p-bltzmc10 951-1036-hr 26-11-2008 16:23 page 1015 section...

How many hours of sleep do you typicallyget each night? Table 10.3 indicates that

75 million out of 300 million Americans aregetting six hours of sleep on a typical night.The probability of an American getting sixhours of sleep on a typical night is This fraction can be reduced to orexpressed as 0.25, or 25%.Thus, 25% of Americans get six hours of sleep each night.

We find a probability by dividing one number by another. Probabilities areassigned to an event, such as getting six hours of sleep on a typical night. Eventsthat are certain to occur are assigned probabilities of 1, or 100%. For example, theprobability that a given individual will eventually die is 1. Although Woody Allenwhined, “I don’t want to achieve immortality through my work. I want to achieveit through not dying,” death (and taxes) are always certain. By contrast, if an eventcannot occur, its probability is 0. Regrettably, the probability that Elvis will returnand serenade us with one final reprise of “Don’t Be Cruel” (and we hope we’renot) is 0.

Probabilities of events are expressed as numbers ranging from 0 to 1, or 0% to100%.The closer the probability of a given event is to 1, the more likely it is that theevent will occur.The closer the probability of a given event is to 0, the less likely it isthat the event will occur.

Empirical ProbabilityEmpirical probability applies to situations in which we observe how frequently anevent occurs. We use the following formula to compute the empirical probability ofan event:

14 ,75

300 .

Section 10.7 Probability 1015

Objectives

� Compute empiricalprobability.

� Compute theoreticalprobability.

� Find the probability that anevent will not occur.

� Find the probability of oneevent or a second eventoccurring.

� Find the probability of oneevent and a second eventoccurring.

ProbabilitySec t i on 10.7

Table 10.3 The Hours of Sleep Americans Get on a Typical Night

Hours of Sleep

Number of Americans,in millions

4 or less 12

5 27

6 75

7 90

8 81

9 9

10 or more 6

Total: 300

Source: Discovery Health Media

Computing Empirical ProbabilityThe empirical probability of event denoted by is

P1E2 =

observed number of times E occurstotal number of observed occurrences

.

P1E2,E,

Possible Values for Probabilities

100% or 1

50% or q

0% or 0

Certain

Likely

50-50 Chance

Unlikely

Impossible

� Compute empirical probability.

Empirical Probabilities with Real-World Data

When women turn 40, their gynecologists typically remind them that it is time toundergo mammography screening for breast cancer. The data in Table 10.4 on thenext page are based on 100,000 U.S. women, ages 40 to 50, who participated inmammography screening.

EXAMPLE 1

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a. Use Table 10.4 to find the probability that a woman aged 40 to 50 has breastcancer.

b. Among women without breast cancer, find the probability of a positivemammogram.

c. Among women with positive mammograms, find the probability of not havingbreast cancer.

Solution

a. We begin with the probability that a woman aged 40 to 50 has breast cancer.The probability of having breast cancer is the number of women with breastcancer divided by the total number of women.

The empirical probability that a woman aged 40 to 50 has breast cancer is or 0.008.

b. Now, we find the probability of a positive mammogram among women withoutbreast cancer. Thus, we restrict the data to women without breast cancer:

1125 ,

=

800100,000

=

1125

= 0.008

P1breast cancer2 =

number of women with breast cancertotal number of women

1016 Chapter 10 Sequences, Induction, and Probability

No Breast Cancer

720

80

6944

92,256

Breast Cancer

Positive Mammogram

Negative Mammogram

720 + 6944 = 7664women have positive

mammograms.

720 + 80 = 800women have

breast cancer.

6944 + 92,256 = 99,200women do not have

breast cancer.

80 + 92,256 = 92,336women have negative

mammograms.

Table 10.4 Mammography Screening on 100,000 U.S. Women, Ages 40 to 50

Source: Gerd Gigerenzer, Calculated Risks, Simon and Schuster, 2002

No Breast Cancer

Positive Mammogram 6944

Negative Mammogram 92,256

Within the restricted data, the probability of a positive mammogram is thenumber of women with positive mammograms divided by the total number ofwomen.

