**Select the best answer.**

**9.**The probability that an event happens is 0.42. What is the probability that the event won’t happen?

*a*.–0.42

*b*.0.58

*c*.0

*d*.1

**10.**When a meteorologist says that there is a 30% chance of showers, what type of probability is the person using?

*a*.Classical

*b*.Empirical

*c*.Relative

*d*.Subjective

**11.**The sample space for tossing 3 coins consists of how many outcomes?

*a*.2

*b*.4

*c*.6

*d*.8

**12.**The complement of guessing 5 correct answers on a 5-question true/false exam is

*a*.Guessing 5 incorrect answers

*b*.Guessing at least 1 incorrect answer

*c*.Guessing at least 1 correct answer

*d*.Guessing no incorrect answers

**13.**When two dice are rolled, the sample space consists of how many events?

*a*.6

*b*.12

*c*.36

*d*.54

**14.**What is _{n}P_{0}?

*a*.0

*b*.1

*c*.*n*

*d*.It cannot be determined.

**15.**What is the number of permutations of 6 different objects taken all together?

*a*. 0

*b*.1

*c*.36

*d*.720

**16.**What is 0!?

*a*.0

*b*.1

*c*.Undefined

*d*.10

**17.**What is * _{n}C_{n}*?

*a*.0

*b*.1

*c*.*n*

*d*.It cannot be determined.

**Complete the following statements with the best answer.**

**18.**The set of all possible outcomes of a probability experiment is called the______.

**19.**The probability of an event can be any number between and including ______ and ______.

**20.**If an event cannot occur, its probability is ______.

**21.**The sum of the probabilities of the events in the sample space is ______.

**22.**When two events cannot occur at the same time, they are said to be ______.

**23.**When a card is drawn, find the probability of getting

*a*.A jack

*b*.A 4

*c*.A card less than 6 (an ace is considered above 6)

**24.Selecting a Card** When a card is drawn from a deck, find the probability of getting

*a*.A diamond

*b*.A 5 or a heart

*c*.A 5 and a heart

*d*.A king

*e*.A red card

**25.Selecting a Sweater** At a men’s clothing store, 12 men purchased blue golf sweaters, 8 purchased green sweaters, 4 purchased gray sweaters, and 7 bought black sweaters. If a customer is selected at random, find the probability that he purchased

*a*.A blue sweater

*b*.A green or gray sweater

*c*.A green or black or blue sweater

*d*.A sweater that was not black

**26.Rolling Dice** When 2 dice are rolled, find the probability of getting

*a*.A sum of 6 or 7

*b*.A sum greater than 8

*c*.A sum less than 3 or greater than 8

*d*.A sum that is divisible by 3

*e*.A sum of 16

*f*.A sum less than 11

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**27.****Appliance Ownership** The probability that a person owns a microwave oven is 0.75, that a person owns a compact disk player is 0.25, and that a person owns both a microwave and a CD player is 0.16. Find the probability that a person owns either a microwave or a CD player, but not both.

**28.****Starting Salaries** Of the physics graduates of a university, 30% received a starting salary of $30,000 or more. If 5 of the graduates are selected at random, find the probability that all had a starting salary of $30,000 or more.

**29.Selecting Cards** Five cards are drawn from an ordinary deck *without* replacement. Find the probability of getting

*a*.All red cards

*b*.All diamonds

*c*.All aces

**30.****Scholarships** The probability that Samantha will be accepted by the college of her choice and obtain a scholarship is 0.35. If the probability that she is accepted by the college is 0.65, find the probability that she will obtain a scholarship given that she is accepted by the college.

**31.****New Car Warranty** The probability that a customer will buy a car and an extended warranty is 0.16. If the probability that a customer will purchase a car is 0.30, find the probability that the customer will also purchase the extended warranty.

**32.****Bowling and Club Membership** Of the members of the Spring Lake Bowling Lanes, 57% have a lifetime membership and bowl regularly (three or more times a week). If 70% of the club members bowl regularly, find the probability that a randomly selected member is a lifetime member, given that he or she bowls regularly.

**33.****Work and Weather** The probability that Mike has to work overtime and it rains is 0.028. Mike hears the weather forecast, and there is a 50% chance of rain. Find the probability that he will have to work overtime, given that it rains.

**34.Education of Factory Employees** At a large factory, the employees were surveyed and classified according to their level of education and whether they attend a sports event at least once a month. The data are shown in the table.

If an employee is selected at random, find the probability that

*a*.The employee attends sports events regularly, given that he or she graduated from college (2- or 4-year degree)

*b*.Given that the employee is a high school graduate, he or she does not attend sports events regularly

**35.****Heart Attacks** In a certain high-risk group, the chances of a person having suffered a heart attack are 55%. If 6 people are chosen, find the probability that at least 1 will have had a heart attack.

**36.****Rolling a Die** A single die is rolled 4 times. Find the probability of getting at least one 5.

**37.****Eye Color** If 85% of all people have brown eyes and 6 people are selected at random, find the probability that at least 1 of them has brown eyes.

**38.****Singer Selection** How many ways can 5 sopranos and 4 altos be selected from 7 sopranos and 9 altos?

**39.****Speaker Selection** How many different ways can 8 speakers be seated on a stage?

**40.****Stocking Machines** A soda machine servicer must restock and collect money from 15 machines, each one at a different location. How many ways can she select 4 machines to service in 1 day?

**41.****ID Cards** One company’s ID cards consist of 5 letters followed by 2 digits. How many cards can be made if repetitions are allowed? If repetitions are not allowed?

**42.**How many different arrangements of the letters in the word *number* can be made?

**43.****Physics Test** A physics test consists of 25 true/false questions. How many different possible answer keys can be made?

**44.****Cellular Telephones** How many different ways can 5 cellular telephones be selected from 8 cellular phones?

**45.****Fruit Selection** On a lunch counter, there are 3 oranges, 5 apples, and 2 bananas. If 3 pieces of fruit are selected, find the probability that 1 orange, 1 apple, and 1 banana are selected.

**46.****Cruise Ship Activities** A cruise director schedules 4 different movies, 2 bridge games, and 3 tennis games for a 2-day period. If a couple selects 3 activities, find the probability that they attend 2 movies and 1 tennis game.

**47.****Committee Selection** At a sorority meeting, there are 6 seniors, 4 juniors, and 2 sophomores. If a committee of 3 is to be formed, find the probability that 1 of each will be selected.

**48.****Banquet Meal Choices** For a banquet, a committee can select beef, pork, chicken, or veal; baked potatoes or mashed potatoes; and peas or green beans for a vegetable. Draw a tree diagram for all possible choices of a meat, a potato, and a vegetable.

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**Critical Thinking Challenges**

**1.Con Man Game** Consider this problem: A con man has 3 coins. One coin has been specially made and has a head on each side. A second coin has been specially made, and on each side it has a tail. Finally, a third coin has a head and a tail on it. All coins are of the same denomination. The con man places the 3 coins in his pocket, selects one, and shows you one side. It is heads. He is willing to bet you even money that it is the two-headed coin. His reasoning is that it can’t be the two-tailed coin since a head is showing; therefore, there is a 50-50 chance of it being the two-headed coin. Would you take the bet? *(Hint:* See Exercise 1 in Data Projects.)

**2.de Méré Dice Game** Chevalier de Méré won money when he bet unsuspecting patrons that in 4 rolls of 1 die, he could get at least one 6, but he lost money when he bet that in 24 rolls of 2 dice, he could get at least a double 6. Using the probability rules, find the probability of each event and explain why he won the majority of the time on the first game but lost the majority of the time when playing the second game. *(Hint:* Find the probabilities of losing each game and subtract from 1.)

**3.Classical Birthday Problem** How many people do you think need to be in a room so that 2 people will have the same birthday (month and day)? You might think it is 366. This would, of course, guarantee it (excluding leap year), but how many people would need to be in a room so that there would be a 90% probability that 2 people would be born on the same day? What about a 50% probability?

Actually, the number is much smaller than you may think. For example, if you have 50 people in a room, the probability that 2 people will have the same birthday is 97%. If you have 23 people in a room, there is a 50% probability that 2 people were born on the same day!

The problem can be solved by using the probability rules. It must be assumed that all birthdays are equally likely, but this assumption will have little effect on the answers. The way to find the answer is by using the complementary event rule as *P(2* people having the same birthday) = 1 – *P*(all have different birthdays).

For example, suppose there were 3 people in the room. The probability that each had a different birthday would be

Hence, the probability that at least 2 of the 3 people will have the same birthday will be

1 – 0.992 = 0.008

Hence, for *k* people, the formula is

*P*(at least 2 people have the same birthday)

Using your calculator, complete the table and verify that for at least a 50% chance of 2 people having the same birthday, 23 or more people will be needed.

Number of people | Probability that at least 2 have same birthday |

1 | 0.000 |

2 | 0.003 |

5 | 0.027 |

10 | |

15 | |

20 | |

21 | |

22 | |

23 |

**4.**We know that if the probability of an event happening is 100%, then the event is a certainty. Can it be concluded that if there is a 50% chance of contracting a communicable disease through contact with an infected person, there would be a 100% chance of contracting the disease if 2 contacts were made with the infected person? Explain your answer.

** Data Projects**

**1.Business and Finance** Select a pizza restaurant and a sandwich shop. For the pizza restaurant look at the menu to determine how many sizes, crust types, and toppings are available. How many different pizza types are possible? For the sandwich shop determine how many breads, meats, veggies, cheeses, sauces, and condiments are available. How many different sandwich choices are possible?

**2.Sports and Leisure** When poker games are shown on television, there are often percentages displayed that show how likely it is that a certain hand will win. Investigate how these percentages are determined. Show an example with two competing hands in a Texas Hold ’em game. Include the percentages that each hand will win after the deal, the flop, the turn, and the river.

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**3.Technology** A music player or music organization program can keep track of how many different artists are in a library. First note how many different artists are in your music library. Then find the probability that if 25 songs are selected at random, none will have the same artist.

**4.Health and Wellness** Assume that the gender distribution of babies is such that one-half the time females are born and one-half the time males are born. In a family of 3 children, what is the probability that all are girls? In a family of 4? Is it unusual that in a family with 4 children all would be girls? In a family of 5?

**5.Politics and Economics** Consider the U.S. Senate. Find out about the composition of any three of the Senate’s standing committees. How many different committees of Senators are possible, knowing the party composition of the Senate and the number of committee members from each party for each committee?

**6.Your Class** Research the famous Monty Hall probability problem. Conduct a simulation of the Monty Hall problem online using a simulation program or in class using live “contestants.” After 50 simulations compare your results to those stated in the research you did. Did your simulation support the conclusions?

Answers to Applying the Concepts

**Section 4–1 Tossing a Coin**

**1.**The sample space is the listing of all possible outcomes of the coin toss.

**2.**The possible outcomes are heads or tails.

**3.**Classical probability says that a fair coin has a 50–50 chance of coming up heads or tails.

**4.**The law of large numbers says that as you increase the number of trials, the overall results will approach the theoretical probability. However, since the coin has no “memory,” it still has a 50–50 chance of coming up heads or tails on the next toss. Knowing what has already happened should not change your opinion on what will happen on the next toss.

**5.**The empirical approach to probability is based on running an experiment and looking at the results. You cannot do that at this time.

**6.**Subjective probabilities could be used if you believe the coin is biased.

**7.**Answers will vary; however, they should address that a fair coin has a 50-50 chance of coming up heads or tails on the next flip.