P1positive mammogram2 =

number of women with positive mammogramstotal number of women in the restricted data

69446944+92,256

694499,200

= = =0.07

This is the total number ofwomen without breast cancer.

Among women without breast cancer, the empirical probability of a positivemammogram is or 0.07.

c. Now, we find the probability of not having breast cancer among women withpositive mammograms. Thus, we restrict the data to women with positivemammograms:

694499,200 ,

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Within the restricted data, the probability of not having breast cancer is thenumber of women with no breast cancer divided by the total number ofwomen.

P1no breast cancer2 =

number of women with no breast cancertotal number of women in the restricted data

Breast Cancer No Breast Cancer

Positive Mammogram 720 6944

Computing Theoretical ProbabilityIf an event has equally likely outcomes and its sample space has equally likely outcomes, the theoretical probability of event denoted by is

The sum of the theoretical probabilities of all possible outcomes in the samplespace is 1.

P1E2 =

number of outcomes in event Enumber of outcomes in samples space S

=

n1E2

n1S2.

P1E2,E,n1S2Sn1E2E

� Compute theoretical probability.

6944720+6944

69447664

= = ≠0.906

This is the total number ofwomen with positive mammograms.

Among women with positive mammograms, the probability of not havingbreast cancer is or approximately 0.906.

Check Point 1 Use the data in Table 10.4 to solve this exercise. Expressprobabilities as fractions and as decimals rounded to three decimal places.

a. Find the probability that a woman aged 40 to 50 has a positive mammogram.

b. Among women with breast cancer, find the probability of a positivemammogram.

c. Among women with positive mammograms, find the probability of havingbreast cancer.

Theoretical ProbabilityYou toss a coin. Although it is equally likely to land either heads up, denoted by or tails up, denoted by the actual outcome is uncertain. Any occurrence for whichthe outcome is uncertain is called an experiment.Thus, tossing a coin is an example ofan experiment.The set of all possible outcomes of an experiment is the sample spaceof the experiment, denoted by The sample space for the coin-tossing experiment is

We can define an event more formally using these concepts.An event, denoted by is any subcollection, or subset, of a sample space. For example, the subset isthe event of landing tails up when a coin is tossed.

Theoretical probability applies to situations like this, in which the samplespace only contains equally likely outcomes, all of which are known.To calculate thetheoretical probability of an event, we divide the number of outcomes resulting inthe event by the number of outcomes in the sample space.

E = 5T6E,

S={H, T}.

Lands heads up Lands tails up

S.

T,H,

69447664 ,

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How can we use this formula to compute the probability of a coin landing tailsup? We use the following sets:

The probability of a coin landing tails up is

Theoretical probability applies to many games of chance, including rolling dice,lotteries, card games, and roulette. The next example deals with the experiment ofrolling a die. Figure 10.10 illustrates that when a die is rolled, there are six equallylikely outcomes.The sample space can be shown as

Computing Theoretical Probability

A die is rolled. Find the probability of getting a number less than 5.

Solution The sample space of equally likely outcomes is There are six outcomes in the sample space, so

We are interested in the probability of getting a number less than 5. The eventof getting a number less than 5 can be represented by

There are four outcomes in this event, so The probability of rolling a number less than 5 is

Check Point 2 A die is rolled. Find the probability of getting a numbergreater than 4.

Computing Theoretical Probability

Two ordinary six-sided dice are rolled. What is the probability of getting a sum of 8?

Solution Each die has six equally likely outcomes. By the FundamentalCounting Principle, there are or 36, equally likely outcomes in the samplespace. That is, The 36 outcomes are shown below as ordered pairs. Thefive ways of rolling a sum of 8 appear in the green highlighted diagonal.

S = 511, 12, 11, 22, 11, 32, 11, 42,

11, 52, 11, 62, 12, 12, 12, 22,

12, 32, 12, 42, 12, 52, 12, 62,

13, 12, 13, 22, 13, 32, 13, 42,

13, 52, 13, 62, 14, 12, 14, 22,

14, 32, 14, 42, 14, 52, 14, 62,

15, 12, 15, 22, 15, 32, 15, 42,

15, 52, 15, 62, 16, 12, 16, 22,

16, 32, 16, 42, 16, 52, 16, 626

n1S2 = 36.6 # 6,

EXAMPLE 3

P1E2 =

n1E2

n1S2=

46

=

23

.

n1E2 = 4.

E = 51, 2, 3, 46.

n1S2 = 6.S = 51, 2, 3, 4, 5, 66.

EXAMPLE 2

S = 51, 2, 3, 4, 5, 66.

P1E2 =

number of outcomes that result in tails uptotal number of possible outcomes

=

n1E2

n1S2=

12

.

S={H, T}.E={T}

This is the eventof landing tails up.

This is the sample space withall equally likely outcomes.

Second Die

Firs

t Die

(6, 1) (6, 6)(6, 5)(6, 4)(6, 3)(6, 2)

(5, 1) (5, 6)(5, 5)(5, 4)(5, 3)(5, 2)

(4, 1) (4, 6)(4, 5)(4, 4)(4, 3)(4, 2)

(3, 1) (3, 6)(3, 5)(3, 4)(3, 3)(3, 2)

(2, 1) (2, 6)(2, 5)(2, 4)(2, 3)(2, 2)

(1, 1) (1, 6)(1, 5)(1, 4)(1, 3)(1, 2)

Figure 10.10 Outcomes whena die is rolled

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The phrase “getting a sum of 8” describes the event

This event has 5 outcomes, so Thus, the probability of getting a sum of 8 is

Check Point 3 What is the probability of getting a sum of 5 when two six-sideddice are rolled?

Computing Theoretical Probability without Listing an Event and the Sample SpaceIn some situations, we can compute theoretical probability without having to writeout each event and each sample space. For example, suppose you are dealt one cardfrom a standard 52-card deck, illustrated in Figure 10.11. The deck has four suits:Hearts and diamonds are red, and clubs and spades are black. Each suit has 13different face values—A(ace), 2, 3, 4, 5, 6, 7, 8, 9, 10, J(jack), Q(queen), and K(king).Jacks, queens, and kings are called picture cards or face cards.

P1E2 =

n1E2

n1S2=

536

.

n1E2 = 5.

E = 516, 22, 15, 32, 14, 42, 13, 52, 12, 626.

Hearts

Kin

gs

Que

ens

Ace

s

Jack

s

Clubs

DiamondsSuits

Picture cards

SpadesFigure 10.11 A standard 52-card bridge deck

Probability and a Deck of 52 Cards

You are dealt one card from a standard 52-card deck. Find the probability of beingdealt a heart.

Solution Let be the event of being dealt a heart. Because there are 13 heartsin the deck, the event of being dealt a heart can occur in 13 ways. The number ofoutcomes in event is 13: With 52 cards in the deck, the total numberof possible ways of being dealt a single card is 52. The number of outcomes in thesample space is 52: The probability of being dealt a heart is

Check Point 4 If you are dealt one card from a standard 52-card deck, find theprobability of being dealt a king.

If your state has a lottery drawing each week, the probability that someonewill win the top prize is relatively high. If there is no winner this week, it is virtuallycertain that eventually someone will be graced with millions of dollars. So, why areyou so unlucky compared to this undisclosed someone? In Example 5, we providean answer to this question, using the counting principles discussed in Section 10.6.

P1E2 =

n1E2

n1S2=

1352

=

14

.

n1S2 = 52.

n1E2 = 13.E

E

EXAMPLE 4

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Probability and Combinations: Winning the Lottery

Florida’s lottery game, LOTTO, is set up so that each player chooses six differentnumbers from 1 to 53. If the six numbers chosen match the six numbers drawnrandomly, the player wins (or shares) the top cash prize. (As of this writing, the topcash prize has ranged from $7 million to $106.5 million.) With one LOTTO ticket,what is the probability of winning this prize?