**Section 4–2 Which Pain Reliever Is Best?**

**1.**There were 192 + 186 + 188 = 566 subjects in the study.

**2.**The study lasted for 12 weeks.

**3.**The variables are the type of pain reliever and the side effects.

**4.**Both variables are qualitative and nominal.

**5.**The numbers in the table are exact figures.

**6.**The probability that a randomly selected person was receiving a placebo is 192/566 = 0.3392 (about 34%).

**7.**The probability that a randomly selected person was receiving a placebo or drug *A* is (192 + 186)/566 = 378/566 = 0.6678 (about 67%). These are mutually exclusive events. The complement is that a randomly selected person was receiving drug *B*.

**8.**The probability that a randomly selected person was receiving a placebo or experienced a neurological headache is (192 + 55 + 72)/566 = 319/566 = 0. 5636 (about 56%).

**9.**The probability that a randomly selected person was not receiving a placebo or experienced a sinus headache is (186 + 188)/566 + 11/566 = 385/566 = 0.6802 (about 68%).

**Section 4–3 Guilty or Innocent?**

**1.**The probability of another couple with the same characteristics being in that area is , assuming the characteristics are independent of one another.

**2.**You would use the multiplication rule, since we are looking for the probability of multiple events happening together.

**3.**We do not know if the characteristics are dependent or independent, but we assumed independence for the calculation in question 1.

**4.**The probabilities would change if there were dependence among two or more events.

**5.**Answers will vary. One possible answer is that probabilities can be used to explain how unlikely it is to have a set of events occur at the same time (in this case, how unlikely it is to have another couple with the same characteristics in that area).

**6.**Answers will vary. One possible answer is that if the only eyewitness was the woman who was mugged and the probabilities are accurate, it seems very unlikely that a couple matching these characteristics would be in that area at that time. This might cause you to convict the couple.

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**7.**Answers will vary. One possible answer is that our probabilities are theoretical and serve a purpose when appropriate, but that court cases are based on much more than impersonal chance.

**8.**Answers will vary. One possible answer is that juries decide whether to convict a defendant if they find evidence “beyond a reasonable doubt” that the person is guilty. In probability terms, this means that if the defendant was actually innocent, then the chance of seeing the events that occurred is so unlikely as to have occurred by chance. Therefore, the jury concludes that the defendant is guilty.

**Section 4–4 Garage Door Openers**

**1.**Four on/off switches lead to 16 different settings.

**2.**With 5 on/off switches, there are 2^{5} = 32 different settings. With 6 on/off switches, there are 2^{6} = 64 different settings. In general, if there are *k* on/off switches, there are 2* ^{k}* different settings.

**3.**With 8 consecutive on/off switches, there are 2^{8} = 256 different settings.

**4.**It is less likely for someone to be able to open your garage door if you have 8 on/off settings (probability about 0.4%) than if you have 4 on/off switches (probability about 6.0%). Having 8 on/off switches in the opener seems pretty safe.

**5.**Each key blank could be made into 5^{5} = 3125 possible keys.

**6.**If there were 420,000 Dodge Caravans sold in the United States, then any one key could start about 420,000/3125 = 134.4, or about 134, different Caravans.

**7.**Answers will vary.

**Section 4–5 Counting Rules and Probability**

**1.**There are five different events: each multiple-choice question is an event.

**2.**These events are independent.

**3.**If you guess on 1 question, the probability of getting it correct is 0.20. Thus, if you guess on all 5 questions, the probability of getting all of them correct is (0.20)^{5} = 0.00032.

**4.**The probability that a person would guess answer A for a question is 0.20, so the probability that a person would guess answer A for each question is (0.20)^{5} = 0.00032.

**5.**There are five different events: each matching question is an event.

**6.**These are dependent events.

**7.**The probability of getting them all correct if you are guessing is .

**8.**The difference between the two problems is that we are sampling without replacement in the second problem, so the denominator changes in the event probabilities.

^{1}Strictly speaking, a percent is not a probability. However, in everyday language, probabilities are often expressed as percents (i.e., there isa60% chance of rain tomorrow). For this reason, some probabilities will be expressed as percents throughout this book.

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CHAPTER5 | |

Discrete Probability Distributions |

**Objectives**

After completing this chapter, you should be able to

**1**Construct a probability distribution for a random variable.

**2**Find the mean, variance, standard deviation, and expected value for a discrete random variable.

**3**Find the exact probability for *X* successes in *n* trials of a binomial experiment.

**4**Find the mean, variance, and standard deviation for the variable of a binomial distribution.

**Outline**

**Introduction**

**5–1Probability Distributions**

**5–2Mean, Variance, Standard Deviation, and Expectation**

**5–3The Binomial Distribution**

**Summary**

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**Statistics Today**

**Is Pooling Worthwhile?**

Blood samples are used to screen people for certain diseases. When the disease is rare, health care workers sometimes combine or pool the blood samples of a group of individuals into one batch and then test it. If the test result of the batch is negative, no further testing is needed since none of the individuals in the group has the disease. However, if the test result of the batch is positive, each individual in the group must be tested.

Consider this hypothetical example: Suppose the probability of a person having the disease is 0.05, and a pooled sample of 15 individuals is tested. What is the probability that no further testing will be needed for the individuals in the sample? The answer to this question can be found by using what is called the *binomial distribution*. See Statistics Today—Revisited at the end of the chapter.

This chapter explains probability distributions in general and a specific, often used distribution called the binomial distribution. The Poisson, hypergeometric, and multinomial distributions are also explained.

**Introduction**

Many decisions in business, insurance, and other real-life situations are made by assigning probabilities to all possible outcomes pertaining to the situation and then evaluating the results. For example, a saleswoman can compute the probability that she will make 0, 1, 2, or 3 or more sales in a single day. An insurance company might be able to assign probabilities to the number of vehicles a family owns. A self-employed speaker might be able to compute the probabilities for giving 0, 1, 2, 3, or 4 or more speeches each week. Once these probabilities are assigned, statistics such as the mean, variance, and standard deviation can be computed for these events. With these statistics, various decisions can be made. The saleswoman will be able to compute the average number of sales she makes per week, and if she is working on commission, she will be able to approximate her weekly income over a period of time, say, monthly. The public speaker will be able to plan ahead and approximate his average income and expenses. The insurance company can use its information to design special computer forms and programs to accommodate its customers’ future needs.

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This chapter explains the concepts and applications of what is called a *probability distribution*. In addition, a special probability distribution, the *binomial distribution*, is explained.

Objective 1

Construct a probability distribution for a random variable.

**5–1Probability Distributions**

Before probability distribution is defined formally, the definition of a variable is reviewed. In Chapter 1, a *variable* was defined as a characteristic or attribute that can assume different values. Various letters of the alphabet, such as *X, Y,* or *Z,* are used to represent variables. Since the variables in this chapter are associated with probability, they are called *random variables*.

For example, if a die is rolled, a letter such as *X* can be used to represent the outcomes. Then the value that *X* can assume is 1, 2, 3, 4, 5, or 6, corresponding to the outcomes of rolling a single die. If two coins are tossed, a letter, say *Y,* can be used to represent the number of heads, in this case 0, 1, or 2. As another example, if the temperature at 8:00 A.M. is 43° and at noon it is 53°, then the values *T* that the temperature assumes are said to be random, since they are due to various atmospheric conditions at the time the temperature was taken.

A **random variable** is a variable whose values are determined by chance.

Also recall from Chapter 1 that you can classify variables as discrete or continuous by observing the values the variable can assume. If a variable can assume only a specific number of values, such as the outcomes for the roll of a die or the outcomes for the toss of a coin, then the variable is called a *discrete variable*.

*Discrete variables* have a finite number of possible values or an infinite number of values that can be counted. The word *counted* means that they can be enumerated using the numbers 1, 2, 3, etc. For example, the number of joggers in Riverview Park each day and the number of phone calls received after a TV commercial airs are examples of discrete variables, since they can be counted.

Variables that can assume all values in the interval between any two given values are called *continuous variables*. For example, if the temperature goes from 62 to 78° in a 24-hour period, it has passed through every possible number from 62 to 78. *Continuous random variables are obtained from data that can be measured rather than counted*. Continuous random variables can assume an infinite number of values and can be decimal and fractional values. On a continuous scale, a person’s weight might be exactly 183.426 pounds if a scale could measure weight to the thousandths place; however, on a digital scale that measures only to tenths of pounds, the weight would be 183.4 pounds. Examples of continuous variables are heights, weights, temperatures, and time. In this chapter only discrete random variables are used; Chapter 6 explains continuous random variables.

The procedure shown here for constructing a probability distribution for a discrete random variable uses the probability experiment of tossing three coins. Recall that when three coins are tossed, the sample space is represented as TTT, TTH, THT, HTT, HHT, HTH, THH, HHH; and if *X* is the random variable for the number of heads, then *X* assumes the value 0, 1, 2, or 3.

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Probabilities for the values of *X* can be determined as follows:

Hence, the probability of getting no heads is , one head is , two heads is , and three heads is . From these values, a probability distribution can be constructed by listing the outcomes and assigning the probability of each outcome, as shown here.

A **discrete probability distribution** consists of the values a random variable can assume and the corresponding probabilities of the values. The probabilities are determined theoretically or by observation.

Discrete probability distributions can be shown by using a graph or a table. Probability distributions can also be represented by a formula. See Exercises 31–36 at the end of this section for examples.

Example 5–1

**Rolling a Die**

Construct a probability distribution for rolling a single die.

Solution

Since the sample space is 1, 2, 3, 4, 5, 6 and each outcome has a probability of , the distribution is as shown.

Probability distributions can be shown graphically by representing the values of *X* on the *x* axis and the probabilities *P*(*X*) on the *y* axis.

Example 5–2

**Tossing Coins**

Represent graphically the probability distribution for the sample space for tossing three coins.

Solution

The values that *X* assumes are located on the *x* axis, and the values for *P*(*X*) are located on the *y* axis. The graph is shown in Figure 5–1.

Note that for visual appearances, it is not necessary to start with 0 at the origin.

Examples 5–1 and 5–2 are illustrations of *theoretical* probability distributions. You did not need to actually perform the experiments to compute the probabilities. In contrast, to construct actual probability distributions, you must observe the variable over a period of time. They are empirical, as shown in Example 5–3.

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Figure 5–1

**Probability Distribution for ****Example 5–2**

Example 5–3

**Baseball World Series**

The baseball World Series is played by the winner of the National League and the American League. The first team to win four games wins the World Series. In other words, the series will consist of four to seven games, depending on the individual victories. The data shown consist of the number of games played in the World Series from 1965 through 2005. (There was no World Series in 1994.) The number of games played is represented by the variable *X.* Find the probability *P*(*X*) for each *X,* construct a probability distribution, and draw a graph for the data.

X | Number of times played |

4 | 8 |

5 | 7 |

6 | 9 |

7 | 16 |

40 |

Solution

The probability *P*(*X*) can be computed for each *X* by dividing the number of games *X* by the total.

For 4 games, = 0.200 | For 6 games, = 0.225 |

For 5 games, = 0.175 | For 7 games, = 0.400 |

The probability distribution is

The graph is shown in Figure 5–2.

Figure 5–2

**Probability Distribution for ****Example 5–3**

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*Speaking of* Statistics

**Coins, Births, and Other Random (?) Events**

Examples of random events such as tossing coins are used in almost all books on probability. But is flipping a coin really a random event?

Tossing coins dates back to ancient Roman times when the coins usually consisted of the Emperor’s head on one side (i.e., heads) and another icon such as a ship on the other side (i.e., ships). Tossing coins was used in both fortune telling and ancient Roman games.