Solution Because the order of the six numbers does not matter, this is a situationinvolving combinations. Let be the event of winning the lottery with one ticket.With one LOTTO ticket, there is only one way of winning. Thus,The sample space is the set of all possible six-number combinations. We can use thecombinations formula

to find the total number of possible combinations. We are selecting numbersfrom a collection of numbers.

There are nearly 23 million number combinations possible in LOTTO. If a personbuys one LOTTO ticket, the probability of winning is

The probability of winning the top prize with one LOTTO ticket is or about

1 in 23 million.

Suppose that a person buys 5000 different tickets in Florida’s LOTTO.Because that person has selected 5000 different combinations of the six numbers,the probability of winning is

The chances of winning top prize are about 218 in a million.At $1 per LOTTO ticket,it is highly probable that our LOTTO player will be $5000 poorer. Knowing a littleprobability helps a lotto.

Check Point 5 People lose interest when they do not win at games of chance,including Florida’s LOTTO. With drawings twice weekly instead of once, thegame described in Example 5 was brought in to bring back lost players andincrease ticket sales. The original LOTTO was set up so that each player chose sixdifferent numbers from 1 to 49, rather than from 1 to 53, with a lottery drawingonly once a week. With one LOTTO ticket, what was the probability of winningthe top cash prize in Florida’s original LOTTO? Express the answer as a fractionand as a decimal correct to ten places.

Probability of an Event Not OccurringIf we know the probability of an event we can determine the probability thatthe event will not occur, denoted by Because the sum of the probabilities ofall possible outcomes in any situation is 1,

We now solve this equation for the probability that event will notoccur, by subtracting from both sides.The resulting formula is given in the boxon the next page.

P1E2EP1not E2,

P1E2 + P1not E2 = 1.

P1not E2.E,P1E2,

500022,957,480

L 0.000218.

122,957,480 ,

P1E2 =

n1E2

n1S2=

122,957,480

L 0.0000000436.

53C6 =

53!153 - 62!6!

=

53!47!6!

=

53 # 52 # 51 # 50 # 49 # 48 # 47!

47! # 6 # 5 # 4 # 3 # 2 # 1= 22,957,480

n = 53r = 6

nCr =

n!1n - r2!r!

n1E2 = 1.E

EXAMPLE 5

Comparing theProbability ofDying to theProbability ofWinning Florida’sLOTTO

As a healthy nonsmoking 30-year-old, your probability of dyingthis year is approximately 0.001.Divide this probability by theprobability of winning LOTTOwith one ticket:

A healthy 30-year-old is nearly23,000 times more likely to die thisyear than to win Florida’s lottery.

0.0010.0000000436

L 22,936.

� Find the probability that an eventwill not occur.

State lotteries keep 50 cents on thedollar, resulting in $10 billion a yearfor public funding.© Damon Higgins/The Palm Beach Post

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The Probability of an Event Not OccurringThe probability that an event will not occur is equal to 1 minus the probabilitythat it will occur.

P1not E2 = 1 - P1E2

E

The Probability of an Event Not Occurring

The circle graph in Figure10.12 shows the distribu-tion, by age group, of the191 million car drivers inthe United States, with allnumbers rounded to thenearest million. If onedriver is randomly selectedfrom this population, findthe probability that theperson is not in the 20–29age group. Express theprobability as a simplifiedfraction.

EXAMPLE 6

Figure 10.12 Source: U.S. Census Bureau

≤ 199 million

20–2933 million

30–3940 million

40–4940 million

50–5931 million

60–6919 million

70–7913 million

≥ 806 million

Number of U.S. Car Drivers, by Age Group

� Find the probability of one eventor a second event occurring.

Solution We use the probability that the selected person is in the 20–29 age group tofind the probability that the selected person is not in this age group.

P(not in 20–29 age group)

The probability that a randomly selected driver is not in the 20–29 age group is

Check Point 6 If one driver is randomly selected from the population representedin Figure 10.12, find the probability, expressed as a simplified fraction, that theperson is not in the 50–59 age group.