A Chinese form of divination called the *I-Ching* (pronounced E-Ching) is thought to be at least 4000 years old. It consists of 64 hexagrams made up of six horizontal lines. Each line is either broken or unbroken, representing the yin and the yang. These 64 hexagrams are supposed to represent all possible situations in life. To consult the I-Ching, a question is asked and then three coins are tossed six times. The way the coins fall, either heads up or heads down, determines whether the line is broken (yin) or unbroken (yang). Once the hexagon is determined, its meaning is consulted and interpreted to get the answer to the question. (*Note:* Another method used to determine the hexagon employs yarrow sticks.)

In the 16th century, a mathematician named Abraham DeMoivre used the outcomes of tossing coins to study what later became known as the normal distribution; however, his work at that time was not widely known.

Mathematicians usually consider the outcomes of a coin toss a random event. That is, each probability of getting a head is , and the probability of getting a tail is . Also, it is not possible to predict with 100% certainty which outcome will occur. But new studies question this theory. During World War II a South African mathematician named John Kerrich tossed a coin 10,000 times while he was interned in a German prison camp. Unfortunately, the results of his experiment were never recorded, so we don’t know the number of heads that occurred.

Several studies have shown that when a coin-tossing device is used, the probability that a coin will land on the same side on which it is placed on the coin-tossing device is about 51%. It would take about 10,000 tosses to become aware of this bias. Furthermore, researchers showed that when a coin is spun on its edge, the coin falls tails up about 80% of the time since there is more metal on the heads side of a coin. This makes the coin slightly heavier on the heads side than on the tails side.

Another assumption commonly made in probability theory is that the number of male births is equal to the number of female births and that the probability of a boy being born is and the probability of a girl being born is . We know this is not exactly true.

In the later 1700s, a French mathematician named Pierre Simon Laplace attempted to prove that more males than females are born. He used records from 1745 to 1770 in Paris and showed that the percentage of females born was about 49%. Although these percentages vary somewhat from location to location, further surveys show they are generally true worldwide. Even though there are discrepancies, we generally consider the outcomes to be 50-50 since these discrepancies are relatively small.

Based on this article, would you consider the coin toss at the beginning of a football game fair?

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**Two Requirements for a Probability Distribution**

1.The sum of the probabilities of all the events in the sample space must equal 1; that is, Σ*P*(*X*) = 1.

2.The probability of each event in the sample space must be between or equal to 0 and 1. That is, 0 ≤ *P*(*X*) ≤ 1.

The first requirement states that the sum of the probabilities of all the events must be equal to 1. This sum cannot be less than 1 or greater than 1 since the sample space includes *all* possible outcomes of the probability experiment. The second requirement states that the probability of any individual event must be a value from 0 to 1. The reason (as stated in Chapter 4) is that the range of the probability of any individual value can be 0, 1, or any value between 0 and 1. A probability cannot be a negative number or greater than 1.

Example 5–4

**Probability Distributions**

Determine whether each distribution is a probability distribution.

Solution

*a*.Yes, it is a probability distribution.

*b*.No, it is not a probability distribution, since *P*(*X*) cannot be 1.5 or –1.0.

*c*.Yes, it is a probability distribution.

*d*.No, it is not, since Σ*P*(*X*) = 1.2.

Many variables in business, education, engineering, and other areas can be analyzed by using probability distributions. Section 5–2 shows methods for finding the mean and standard deviation for a probability distribution.

*Applying the Concepts* 5–1

**Dropping College Courses**

Use the following table to answer the questions.

Reason for Dropping a College Course | Frequency | Percentage |

Too difficult | 45 | |

Illness | 40 | |

Change in work schedule | 20 | |

Change of major | 14 | |

Family-related problems | 9 | |

Money | 7 | |

Miscellaneous | 6 | |

No meaningful reason | 3 |

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1.What is the variable under study? Is it a random variable?

2.How many people were in the study?

3.Complete the table.

4.From the information given, what is the probability that a student will drop a class because of illness? Money? Change of major?

5.Would you consider the information in the table to be a probability distribution?

6.Are the categories mutually exclusive?

7.Are the categories independent?

8.Are the categories exhaustive?

9.Are the two requirements for a discrete probability distribution met?

See page 295 for the answers.

Exercises 5–1 |

**1.**Define and give three examples of a random variable.

**2.**Explain the difference between a discrete and a continuous random variable.

**3.**Give three examples of a discrete random variable.

**4.**Give three examples of a continuous random variable.

**5.**What is a probability distribution? Give an example.

**For Exercises 6 through 11, determine whether the distribution represents a probability distribution. If it does not, state why.**

**6.**

**7.**

**8.**

**9.**

**10.**

**11.**

**For Exercises 12 through 18, state whether the variable is discrete or continuous.**

**12.**The speed of a jet airplane

**13.**The number of cheeseburgers a fast-food restaurant serves each day

**14.**The number of people who play the state lottery each day

**15.**The weight of a Siberian tiger

**16.**The time it takes to complete a marathon

**17.**The number of mathematics majors in your school

**18.**The blood pressures of all patients admitted to a hospital on a specific day

**For Exercises 19 through 26, construct a probability distribution for the data and draw a graph for the distribution.**

**19.Medical Tests** The probabilities that a patient will have 0, 1, 2, or 3 medical tests performed on entering a hospital are , and , respectively.

**20.Student Volunteers** The probabilities that a student volunteer hosts 1, 2, 3, or 4 prospective first-year students are 0.4, 0.3, 0.2, and 0.1, respectively.

**21.Birthday Cake Sales** The probabilities that a bakery has a demand for 2, 3, 5, or 7 birthday cakes on any given day are 0.35, 0.41, 0.15, and 0.09, respectively.

**22.DVD Rentals** The probabilities that a customer will rent 0, 1, 2, 3, or 4 DVDs on a single visit to the rental store are 0.15, 0.25, 0.3, 0.25, and 0.05, respectively.

**23.Loaded Die** A die is loaded in such a way that the probabilities of getting 1, 2, 3, 4, 5, and 6 are , , and , respectively.

**24.Item Selection** The probabilities that a customer selects 1, 2, 3, 4, and 5 items at a convenience store are 0.32, 0.12, 0.23, 0.18, and 0.15, respectively.

**25.Student Classes** The probabilities that a student is registered for 2, 3, 4, or 5 classes are 0.01, 0.34, 0.62, and 0.03, respectively.

**26.Garage Space** The probabilities that a randomly selected home has garage space for 0, 1, 2, or 3 cars are 0.22, 0.33, 0.37, and 0.08, respectively.

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**27.Selecting a Monetary Bill** A box contains three $1 bills, two $5 bills, five $10 bills, and one $20 bill. Construct a probability distribution for the data if *x* represents the value of a single bill drawn at random and then replaced.

**28.Family with Children** Construct a probability distribution for a family of three children. Let *X* represent the number of boys.

**29.Drawing a Card** Construct a probability distribution for drawing a card from a deck of 40 cards consisting of 10 cards numbered 1, 10 cards numbered 2, 15 cards numbered 3, and 5 cards numbered 4.

**30.Rolling Two Dice** Using the sample space for tossing two dice, construct a probability distribution for the sums 2 through 12.

**Extending the Concepts**

A probability distribution can be written in formula notation such as *P*(*X*) = 1/*X*, where *X* = 2, 3, 6. The distribution is shown as follows:

**For Exercises 31 through 36, write the distribution for the formula and determine whether it is a probability distribution.**

**31.***P*(*X*) = *X*/6 for *X* = 1, 2, 3

**32.***P*(*X*) = *X* for *X* = 0.2, 0.3, 0.5

**33.***P*(*X*) = *X*/6 for *X* = 3, 4, 7

**34.***P*(*X*) = *X* + 0.1 for *X* = 0.1, 0.02, 0.04

**35.***P*(*X*) = *X*/7 for *X* = 1, 2, 4

**36.***P*(*X*) = *X*/(*X* + 2) for *X* = 0, 1, 2

Objective 2

Find the mean, variance, standard deviation, and expected value for a discrete random variable.

**5–2Mean, Variance, Standard Deviation, and Expectation**

The mean, variance, and standard deviation for a probability distribution are computed differently from the mean, variance, and standard deviation for samples. This section explains how these measures—as well as a new measure called the *expectation*—are calculated for probability distributions.

**Mean**

In Chapter 3, the mean for a sample or population was computed by adding the values and dividing by the total number of values, as shown in these formulas:

But how would you compute the mean of the number of spots that show on top when a die is rolled? You could try rolling the die, say, 10 times, recording the number of spots, and finding the mean; however, this answer would only approximate the true mean. What about 50 rolls or 100 rolls? Actually, the more times the die is rolled, the better the approximation. You might ask, then, How many times must the die be rolled to get the exact answer? *It must be rolled an infinite number of times.* Since this task is impossible, the previous formulas cannot be used because the denominators would be infinity. Hence, a new method of computing the mean is necessary. This method gives the exact theoretical value of the mean as if it were possible to roll the die an infinite number of times.

Before the formula is stated, an example will be used to explain the concept. Suppose two coins are tossed repeatedly, and the number of heads that occurred is recorded. What will be the mean of the number of heads? The sample space is

HH, HT, TH, TT

and each outcome has a probability of . Now, in the long run, you would *expect* two heads (HH) to occur approximately of the time, one head to occur approximately of the time (HT or TH), and no heads (TT) to occur approximately of the time. Hence, on average, you would expect the number of heads to be

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That is, if it were possible to toss the coins many times or an infinite number of times, the *average* of the number of heads would be 1.

Hence, to find the mean for a probability distribution, you must multiply each possible outcome by its corresponding probability and find the sum of the products.

Historical Note

A professor, Augustin Louis Cauchy (1789–1857), wrote a book on probability. While he was teaching at the Military School of Paris, one of his students was Napoleon Bonaparte.

**Formula for the Mean of a Probability Distribution**

The mean of a random variable with a discrete probability distribution is

µ | = | X_{1} · P(X_{1}) + X_{2} · P(X_{2}) + X_{3} · P(X_{3}) + … + X · _{n}P(X)_{n} |

= | ΣX · P(X) |

where *X*_{1}, *X*_{2}, *X*_{3}, … , *X _{n}* are the outcomes and

*P*(

*X*

_{1}),

*P*(

*X*

_{2}),

*P*(

*X*

_{3}), … ,

*P*(

*X*) are the corresponding probabilities.

_{n}*Note:* Σ*X* · *P*(*X*) means to sum the products.

Rounding Rule for the Mean, Variance, and Standard Deviation for a Probability Distribution The rounding rule for the mean, variance, and standard deviation for variables of a probability distribution is this: The mean, variance, and standard deviation should be rounded to one more decimal place than the outcome *X.* When fractions are used, they should be reduced to lowest terms.

Examples 5–5 through 5–8 illustrate the use of the formula.

Example 5–5

**Rolling a Die**

Find the mean of the number of spots that appear when a die is tossed.

Solution

In the toss of a die, the mean can be computed thus.

That is, when a die is tossed many times, the theoretical mean will be 3.5. Note that even though the die cannot show a 3.5, the theoretical average is 3.5.

The reason why this formula gives the theoretical mean is that in the long run, each outcome would occur approximately of the time. Hence, multiplying the outcome by its corresponding probability and finding the sum would yield the theoretical mean. In other words, outcome 1 would occur approximately of the time, outcome 2 would occur approximately of the time, etc.

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Example 5–6

**Children in a Family**

In a family with two children, find the mean of the number of children who will be girls.