Or Probabilities with Mutually Exclusive EventsSuppose that you randomly select one card from a deck of 52 cards. Let be the eventof selecting a king and let be the event of selecting a queen. Only one card is selected,so it is impossible to get both a king and a queen. The events of selecting a king and aqueen cannot occur simultaneously. They are called mutually exclusive events. If it isimpossible for any two events, and to occur simultaneously, they are said to bemutually exclusive. If and are mutually exclusive events, the probability that either

or will occur is determined by adding their individual probabilities.BABA

B,A

BA

158191 .

The graph shows 33 milliondrivers in the 20–29 age group.

This number, 191 million drivers,was given, but can be obtained by

adding the numbers in the eight sectors.

33191

=1-

191191

33191

= -158191

=

=1-P(in 20–29 age group)

Or Probabilities with Mutually Exclusive EventsIf and are mutually exclusive events, then

Using set notation, P1A ´ B2 = P1A2 + P1B2.

P1A or B2 = P1A2 + P1B2.

BA

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13 Clubs

13 Spades

13 Diamonds

13 Hearts

Figure 10.13 A deck of 52 cards

A

3

5

7910

2

4

6

8 K

Q

J

K

K

KQ

Q

QJ

J

J

Diamonds

Picture cards

3 diamonds thatare picture cards

Figure 10.15

The Probability of Either of Two Mutually Exclusive Events Occurring

If one card is randomly selected from a deck of cards, what is the probability ofselecting a king or a queen?

Solution We find the probability that either of these mutually exclusive eventswill occur by adding their individual probabilities.

The probability of selecting a king or a queen is

Check Point 7 If you roll a single, six-sided die, what is the probability of gettingeither a 4 or a 5?

Or Probabilities with Events That Are Not Mutually ExclusiveConsider the deck of 52 cards shown in Figure 10.13. Suppose that these cards areshuffled and you randomly select one card from the deck. What is the probability ofselecting a diamond or a picture card (jack, queen, king)? Begin by adding theirindividual probabilities.

However, this sum is not the probability of selecting a diamondor a picture card.The problem is that there are three cards thatare simultaneously diamonds and picture cards, shown inFigure 10.14.The events of selecting a diamond and selecting apicture card are not mutually exclusive. It is possible to select acard that is both a diamond and a picture card.

The situation is illustrated in the diagram in Figure 10.15.Why can’t we find the probability of selecting a diamond or apicture card by adding their individual probabilities? The dia-gram shows that three of the cards, the three diamonds thatare picture cards, get counted twice when we add the individual probabilities. First thethree cards get counted as diamonds and then they get counted as picture cards. Inorder to avoid the error of counting the three cards twice, we need to subtract theprobability of getting a diamond and a picture card, as follows:

Thus, the probability of selecting a diamond or a picture card is In general, if and are events that are not mutually exclusive, the proba-

bility that or will occur is determined by adding their individual probabilitiesand then subtracting the probability that and occur simultaneously.BA

BABA

1126 .

=

1352

+

1252

-

352

=

13 + 12 - 352

=

2252

=

1126

.

= P1diamond2 + P1picture card2 - P1diamond and picture card2

P1diamond or picture card2

352 ,

1352

1252

P(diamond)+P(picture card)= +

There are 13 diamondsin the deck of 52 cards.

There are 12 picture cardsin the deck of 52 cards.

213 .

P1king or queen2 = P1king2 + P1queen2 =

452

+

452

=

852

=

213

EXAMPLE 7

Figure 10.14 Threediamonds are picture cards.

Or Probabilities with Events That Are Not Mutually ExclusiveIf and are not mutually exclusive events, then

Using set notation,

P1A ´ B2 = P1A2 + P1B2 - P1A ¨ B2.

P1A or B2 = P1A2 + P1B2 - P1A and B2.

BA

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1

2

3

45

6

7

8

Figure 10.16 It is equallyprobable that the pointer will landon any one of the eight regions.