Solution

The probability distribution is as follows:

Hence, the mean is

Example 5–7

**Tossing Coins**

If three coins are tossed, find the mean of the number of heads that occur. (See the table preceding Example 5–1.)

Solution

The probability distribution is

The mean is

The value 1.5 cannot occur as an outcome. Nevertheless, it is the long-run or theoretical average.

Example 5–8

**Number of Trips of Five Nights or More**

The probability distribution shown represents the number of trips of five nights or more that American adults take per year. (That is, 6% do not take any trips lasting five nights or more, 70% take one trip lasting five nights or more per year, etc.) Find the mean.

Solution

µ | = | ΣX · P(X) |

= | (0)(0.06) + (1)(0.70) + (2)(0.20) + (3)(0.03) + (4)(0.01) | |

= | 0 + 0.70 + 0.40 + 0.09 + 0.04 | |

= | 1.23 ≈ 1.2 |

Hence, the mean of the number of trips lasting five nights or more per year taken by American adults is 1.2.

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Historical Note

Fey Manufacturing Co., located in San Francisco, invented the first three-reel, automatic payout slot machine in 1895.

**Variance and Standard Deviation**

For a probability distribution, the mean of the random variable describes the measure of the so-called long-run or theoretical average, but it does not tell anything about the spread of the distribution. Recall from Chapter 3 that in order to measure this spread or variability, statisticians use the variance and standard deviation. These formulas were used:

These formulas cannot be used for a random variable of a probability distribution since *N* is infinite, so the variance and standard deviation must be computed differently.

To find the variance for the random variable of a probability distribution, subtract the theoretical mean of the random variable from each outcome and square the difference. Then multiply each difference by its corresponding probability and add the products. The formula is

Finding the variance by using this formula is somewhat tedious. So for simplified computations, a shortcut formula can be used. This formula is algebraically equivalent to the longer one and is used in the examples that follow.

**Formula for the Variance of a Probability Distribution**

Find the variance of a probability distribution by multiplying the square of each outcome by its corresponding probability, summing those products, and subtracting the square of the mean. The formula for the variance of a probability distribution is

*σ* = Σ[*X*^{2} · *P*(*X*)] – *µ*^{2}

The standard deviation of a probability distribution is

Remember that the variance and standard deviation cannot be negative.

Example 5–9

**Rolling a Die**

Compute the variance and standard deviation for the probability distribution in Example 5–5.

Solution

Recall that the mean is *µ* = 3.5, as computed in Example 5–5. Square each outcome and multiply by the corresponding probability, sum those products, and then subtract the square of the mean.

To get the standard deviation, find the square root of the variance.

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Example 5–10

**Selecting Numbered Balls**

A box contains 5 balls. Two are numbered 3, one is numbered 4, and two are numbered 5. The balls are mixed and one is selected at random. After a ball is selected, its number is recorded. Then it is replaced. If the experiment is repeated many times, find the variance and standard deviation of the numbers on the balls.

Solution

Let *X* be the number on each ball. The probability distribution is

The mean is

The variance is

The standard deviation is

The mean, variance, and standard deviation can also be found by using vertical columns, as shown.

Find the mean by summing the Σ*X* · *P*(*X*) column, and find the variance by summing the *X*^{2} · *P*(*X*) column and subtracting the square of the mean.

*σ*^{2} = 16.8 – 4^{2} = 16.8 – 16 = 0.8

and

Example 5–11

**On Hold for Talk Radio**

A talk radio station has four telephone lines. If the host is unable to talk (i.e., during a commercial) or is talking to a person, the other callers are placed on hold. When all lines are in use, others who are trying to call in get a busy signal. The probability that 0, 1, 2, 3, or 4 people will get through is shown in the distribution. Find the variance and standard deviation for the distribution.

Should the station have considered getting more phone lines installed?

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Solution

The mean is

µ | = | ΣX · P(X) |

= | 0 · (0.18) + 1 · (0.34) + 2 · (0.23) + 3 · (0.21) + 4 · (0.04) | |

= | 1.6 |

The variance is

σ^{2} | = | Σ[X^{2} · P(X)] – µ^{2} |

= | [0^{2} · (0.18) + 1^{2} · (0.34) + 2^{2} · (0.23) + 3^{2} · (0.21) + 4^{2} · (0.04)] – 1.6^{2} | |

= | [0 + 0.34 + 0.92 + 1.89 + 0.64] – 2.56 | |

= | 3.79 – 2.56 = 1.23 | |

= | 1.2 (rounded) |

The standard deviation is , or .

No. The mean number of people calling at any one time is 1.6. Since the standard deviation is 1.1, most callers would be accommodated by having four phone lines because *µ* + 2*σ* would be 1.6 + 2(1.1) = 1.6 + 2.2 = 3.8. Very few callers would get a busy signal since at least 75% of the callers would either get through or be put on hold. (See Chebyshev’s theorem in Section 3–2.)

**Expectation**

Another concept related to the mean for a probability distribution is that of expected value or expectation. Expected value is used in various types of games of chance, in insurance, and in other areas, such as decision theory.

The **expected value** of a discrete random variable of a probability distribution is the theoretical average of the variable. The formula is

*µ* = *E*(*X*) = Σ*X* · *P*(*X*)

The symbol *E*(*X*) is used for the expected value.

The formula for the expected value is the same as the formula for the theoretical mean. The expected value, then, is the theoretical mean of the probability distribution. That is, *E*(*X*) = *µ*.

When expected value problems involve money, it is customary to round the answer to the nearest cent.

Example 5–12

**Winning Tickets**

One thousand tickets are sold at $1 each for a color television valued at $350. What is the expected value of the gain if you purchase one ticket?

Solution

The problem can be set up as follows:

Win | Lose | |

Gain X | $349 | –$1 |

Probability P(X) |

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Two things should be noted. First, for a win, the net gain is $349, since you do not get the cost of the ticket ($1) back. Second, for a loss, the gain is represented by a negative number, in this case –$1. The solution, then, is

Expected value problems of this type can also be solved by finding the overall gain (i.e., the value of the prize won or the amount of money won, not considering the cost of the ticket for the prize or the cost to play the game) and subtracting the cost of the tickets or the cost to play the game, as shown:

Here, the overall gain ($350) must be used.

Note that the expectation is –$0.65. This does not mean that you lose $0.65, since you can only win a television set valued at $350 or lose $1 on the ticket. What this expectation means is that the average of the losses is $0.65 for each of the 1000 ticket holders. Here is another way of looking at this situation: If you purchased one ticket each week over a long time, the average loss would be $0.65 per ticket, since theoretically, on average, you would win the set once for each 1000 tickets purchased.

Example 5–13

**Special Die**

A special six-sided die is made in which 3 sides have 6 spots, 2 sides have 4 spots, and 1 side has 1 spot. If the die is rolled, find the expected value of the number of spots that will occur.

Solution

Since there are 3 sides with 6 spots, the probability of getting a 6 is = . Since there are 2 sides with 4 spots, the probability of getting 4 spots is = . The probability of getting 1 spot is since 1 side has 1 spot.

Notice you can only get 1, 4, or 6 spots; but if you rolled the die a large number of times and found the average, it would be about .

Example 5–14

**Bond Investment**

A financial adviser suggests that his client select one of two types of bonds in which to invest $5000. Bond *X* pays a return of 4% and has a default rate of 2%. Bond *Y* has a % return and a default rate of 1%. Find the expected rate of return and decide which bond would be a better investment. When the bond defaults, the investor loses all the investment.

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Solution

The return on bond *X* is $5000 · 4% = $200. The expected return then is

*E*(*X*) = $200(0.98) – $5000(0.02) = $96

The return on bond *Y* is $5000 · % = $125. The expected return then is

*E*(*X*) = $125(0.99) – $5000(0.01) = $73.75

Hence, bond *X* would be a better investment since the expected return is higher.

In gambling games, if the expected value of the game is zero, the game is said to be fair. If the expected value of a game is positive, then the game is in favor of the player. That is, the player has a better than even chance of winning. If the expected value of the game is negative, then the game is said to be in favor of the house. That is, in the long run, the players will lose money.

In his book *Probabilities in Everyday Life* (Ivy Books, 1986), author John D. McGervy gives the expectations for various casino games. For keno, the house wins $0.27 on every $1.00 bet. For Chuck-a-Luck, the house wins about $0.52 on every $1.00 bet. For roulette, the house wins about $0.90 on every $1.00 bet. For craps, the house wins about $0.88 on every $1.00 bet. The bottom line here is that if you gamble long enough, sooner or later you will end up losing money.

*Applying the Concepts* 5–2

**Expected Value**

On March 28, 1979, the nuclear generating facility at Three Mile Island, Pennsylvania, began discharging radiation into the atmosphere. People exposed to even low levels of radiation can experience health problems ranging from very mild to severe, even causing death. A local newspaper reported that 11 babies were born with kidney problems in the three-county area surrounding the Three Mile Island nuclear power plant. The expected value for that problem in infants in that area was 3. Answer the following questions.

1.What does *expected value* mean?

2.Would you expect the exact value of 3 all the time?

3.If a news reporter stated that the number of cases of kidney problems in newborns was nearly four times as much as was usually expected, do you think pregnant mothers living in that area would be overly concerned?

4.Is it unlikely that 11 occurred by chance?

5.Are there any other statistics that could better inform the public?

6.Assume that 3 out of 2500 babies were born with kidney problems in that three-county area the year before the accident. Also assume that 11 out of 2500 babies were born with kidney problems in that three-county area the year after the accident. What is the real percent of increase in that abnormality?

7.Do you think that pregnant mothers living in that area should be overly concerned after looking at the results in terms of rates?

See page 296 for the answers.

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Exercises 5–2 |

**1.Defective Transistors** From past experience, a company has found that in cartons of transistors, 92% contain no defective transistors, 3% contain one defective transistor, 3% contain two defective transistors, and 2% contain three defective transistors. Find the mean, variance, and standard deviation for the defective transistors.

About how many extra transistors per day would the company need to replace the defective ones if it used 10 cartons per day?

**2.Suit Sales** The number of suits sold per day at a retail store is shown in the table, with the corresponding probabilities. Find the mean, variance, and standard deviation of the distribution.

If the manager of the retail store wants to be sure that he has enough suits for the next 5 days, how many should the manager purchase?

**3.Number of Credit Cards** A bank vice president feels that each savings account customer has, on average, three credit cards. The following distribution represents the number of credit cards people own. Find the mean, variance, and standard deviation. Is the vice president correct?

**4.Trivia Quiz** The probabilities that a player will get 5 to 10 questions right on a trivia quiz are shown below. Find the mean, variance, and standard deviation for the distribution.

**5.Cellular Phone Sales** The probability that a cellular phone company kiosk sells *X* number of new phone contracts per day is shown below. Find the mean, variance, and standard deviation for this probability distribution.

What is the probability that they will sell 6 or more contracts three days in a row?

**6.Animal Shelter Adoptions** The local animal shelter adopts out cats and dogs each week with the following probabilities.

Find the mean, variance, and standard deviation for the number of animals adopted each week. What is the probability that they find homes for more than 5 animals in a given week?

**7.Commercials During Children’s TV Programs** A concerned parents group determined the number of commercials shown in each of five children’s programs over a period of time. Find the mean, variance, and standard deviation for the distribution shown.

**8.Number of Televisions per Household** A study conducted by a TV station showed the number of televisions per household and the corresponding probabilities for each. Find the mean, variance, and standard deviation.

If you were taking a survey on the programs that were watched on television, how many program diaries would you send to each household in the survey?

**9.Students Using the Math Lab** The number of students using the Math Lab per day is found in the distribution below. Find the mean, variance, and standard deviation for this probability distribution.

What is the probability that fewer than 8 or more than 12 use the lab in a given day?

**10.Pizza Deliveries** A pizza shop owner determines the number of pizzas that are delivered each day. Find the mean, variance, and standard deviation for the distribution shown. If the manager stated that 45 pizzas were delivered on one day, do you think that this is a believable claim?

**11.Insurance** An insurance company insures a person’s antique coin collection worth $20,000 for an annual premium of $300. If the company figures that the probability of the collection being stolen is 0.002, what will be the company’s expected profit?

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**12.Job Bids** A landscape contractor bids on jobs where he can make $3000 profit. The probabilities of getting 1, 2, 3, or 4 jobs per month are shown.

Find the contractor’s expected profit per month.

**13.Rolling Dice** If a person rolls doubles when she tosses two dice, she wins $5. For the game to be fair, how much should she pay to play the game?

**14.Dice Game** A person pays $2 to play a certain game by rolling a single die once. If a 1 or a 2 comes up, the person wins nothing. If, however, the player rolls a 3, 4, 5, or 6, he or she wins the difference between the number rolled and $2. Find the expectation for this game. Is the game fair?

**15.Lottery Prizes** A lottery offers one $1000 prize, one $500 prize, and five $100 prizes. One thousand tickets are sold at $3 each. Find the expectation if a person buys one ticket.

**16.**In Exercise 15, find the expectation if a person buys two tickets. Assume that the player’s ticket is replaced after each draw and that the same ticket can win more than one prize.

**17.Winning the Lottery** For a daily lottery, a person selects a three-digit number. If the person plays for $1, she can win $500. Find the expectation. In the same daily lottery, if a person boxes a number, she will win $80. Find the expectation if the number 123 is played for $1 and boxed. (When a number is “boxed,” it can win when the digits occur in any order.)

**18.Life Insurance** A 35-year-old woman purchases a $100,000 term life insurance policy for an annual payment of $360. Based on a period life table for the U.S. government, the probability that she will survive the year is 0.999057. Find the expected value of the policy for the insurance company.

**19.Raffle Ticket Sales** A civic group sells 1000 raffle tickets to raise $2500 for its namesake charity. First prize is $1000, second prize is $300, and third prize is $200. How much should the group charge for each ticket?

**Extending the Concepts**

**20.Rolling Dice** Construct a probability distribution for the sum shown on the faces when two dice are rolled. Find the mean, variance, and standard deviation of the distribution.

**21.Rolling a Die** When one die is rolled, the expected value of the number of spots is 3.5. In Exercise 20, the mean number of spots was found for rolling two dice. What is the mean number of spots if three dice are rolled?

**22.**The formula for finding the variance for a probability distribution is

*σ* = Σ[(*X* – *µ*)^{2} · *P*(*X*)]

Verify algebraically that this formula gives the same result as the shortcut formula shown in this section.

**23.Rolling a Die** Roll a die 100 times. Compute the mean and standard deviation. How does the result compare with the theoretical results of Example 5–5?

**24.Rolling Two Dice** Roll two dice 100 times and find the mean, variance, and standard deviation of the sum of the spots. Compare the result with the theoretical results obtained in Exercise 20.

**25.Extracurricular Activities** Conduct a survey of the number of extracurricular activities your classmates are enrolled in. Construct a probability distribution and find the mean, variance, and standard deviation.

**26.Promotional Campaign** In a recent promotional campaign, a company offered these prizes and the corresponding probabilities. Find the expected value of winning. The tickets are free.

Number of prizes | Amount | Probability |

1 | $100,000 | |

2 | 10,000 | |

5 | 1,000 | |

10 | 100 |

If the winner has to mail in the winning ticket to claim the prize, what will be the expectation if the cost of the stamp is considered? Use the current cost of a stamp for a first-class letter.

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*Speaking of* Statistics

This study shows that a part of the brain reacts to the impact of losing, and it might explain why people tend to increase their bets after losing when gambling. Explain how this type of split decision making may influence fighter pilots, firefighters, or police officers, as the article states.

*Source: Psychology Today,* August 2002, p. 22. Used with permission.

**Technology Step by Step**

**TI-83 Plus or TI-84 Plus**

**Step by Step**

To calculate the mean and variance for a discrete random variable by using the formulas:

**1.**Enter the *x* values into L_{1} and the probabilities into L_{2}.

**2.**Move the cursor to the top of the L_{3} column so that L_{3} is highlighted.

**3.**Type **L _{1}** multiplied by

**L**then press

_{2},**ENTER.**

**4.**Move the cursor to the top of the L_{4} column so that L_{4} is highlighted.

**5.**Type **L _{1}** followed by the

*x*

^{2}key multiplied by

**L**then press

_{2},**ENTER.**

**6.**Type **2nd QUIT** to return to the home screen.

**7.**Type **2nd LIST,** move the cursor to MATH, type **5** for sum, then type **L _{3},** then press

**ENTER.**

**8.**Type **2nd ENTER,** move the cursor to L_{3}, type L_{4}, then press **ENTER.**

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Using the data from Example 5–10 gives the following:

To calculate the mean and standard deviation for a discrete random variable without using the formulas, modify the procedure to calculate the mean and standard deviation from grouped data (Chapter 3) by entering the *x* values into L_{1} and the probabilities into L_{2}.

Objective 3

Find the exact probability for *X* successes in *n* trials of a binomial experiment.

**5–3The Binomial Distribution**

Many types of probability problems have only two outcomes or can be reduced to two outcomes. For example, when a coin is tossed, it can land heads or tails. When a baby is born, it will be either male or female. In a basketball game, a team either wins or loses. A true/false item can be answered in only two ways, true or false. Other situations can be reduced to two outcomes. For example, a medical treatment can be classified as effective or ineffective, depending on the results. A person can be classified as having normal or abnormal blood pressure, depending on the measure of the blood pressure gauge. A multiple-choice question, even though there are four or five answer choices, can be classified as correct or incorrect. Situations like these are called *binomial experiments.*

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Historical Note

In 1653, Blaise Pascal created a triangle of numbers called *Pascal’s triangle* that can be used in the binomial distribution.

A **binomial experiment** is a probability experiment that satisfies the following four requirements:

1.There must be a fixed number of trials.

2.Each trial can have only two outcomes or outcomes that can be reduced to two outcomes. These outcomes can be considered as either success or failure.

3.The outcomes of each trial must be independent of one another.

4.The probability of a success must remain the same for each trial.

A binomial experiment and its results give rise to a special probability distribution called the *binomial distribution.*

The outcomes of a binomial experiment and the corresponding probabilities of these outcomes are called a **binomial distribution.**

In binomial experiments, the outcomes are usually classified as successes or failures. For example, the correct answer to a multiple-choice item can be classified as a success, but any of the other choices would be incorrect and hence classified as a failure. The notation that is commonly used for binomial experiments and the binomial distribution is defined now.

**Notation for the Binomial Distribution**

P(S) | The symbol for the probability of success |

P(F) | The symbol for the probability of failure |

p | The numerical probability of a success |

q | The numerical probability of a failure |

P(S) = p and P(F) = 1 – p = q | |

n | The number of trials |

X | The number of successes in n trials |

Note that 0 ≤ *X* ≤ *n* and *X* = 0, 1, 2, 3, … , *n*.

The probability of a success in a binomial experiment can be computed with this formula.

**Binomial Probability Formula**

In a binomial experiment, the probability of exactly *X* successes in *n* trials is

An explanation of why the formula works is given following Example 5–15.

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Example 5–15

**Tossing Coins**

A coin is tossed 3 times. Find the probability of getting exactly two heads.

Solution

This problem can be solved by looking at the sample space. There are three ways to get two heads.

HHH, HHT, HTH, THH, TTH, THT, HTT, TTT

The answer is , or 0.375.

Looking at the problem in Example 5–15 from the standpoint of a binomial experiment, one can show that it meets the four requirements.

**1.**There are a fixed number of trials (three).

**2.**There are only two outcomes for each trial, heads or tails.

**3.**The outcomes are independent of one another (the outcome of one toss in no way affects the outcome of another toss).

**4.**The probability of a success (heads) is in each case.

In this case, *n* = 3, *X* = 2, *p* = , and *q* = . Hence, substituting in the formula gives

which is the same answer obtained by using the sample space.

The same example can be used to explain the formula. First, note that there are three ways to get exactly two heads and one tail from a possible eight ways. They are HHT, HTH, and THH. In this case, then, the number of ways of obtaining two heads from three coin tosses is _{3}*C*_{2}, or 3, as shown in Chapter 4. In general, the number of ways to get *X* successes from *n* trials without regard to order is

This is the first part of the binomial formula. (Some calculators can be used for this.)

Next, each success has a probability of and can occur twice. Likewise, each failure has a probability of and can occur once, giving the part of the formula. To generalize, then, each success has a probability of *p* and can occur *X* times, and each failure has a probability of *q* and can occur *n* – *X* times. Putting it all together yields the binomial probability formula.

Example 5–16

**Survey on Doctor Visits**

A survey found that one out of five Americans say he or she has visited a doctor in any given month. If 10 people are selected at random, find the probability that exactly 3 will have visited a doctor last month.

Source: *Reader’s Digest.*

Solution

In this case, *n* = 10, *X* = 3, *p* = , and *q* = . Hence,

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Example 5–17

**Survey on Employment**

A survey from Teenage Research Unlimited (Northbrook, Illinois) found that 30% of teenage consumers receive their spending money from part-time jobs. If 5 teenagers are selected at random, find the probability that at least 3 of them will have part-time jobs.

Solution

To find the probability that at least 3 have part-time jobs, it is necessary to find the individual probabilities for 3, or 4, or 5, and then add them to get the total probability.

Hence,

*P*(at least three teenagers have part-time jobs)

= 0.132 + 0.028 + 0.002 = 0.162

Computing probabilities by using the binomial probability formula can be quite tedious at times, so tables have been developed for selected values of *n* and *p*. Table B in Appendix C gives the probabilities for individual events. Example 5–18 shows how to use Table B to compute probabilities for binomial experiments.

Example 5–18

**Tossing Coins**

Solve the problem in Example 5–15 by using Table B.

Solution

Since *n* = 3, *X* = 2, and *p* = 0.5, the value 0.375 is found as shown in Figure 5–3.

Figure 5–3

**Using ****Table B**** for ****Example 5–18**

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Example 5–19

**Survey on Fear of Being Home Alone at Night**

*Public Opinion* reported that 5% of Americans are afraid of being alone in a house at night. If a random sample of 20 Americans is selected, find these probabilities by using the binomial table.

*a*.There are exactly 5 people in the sample who are afraid of being alone at night.

*b*.There are at most 3 people in the sample who are afraid of being alone at night.

*c*.There are at least 3 people in the sample who are afraid of being alone at night.

Source: *100% American* by Daniel Evan Weiss.

Solution

*a*.*n* = 20, *p* = 0.05, and *X* = 5. From the table, we get 0.002.

*b*.*n* = 20 and *p* = 0.05. “At most 3 people” means 0, or 1, or 2, or 3.

Hence, the solution is

P(0) + P(1) + P(2) + P(3) | = | 0.358 + 0.377 + 0.189 + 0.060 |

= | 0.984 |

*c*.*n* = 20 and *p* = 0.05. “At least 3 people” means 3, 4, 5, … , 20. This problem can best be solved by finding *P*(0) + *P*(1) + *P*(2) and subtracting from 1.

P(0) + P(1) + P(2) | = | 0.358 + 0.377 + 0.189 = 0.924 |

1 – 0.924 | = | 0.076 |

Example 5–20

**Driving While Intoxicated**

A report from the Secretary of Health and Human Services stated that 70% of single-vehicle traffic fatalities that occur at night on weekends involve an intoxicated driver. If a sample of 15 single-vehicle traffic fatalities that occur at night on a weekend is selected, find the probability that exactly 12 involve a driver who is intoxicated.

Source: *100% American* by Daniel Evan Weiss.

Solution

Now, *n* = 15, *p* = 0.70, and *X* = 12. From Table B, *P*(12) = 0.170. Hence, the probability is 0.17.

Remember that in the use of the binomial distribution, the outcomes must be independent. For example, in the selection of components from a batch to be tested, each component must be replaced before the next one is selected. Otherwise, the outcomes are not independent. However, a dilemma arises because there is a chance that the same component could be selected again. This situation can be avoided by not replacing the component and using a distribution called the hypergeometric distribution to calculate the probabilities. The hypergeometric distribution is beyond the scope of this book.

Objective 4

Find the mean, variance, and standard deviation for the variable of a binomial distribution.

**Mean, Variance, and Standard Deviation for the Binomial Distribution**

The mean, variance, and standard deviation of a variable that has the *binomial distribution* can be found by using the following formulas.

Mean: µ = n · p | Variance: σ^{2} = n · p · q | Standard deviation: σ = |

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These formulas are algebraically equivalent to the formulas for the mean, variance, and standard deviation of the variables for probability distributions, but because they are for variables of the binomial distribution, they have been simplified by using algebra. The algebraic derivation is omitted here, but their equivalence is shown in Example 5–21.

Example 5–21

**Tossing a Coin**

A coin is tossed 4 times. Find the mean, variance, and standard deviation of the number of heads that will be obtained.

Solution

With the formulas for the binomial distribution and *n* = 4, *p* = , and *q* = , the results are

From Example 5–21, when four coins are tossed many, many times, the average of the number of heads that appear is 2, and the standard deviation of the number of heads is 1. Note that these are theoretical values.

As stated previously, this problem can be solved by using the formulas for expected value. The distribution is shown.

Hence, the simplified binomial formulas give the same results.

Example 5–22

**Rolling a Die**

A die is rolled 360 times. Find the mean, variance, and standard deviation of the number of 4s that will be rolled.

Solution

This is a binomial experiment since getting a 4 is a success and not getting a 4 is considered a failure. Hence *n* = 360, *p* = , and *q* = .

On average, sixty 4s will be rolled. The standard deviation is 7.07.

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Example 5–23

**Likelihood of Twins**

The *Statistical Bulletin* published by Metropolitan Life Insurance Co. reported that 2% of all American births result in twins. If a random sample of 8000 births is taken, find the mean, variance, and standard deviation of the number of births that would result in twins.

Source: *100% American* by Daniel Evan Weiss.

Solution

This is a binomial situation, since a birth can result in either twins or not twins (i.e., two outcomes).

For the sample, the average number of births that would result in twins is 160, the variance is 156.8, or 157, and the standard deviation is 12.5, or 13 if rounded.

*Applying the Concepts* 5–3

**Unsanitary Restaurants**

Health officials routinely check sanitary conditions of restaurants. Assume you visit a popular tourist spot and read in the newspaper that in 3 out of every 7 restaurants checked, there were unsatisfactory health conditions found. Assuming you are planning to eat out 10 times while you are there on vacation, answer the following questions.

1.How likely is it that you will eat at three restaurants with unsanitary conditions?

2.How likely is it that you will eat at four or five restaurants with unsanitary conditions?

3.Explain how you would compute the probability of eating in at least one restaurant with unsanitary conditions. Could you use the complement to solve this problem?

4.What is the most likely number to occur in this experiment?

5.How variable will the data be around the most likely number?

6.Is this a binomial distribution?

7.If it is a binomial distribution, does that mean that the likelihood of a success is always 50% since there are only two possible outcomes?

Check your answers by using the following computer-generated table.

Mean = 4.3 | Std. dev. = 1.56557 |

X | P(X) | Cum. Prob. |

0 | 0.00362 | 0.00362 |

1 | 0.02731 | 0.03093 |

2 | 0.09272 | 0.12365 |

3 | 0.18651 | 0.31016 |

4 | 0.24623 | 0.55639 |

5 | 0.22291 | 0.77930 |

6 | 0.14013 | 0.91943 |

7 | 0.06041 | 0.97983 |

8 | 0.01709 | 0.99692 |

9 | 0.00286 | 0.99979 |

10 | 0.00022 | 1.00000 |

See page 296 for the answers.

Page 285

Exercises 5–3 |

**1.**Which of the following are binomial experiments or can be reduced to binomial experiments?

*a*.Surveying 100 people to determine if they like Sudsy Soap

*b*.Tossing a coin 100 times to see how many heads occur

*c*.Drawing a card with replacement from a deck and getting a heart

*d*.Asking 1000 people which brand of cigarettes they smoke

*e*.Testing four different brands of aspirin to see which brands are effective

*f*.Testing one brand of aspirin by using 10 people to determine whether it is effective

*g*.Asking 100 people if they smoke

*h*.Checking 1000 applicants to see whether they were admitted to White Oak College

*i*.Surveying 300 prisoners to see how many different crimes they were convicted of

*j*.Surveying 300 prisoners to see whether this is their first offense

**2.(ans)** Compute the probability of *X* successes, using Table B in Appendix C.

*a*.*n* = 2, *p* = 0.30, *X* = 1

*b*.*n* = 4, *p* = 0.60, *X* = 3

*c*.*n* = 5, *p* = 0.10, *X* = 0

*d*.*n* = 10, *p* = 0.40, *X* = 4

*e*.*n* = 12, *p* = 0.90, *X* = 2

*f*.*n* = 15, *p* = 0.80, *X* = 12

*g*.*n* = 17, *p* = 0.05, *X* = 0

*h*.*n* = 20, *p* = 0.50, *X* = 10

*i*.*n* = 16, *p* = 0.20, *X* = 3

**3.**Compute the probability of *X* successes, using the binomial formula.

*a*.*n* = 6, *X* = 3, *p* = 0.03

*b*.*n* = 4, *X* = 2, *p* = 0.18

*c*.*n* = 5, *X* = 3, *p* = 0.63

*d*.*n* = 9, *X* = 0, *p* = 0.42

*e*.*n* = 10, *X* = 5, *p* = 0.37

**For Exercises 4 through 13, assume all variables are binomial. ( Note: If values are not found in **

**Table B**

**of**

**Appendix C**

**, use the binomial formula.)**

**4.Burglar Alarms** A burglar alarm system has six fail-safe components. The probability of each failing is 0.05. Find these probabilities.

*a*.Exactly three will fail.

*b*.Fewer than two will fail.

*c*.None will fail.

*d*.Compare the answers for parts *a*, *b*, and *c*, and explain why the results are reasonable.

**5.True/False Exam** A student takes a 20-question, true/false exam and guesses on each question. Find the probability of passing if the lowest passing grade is 15 correct out of 20. Would you consider this event likely to occur? Explain your answer.

**6.Multiple-Choice Exam** A student takes a 20-question, multiple-choice exam with five choices for each question and guesses on each question. Find the probability of guessing at least 15 out of 20 correctly. Would you consider this event likely or unlikely to occur? Explain your answer.

**7.Driving to Work Alone** It is reported that 77% of workers aged 16 and over drive to work alone. Choose 8 workers at random. Find the probability that

*a*.All drive to work alone

*b*.More than one-half drive to work alone

*c*.Exactly 3 drive to work alone

Source: www.factfinder.census.gov

**8.High School Dropouts** Approximately 10.3% of American high school students drop out of school before graduation. Choose 10 students entering high school at random. Find the probability that

*a*.No more than two drop out

*b*.At least 6 graduate

*c*.All 10 stay in school and graduate

Source: www.infoplease.com

**9.Survey on Concern for Criminals** In a survey, 3 of 4 students said the courts show “too much concern” for criminals. Find the probability that at most 3 out of 7 randomly selected students will agree with this statement.

Source: *Harper’s Index.*

**10.Labor Force Couples** The percentage of couples where both parties are in the labor force is 52.1. Choose 5 couples at random. Find the probability that

*a*.None of the couples have both persons working

*b*.More than 3 of the couples have both persons in the labor force

*c*.Fewer than 2 of the couples have both parties working

Source: www.bls.gov

**11.College Education and Business World Success** R. H. Bruskin Associates Market Research found that 40% of Americans do not think that having a college education is important to succeed in the business world. If a random sample of five Americans is selected, find these probabilities.

*a*.Exactly 2 people will agree with that statement.

*b*.At most 3 people will agree with that statement.

*c*.At least 2 people will agree with that statement.

*d*.Fewer than 3 people will agree with that statement.

Source: *100% American* by Daniel Evans Weiss.

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**12.Destination Weddings** Twenty-six percent of couples who plan to marry this year are planning destination weddings. In a random sample of 12 couples who plan to marry, find the probability that

*a*.Exactly 6 couples will have a destination wedding

*b*.At least 6 couples will have a destination wedding

*c*.Fewer than 5 couples will have a destination wedding

Source: *Time* magazine.

**13.People Who Have Some College Education** Fifty-three percent of all persons in the U.S. population have at least some college education. Choose 10 persons at random. Find the probability that

*a*.Exactly one-half have some college education

*b*.At least 5 do not have any college education

*c*.Fewer than 5 have some college education

Source: *New York Times Almanac.*

**14.(ans)** Find the mean, variance, and standard deviation for each of the values of *n* and *p* when the conditions for the binomial distribution are met.

*a*.*n* = 100, *p* = 0.75

*b*.*n* = 300, *p* = 0.3

*c*.*n* = 20, *p* = 0.5

*d*.*n* = 10, *p* = 0.8

*e*.*n* = 1000, *p* = 0.1

*f*.*n* = 500, *p* = 0.25

*g*.*n* = 50, *p* =

*h*.*n* = 36, *p* =

**15.Social Security Recipients** A study found that 1% of Social Security recipients are too young to vote. If 800 Social Security recipients are randomly selected, find the mean, variance, and standard deviation of the number of recipients who are too young to vote.

Source: *Harper’s Index.*

**16.**Find the mean, variance, and standard deviation for the number of heads when 20 coins are tossed.

**17.Defective Calculators** If 3% of calculators are defective, find the mean, variance, and standard deviation of a lot of 300 calculators.

**18.Federal Government Employee E-mail Use** It has been reported that 83% of federal government employees use e-mail. If a sample of 200 federal government employees is selected, find the mean, variance, and standard deviation of the number who use e-mail.

Source: *USA TODAY.*

**19.Watching Fireworks** A survey found that 21% of Americans watch fireworks on television on July 4. Find the mean, variance, and standard deviation of the number of individuals who watch fireworks on television on July 4 if a random sample of 1000 Americans is selected.

Source: USA Snapshot, *USA TODAY.*

**20.Alternate Sources of Fuel** Eighty-five percent of Americans favor spending government money to develop alternative sources of fuel for automobiles. For a random sample of 120 Americans, find the mean, variance, and standard deviation for the number who favor government spending for alternative fuels.

Source: www.pollingreport.com

**21.Survey on Bathing Pets** A survey found that 25% of pet owners had their pets bathed professionally rather than do it themselves. If 18 pet owners are randomly selected, find the probability that exactly 5 people have their pets bathed professionally.

Source: USA Snapshot, *USA TODAY.*

**22.Survey on Answering Machine Ownership** In a survey, 63% of Americans said they own an answering machine. If 14 Americans are selected at random, find the probability that exactly 9 own an answering machine.

Source: USA Snapshot, *USA TODAY.*

**23.Poverty and the Federal Government** One out of every three Americans believes that the U.S. government should take “primary responsibility” for eliminating poverty in the United States. If 10 Americans are selected, find the probability that at most 3 will believe that the U.S. government should take primary responsibility for eliminating poverty.

Source: *Harper’s Index.*

**24.Internet Purchases** Thirty-two percent of adult Internet users have purchased products or services online. For a random sample of 200 adult Internet users, find the mean, variance, and standard deviation for the number who have purchased goods or services online.

Source: www.infoplease.com

**25.Survey on Internet Awareness** In a survey, 58% of American adults said they had never heard of the Internet. If 20 American adults are selected at random, find the probability that exactly 12 will say they have never heard of the Internet.

Source: *Harper’s Index.*

**26.Job Elimination** In the past year, 13% of businesses have eliminated jobs. If 5 businesses are selected at random, find the probability that at least 3 have eliminated jobs during the last year.

Source: *USA TODAY.*

**27.Survey of High School Seniors** Of graduating high school seniors, 14% said that their generation will be remembered for their social concerns. If 7 graduating seniors are selected at random, find the probability that either 2 or 3 will agree with that statement.

Source: *USA TODAY.*

**28.**Is this a binomial distribution? Explain.

Page 287

**Extending the Concepts**

**29.Children in a Family** The graph shown here represents the probability distribution for the number of girls in a family of three children. From this graph, construct a probability distribution

**30.**Construct a binomial distribution graph for the number of defective computer chips in a lot of 4 if *p* = 0.3.

**Technology Step by Step**

**MINITAB**

**Step by Step**

**The Binomial Distribution**

**Calculate a Binomial Probability**

From Example 5–19, it is known that 5% of the population is afraid of being alone at night. If a random sample of 20 Americans is selected, what is the probability that exactly 5 of them are afraid?

*n* = 20 *p* = 0.05 (5%) and *X* = 5 (5 out of 20)

No data need to be entered in the worksheet.

**1.**Select **Calc>Probability Distributions>Binomial.**

**2.**Click the option for Probability.

**3.**Click in the text box for Number of trials:.

**4.**Type in **20,** then Tab to Probability of success, then type **.05.**

**5.**Click the option for Input constant, then type in **5.** Leave the text box for Optional storage empty. If the name of a constant such as K1 is entered here, the results are stored but not displayed in the session window.

**6.**Click [OK]. The results are visible in the session window.

**Probability Density Function**

Binomial with n = 20 and p = 0.05

x f(x)

5 0.0022446

Page 288

**Construct a Binomial Distribution**

These instructions will use *n* = 20 and *p* = 0.05.

**1.**Select **Calc>Make Patterned Data>Simple Set of Numbers.**

**2.**You must enter three items:

a)Enter **X** in the box for Store patterned data in:. MINITAB will use the first empty column of the active worksheet and name it X.

b)Press Tab. Enter the value of 0 for the first value. Press Tab.

c)Enter **20** for the last value. This value should be *n.* In steps of:, the value should be 1.

**3.**Click [OK].

**4.**Select **Calc>Probability Distributions>Binomial.**

**5.**In the dialog box you must enter five items.

a)Click the button for Probability.

b)In the box for Number of trials enter **20.**

c)Enter **.05** in the Probability of success.

d)Check the button for Input columns, then type the column name, **X,** in the text box.

e)Click in the box for Optional storage, then type **Px.**

**6.**Click [OK]. The first available column will be named Px, and the calculated probabilities will be stored in it.

**7.**To view the completed table, click the worksheet icon on the toolbar.

**Graph a Binomial Distribution**

The table must be available in the worksheet.

**1.**Select **Graph>Scatterplot,** then Simple.

a)Double-click on C2 Px for the Y variable and C1 X for the X variable.

b)Click [Data view], then Project lines, then [OK]. Deselect any other type of display that may be selected in this list.

c)Click on [Labels], then Title/Footnotes.

d)Type an appropriate title, such as **Binomial Distribution n = 20, p = .05.**

e)Press Tab to the Subtitle 1, then type in Your Name.

f)Optional: Click [Scales] then [Gridlines] then check the box for Y major ticks.

g)Click [OK] twice.

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The graph will be displayed in a window. Right-click the control box to save, print, or close the graph.

**TI-83 Plus or TI-84 Plus**

**Step by Step**

**Binomial Random Variables**

To find the probability for a binomial variable:

Press **2nd [DISTR]** then 0 for binomial pdf( (Note: On the TI-84 Plus Use A)

The form is binompdf(*n,p,X*).

Example: *n* = 20, *X* = 5, *p* = .05. (Example 5–19*a* from the text) binompdf(20,.05,5)

Example: *n* = 20, *X* = 0, 1, 2, 3, *p* = .05. (Example 5–19*b* from the text) binompdf(20,.05,{0,1,2,3})

The calculator will display the probabilities in a list. Use the arrow keys to view entire display.

To find the cumulative probability for a binomial random variable:

Press **2nd [DISTR]** then **A (ALPHA MATH)** for binomcdf( (Note: On the TI-84 Plus Use B) The form is binomcdf(*n,p,X*). This will calculate the cumulative probability for values from 0 to *X*.

Example: *n* = 20, *X* = 0, 1, 2, 3, *p* = .05 (Example 5–19*b* from the text) binomcdf(20,.05,3)

To construct a binomial probability table:

**1.**Enter the *X* values 0 through *n* into L_{1}.

**2.**Move the cursor to the top of the L_{2} column so that L_{2} is highlighted.

**3.**Type the command binompdf(*n,p,*L_{1}), then press **ENTER.**

Example: *n* = 20, *p* = .05 (Example 5–19 from the text)

Page 290

**Excel**

**Step by Step**

**Creating a Binomial Distribution and Graph**

These instructions will demonstrate how Excel can be used to construct a binomial distribution table for *n* = 20 and *p* = 0.35.

**1.**Type **X** for the binomial variable label in cell A1 of an Excel worksheet.

**2.**Type **P(X)** for the corresponding probabilities in cell B1.

**3.**Enter the integers from 0 to 20 in column A starting at cell A2. Select the Data tab from the toolbar. Then select Data Analysis. Under Analysis Tools, select Random Number Generation and click [OK].

**4.**In the Random Number Generation dialog box, enter the following:

a)Number of Variables: **1**

b)Distribution: Patterned

c)Parameters: From **0** to **20** in steps of **1,** repeating each number: **1** times and repeating each sequence **1** times

d)Output range: **A2:A21**

**5.**Then click [OK].

**6.**To determine the probability corresponding to the first value of the binomial random variable, select cell B2 and type: **=BINOMDIST(0,20,.35,FALSE).** This will give the probability of obtaining 0 successes in 20 trials of a binomial experiment for which the probability of success is 0.35.

**7.**Repeat step **6,** changing the first parameter, for each of the values of the random variable from column A.

*Note:* If you wish to obtain the cumulative probabilities for each of the values in column A, you can type: **=BINOMDIST(0,20,.35,TRUE)** and repeat for each of the values in column A.

To create the graph:

**1.**Select the Insert tab from the toolbar and the Column Chart.

**2.**Select the Clustered Column (the first column chart under the 2-D Column selections).

**3.**You will need to edit the data for the chart.

a)Right-click the mouse on any location of the chart. Click the Select Data option. The Select Data Source dialog box will appear.

b)Click X in the Legend Entries box and click Remove.

c)Click the Edit button under Horizontal Axis Labels to insert a range for the variable X.

d)When the Axis Labels box appears, highlight cells A2 to A21 on the worksheet, then click [OK].

**4.**To change the title of the chart:

a)Left-click once on the current title.

b)Type a new title for the chart, for example, Binomial Distribution (20, .35, .65).

Page 291

**Summary**

Many variables have special probability distributions. This chapter presented several of the most common probability distributions, including the binomial distribution, the multinomial distribution, the Poisson distribution, and the hypergeometric distribution.

The binomial distribution is used when there are only two outcomes for an experiment, there are a fixed number of trials, the probability is the same for each trial, and the outcomes are independent of one another.

A probability distribution can be graphed, and the mean, variance, and standard deviation can be found. The mathematical expectation can also be calculated for a probability distribution. Expectation is used in insurance and games of chance.

**Important Terms**

binomial distribution

binomial experiment

discrete probability distribution

expected value

random variable

**Important Formulas**

Formula for the mean of a probability distribution:

*µ*** = Σ X · P(X)**

Formulas for the variance and standard deviation of a probability distribution:

Formula for expected value:

*E***( X) = ΣX · P(X)**

Binomial probability formula:

Formula for the mean of the binomial distribution:

*µ*** = n · p**

Formulas for the variance and standard deviation of the binomial distribution:

Page 292

**Review Exercises**

**For Exercises 1 through 3, determine whether the distribution represents a probability distribution. If it does not, state why.**

**1.**

**2.**

**3.**

**4.Emergency Calls** The number of emergency calls a local police department receives per 24-hour period is distributed as shown here. Construct a graph for the data.

**5.Credit Cards** A large retail company encourages its employees to get customers to apply for the store credit card. Below is the distribution for the number of credit card applications received per employee for an 8-hour shift.

a.What is the probability that an employee will get 2 or 3 applications during any given shift?

b.Find the mean, variance, and standard deviation for this probability distribution.

**6.Coins in a Box** A box contains 5 pennies, 3 dimes, 1 quarter, and 1 half-dollar. Construct a probability distribution and draw a graph for the data.

**7.****Tie Purchases** At Tyler’s Tie Shop, Tyler found the probabilities that a customer will buy 0, 1, 2, 3, or 4 ties, as shown. Construct a graph for the distribution.

**8.Customers in a Bank** A bank has a drive-through service. The number of customers arriving during a 15-minute period is distributed as shown. Find the mean, variance, and standard deviation for the distribution.

**9.****Museum Visitors** At a small community museum, the number of visitors per hour during the day has the distribution shown here. Find the mean, variance, and standard deviation for the data.

**10.Cans of Paint Purchased** During a recent paint sale at Corner Hardware, the number of cans of paint purchased was distributed as shown. Find the mean, variance, and standard deviation of the distribution.

**11.****Inquiries Received** The number of inquiries received per day for a college catalog is distributed as shown. Find the mean, variance, and standard deviation for the data.

**12.Outdoor Regatta** A producer plans an outdoor regatta for May 3. The cost of the regatta is $8000. This includes advertising, security, printing tickets, entertainment, etc. The producer plans to make $15,000 profit if all goes well. However, if it rains, the regatta will have to be canceled. According to the weather report, the probability of rain is 0.3. Find the producer’s expected profit.

**13.****Card Game** A game is set up as follows: All the diamonds are removed from a deck of cards, and these 13 cards are placed in a bag. The cards are mixed up, and then one card is chosen at random (and then replaced). The player wins according to the following rules.

If the ace is drawn, the player loses $20.

If a face card is drawn, the player wins $10.

If any other card (2–10) is drawn, the player wins $2.

How much should be charged to play this game in order for it to be fair?

**14.**Using Exercise 13, how much should be charged if instead of winning $2 for drawing a 2–10, the player wins the amount shown on the card in dollars?

**15.**Let *x* be a binomial random variable with *n* = 12 and *p* = 0.3. Find the following:

a.*P*(*X* = 8)

b.*P*(*X*< 5)

c.*P*(*X* ≥ 10)

d.*P*(4 < X ≤ 9)

**16.Internet Access via Cell Phone** Fourteen percent of cell phone users use their cell phones to access the Internet. In a random sample of 10 cell phone users, what is the probability that exactly 2 have used their phones to access the Internet? More than 2?

Source: www.infoplease.com

Page 293

**17.****Drug Calculation Test** If 75% of nursing students are able to pass a drug calculation test, find the mean, variance, and standard deviation of the number of students who pass the test in a sample of 180 nursing students.

**18.Flu Shots** It has been reported that 63% of adults aged 65 and over got their flu shots last year. In a random sample of 300 adults aged 65 and over, find the mean, variance, and standard deviation for the number who got their flu shots.

Source: U.S. Center for Disease Control and Prevention.

**19.****U.S. Police Chiefs and the Death Penalty** The chance that a U.S. police chief believes the death penalty “significantly reduces the number of homicides” is 1 in 4. If a random sample of 8 police chiefs is selected, find the probability that at most 3 believe that the death penalty significantly reduces the number of homicides.

Source: *Harper’s Index.*

**20.Household Wood Burning***American Energy Review* reported that 27% of American households burn wood. If a random sample of 500 American households is selected, find the mean, variance, and standard deviation of the number of households that burn wood.

Source: *100% American* by Daniel Evan Weiss.

**21.****Pizza for Breakfast** Three out of four American adults under age 35 have eaten pizza for breakfast. If a random sample of 20 adults under age 35 is selected, find the probability that exactly 16 have eaten pizza for breakfast.

Source: *Harper’s Index.*

**22.Unmarried Women** According to survey records, 75.4% of women aged 20–24 have never been married. In a random sample of 250 young women aged 20–24, find the mean, variance, and standard deviation for the number who are or who have been married.

Source: www.infoplease.com

**Statistics Today**

**Is Pooling Worthwhile?—Revisited**

In the case of the pooled sample, the probability that only one test will be needed can be determined by using the binomial distribution. The question being asked is, In a sample of 15 individuals, what is the probability that no individual will have the disease? Hence, *n* = 15, *p* = 0.05, and *X* = 0. From Table B in Appendix C, the probability is 0.463, or 46% of the time, only one test will be needed. For screening purposes, then, pooling samples in this case would save considerable time, money, and effort as opposed to testing every individual in the population.

**Chapter Quiz**

**Determine whether each statement is true or false. If the statement is false, explain why.**

**1.**The expected value of a random variable can be thought of as a long-run average.

**2.**The number of courses a student is taking this semester is an example of a continuous random variable.

**3.**When the binomial distribution is used, the outcomes must be dependent.

**4.**A binomial experiment has a fixed number of trials.

**Complete these statements with the best answer.**

**5.**Random variable values are determined by _______.

**6.**The mean for a binomial variable can be found by using the formula _______.

**7.**One requirement for a probability distribution is that the sum of all the events in the sample space must equal ______.

**Select the best answer.**

**8.**What is the sum of the probabilities of all outcomes in a probability distribution?

*a.*0

*b.*

*c.*1

*d.*It cannot be determined.

**9.**How many outcomes are there in a binomial experiment?

*a.*0

*b.*1

*c.*2

*d.*It varies.

**10.**The number of trials for a binomial experiment

*a.*Can be infinite

*b.*Is unchanged

*c.*Is unlimited

*d.*Must be fixed

Page 294

**For questions 11 through 14, determine if the distribution represents a probability distribution. If not, state why.**

**11.**

**12.**

**13.**

**14.**

**15.****Calls for a Fire Company** The number of fire calls the Conestoga Valley Fire Company receives per day is distributed as follows:

Construct a graph for the data.

**16.****Telephones per Household** A study was conducted to determine the number of telephones each household has. The data are shown here.

Construct a probability distribution and draw a graph for the data.

**17.****CD Purchases** During a recent CD sale at Matt’s Music Store, the number of CDs customers purchased was distributed as follows:

Find the mean, variance, and standard deviation of the distribution.

**18.****Calls for a Crisis Hot Line** The number of calls received per day at a crisis hot line is distributed as follows:

Find the mean, variance, and standard deviation of the distribution.

**19.****Selecting a Card** There are 6 playing cards placed face down in a box. They are the 4 of diamonds, the 5 of hearts, the 2 of clubs, the 10 of spades, the 3 of diamonds, and the 7 of hearts. A person selects a card. Find the expected value of the draw.

**20.****Selecting a Card** A person selects a card from an ordinary deck of cards. If it is a black card, she wins $2. If it is a red card between or including 3 and 7, she wins $10. If it is a red face card, she wins $25; and if it is a black jack, she wins an extra $100. Find the expectation of the game.

**21.****Carpooling** If 40% of all commuters ride to work in carpools, find the probability that if 8 workers are selected, 5 will ride in carpools.

**22.Employed Women** If 60% of all women are employed outside the home, find the probability that in a sample of 20 women,

*a.*Exactly 15 are employed

*b.*At least 10 are employed

*c.*At most 5 are not employed outside the home

**23.****Driver’s Exam** If 80% of the applicants are able to pass a driver’s proficiency road test, find the mean, variance, and standard deviation of the number of people who pass the test in a sample of 300 applicants.

**24.****Meeting Attendance** A history class has 75 members. If there is a 12% absentee rate per class meeting, find the mean, variance, and standard deviation of the number of students who will be absent from each class.

**Critical Thinking Challenges**

**1.Lottery Numbers** Pennsylvania has a lottery entitled “Big 4.” To win, a player must correctly match four digits from a daily lottery in which four digits are selected. Find the probability of winning.

**2.Lottery Numbers** In the Big 4 lottery, for a bet of $100, the payoff is $5000. What is the expected value of winning? Is it worth it?

**3.Lottery Numbers** If you played the same four-digit number every day (or any four-digit number for that matter) in the Big 4, how often (in years) would you win, assuming you have average luck?

**4.Chuck-a-Luck** In the game Chuck-a-Luck, three dice are rolled. A player bets a certain amount (say $1.00) on a number from 1 to 6. If the number appears on 1 die, the person wins $1.00. If it appears on 2 dice, the person wins $2.00, and if it appears on all 3 dice, the person wins $3.00. What are the chances of winning $1.00? $2.00? $3.00?

**5.Chuck-a-Luck** What is the expected value of the game of Chuck-a-Luck if a player bets $1.00 on one number?

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** Data Projects**

**1.Business and Finance** Assume that a life insurance company would like to make a profit of $250 on a $100,000 policy sold to a person whose probability of surviving the year is 0.9985. What premium should the company charge the customer? If the company would like to make a $250 profit on a $100,000 policy at a premium of $500, what is the lowest life expectancy it should accept for a customer?

**2.Sports and Leisure** Baseball, hockey, and basketball all use a seven-game series to determine their championship. Find the probability that with two evenly matched teams a champion will be found in 4 games. Repeat for 5, 6, and 7 games. Look at the historical results for the three sports. How do the actual results compare to the theoretical?

**3.Technology** Use your most recent itemized phone bill for the data in this problem. Assume that incoming and outgoing calls are equal in the population (why is this a reasonable assumption?). This means assume *p* = 0.5. For the number of calls you made last month, what would be the mean number of outgoing calls in a random selection of calls? Also, compute the standard deviation. Was the number of outgoing calls you made an unusual amount given the above? In a selection of 12 calls, what is the probability that less than 3 were outgoing?

**4.Health and Wellness** Use Red Cross data to determine the percentage of the population with an Rh factor that is positive (A+, B+, AB+ , or O+ blood types). Use that value for *p*. How many students in your class have a positive Rh factor? Is this an unusual amount?

**5.Politics and Economics** Find out what percentage of citizens in your state is registered to vote. Assuming that this is a binomial variable, what would be the mean number of registered voters in a random group of citizens with a sample size equal to the number of students in your class? Also determine the standard deviation. How many students in your class are registered to vote? Is this an unusual number, given the above?

**6.Your Class** Have each student in class toss 4 coins on her or his desk, and note how many heads are showing. Create a frequency table displaying the results. Compare the frequency table to the theoretical probability distribution for the outcome when 4 coins are tossed. Find the mean for the frequency table. How does it compare with the mean for the probability distribution?

**Answers to Applying the Concepts**

**Section 5–1 Dropping College Courses**

**1.**The random variable under study is the reason for dropping a college course.

**2.**There were a total of 144 people in the study.

**3.**The complete table is as follows:

Reason for dropping a college course | Frequency | Percentage |

Too difficult | 45 | 31.25 |

Illness | 40 | 27.78 |

Change in work schedule | 20 | 13.89 |

Change of major | 14 | 9.72 |

Family-related problems | 9 | 6.25 |

Money | 7 | 4.86 |

Miscellaneous | 6 | 4.17 |

No meaningful reason | 3 | 2.08 |

**4.**The probability that a student will drop a class because of illness is about 28%. The probability that a student will drop a class because of money is about 5%. The probability that a student will drop a class because of a change of major is about 10%.

**5.**The information is not itself a probability distribution, but it can be used as one.

**6.**The categories are not necessarily mutually exclusive, but we treated them as such in computing the probabilities.

**7.**The categories are not independent.

**8.**The categories are exhaustive.

**9.**Since all the probabilities are between 0 and 1, inclusive, and the probabilities sum to 1, the requirements for a discrete probability distribution are met.

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**Section 5–2 Expected Value**

**1.**The expected value is the mean in a discrete probability distribution.

**2.**We would expect variation from the expected value of 3.

**3.**Answers will vary. One possible answer is that pregnant mothers in that area might be overly concerned upon hearing that the number of cases of kidney problems in newborns was nearly 4 times what was usually expected. Other mothers (particularly those who had taken a statistics course!) might ask for more information about the claim.

**4.**Answers will vary. One possible answer is that it does seem unlikely to have 11 newborns with kidney problems when we expect only 3 newborns to have kidney problems.

**5.**The public might better be informed by percentages or rates (e.g., rate per 1000 newborns).

**6.**The increase of 8 babies born with kidney problems represents a 0.32% increase (less than %).

**7.**Answers will vary. One possible answer is that the percentage increase does not seem to be something to be overly concerned about.

**Section 5–3 Unsanitary Restaurants**

**1.**The probability of eating at 3 restaurants with unsanitary conditions out of the 10 restaurants is 0.18651.

**2.**The probability of eating at 4 or 5 restaurants with unsanitary conditions out of the 10 restaurants is 0.24623 + 0.22291 = 0.46914.

**3.**To find this probability, you could add the probabilities for eating at 1, 2, … , 10 unsanitary restaurants. An easier way to compute the probability is to subtract the probability of eating at no unsanitary restaurants from 1 (using the complement rule).

**4.**The highest probability for this distribution is 4, but the expected number of unsanitary restaurants that you would eat at is 10 • = 4.3.

**5.**The standard deviation for this distribution is = 1.56.

**6.**This is a binomial distribution. We have two possible outcomes: “success” is eating in an unsanitary restaurant; “failure” is eating in a sanitary restaurant. The probability that one restaurant is unsanitary is independent of the probability that any other restaurant is unsanitary. The probability that a restaurant is unsanitary remains constant at . And we are looking at the number of unsanitary restaurants that we eat at out of 10 “trials.”

**7.**The likelihood of success will vary from situation to situation. Just because we have two possible outcomes, this does not mean that each outcome occurs with probability 0.50.