Table 10.5 Tax Returns Filed and Audited, by Taxable Income, 2006

< $25,000$25,000–$49,999

$50,000–$99,999 » $100,000 Total

Audit 461,729 191,150 163,711 166,839 983,429

No audit 51,509,900 30,637,782 26,300,262 12,726,963 121,174,907

Total 51,971,629 30,828,932 26,463,973 12,893,802 122,158,336

Source: Internal Revenue Service

An Or Probability with Events That Are Not Mutually Exclusive

Figure 10.16 illustrates a spinner. It is equally probable that the pointer will land onany one of the eight regions, numbered 1 through 8. If the pointer lands on a border-line, spin again. Find the probability that the pointer will stop on an even number ora number greater than 5.

Solution It is possible for the pointer to land on a number that is both even andgreater than 5. Two of the numbers, 6 and 8, are even and greater than 5. Theseevents are not mutually exclusive. The probability of landing on a number that iseven or greater than 5 is calculated as follows:

The probability that the pointer will stop on an even number or a number greaterthan 5 is

Check Point 8 Use Figure 10.16 to find the probability that the pointer willstop on an odd number or a number less than 5.

An Or Probability with Real-World Data

Each year the Internal Revenue Service audits a sample of tax forms to verify theiraccuracy. Table 10.5 shows the number of tax returns filed and audited in 2006 bytaxable income.

EXAMPLE 9

58 .

=

4 + 3 - 28

=

58

.

48

38

= +28

-

Four of the eightnumbers, 2, 4, 6,and 8, are even.

Three of the eightnumbers, 6, 7, and 8,are greater than 5.

Two of the eightnumbers, 6 and 8, are

even and greater than 5.

P¢ even orgreater than 5≤ = P1even2 + P1greater than 52 - P¢ even and

greater than 5≤

EXAMPLE 8

If one person is randomly selected from the population represented in Table 10.5,find the probability that

a. the taxpayer had a taxable income less than $25,000 or was audited.

b. the taxpayer had a taxable income less than $25,000 or at least $100,000.

Express probabilities as decimals rounded to the nearest hundredth.

P-BLTZMC10_951-1036-hr 26-11-2008 16:23 Page 1023

Solutiona. It is possible to select a taxpayer who both earned less than $25,000 and was

audited. Thus, these events are not mutually exclusive.

The probability that a taxpayer had a taxable income less than $25,000 or wasaudited is approximately 0.43.

b. A taxable income of at least $100,000 means $100,000 or more. Thus, it is notpossible to select a taxpayer with both a taxable income of less than $25,000and at least $100,000. These events are mutually exclusive.

The probability that a taxpayer had a taxable income less than $25,000 or atleast $100,000 is approximately 0.53.

Check Point 9 If one person is randomly selected from the populationrepresented in Table 10.5, find the probability, expressed as a decimal rounded tothe nearest hundredth, that

a. the taxpayer had a taxable income of at least $100,000 or was not audited.b. the taxpayer had a taxable income less than $25,000 or between $50,000 and

$99,999, inclusive.

And Probabilities with Independent EventsSuppose that you toss a fair coin two times in succession. The outcome of the firsttoss, heads or tails, does not affect what happens when you toss the coin a secondtime. For example, the occurrence of tails on the first toss does not make tails morelikely or less likely to occur on the second toss. The repeated toss of a coin producesindependent events because the outcome of one toss does not influence the outcomeof others. Two events are independent events if the occurrence of either of them hasno effect on the probability of the other.

=

64,865,431122,158,336

L 0.53

51,971,629

122,158,336=

Of the 122,158,336taxpayers, 51,971,629had taxable incomesless than $25,000.

Of the 122,158,336taxpayers, 12,893,802had taxable incomes

of $100,000 or more.

12,893,802

122,158,336+

= P1less than $25,0002 + P1at least $100,0002

P1less than $25,000 or at least $100,0002

=

52,493,329122,158,336

L 0.43

51,971,629

122,158,336=

Of the 122,158,336taxpayers, 51,971,629had taxable incomesless than $25,000.

983,429

122,158,336+

Of the 122,158,336taxpayers, 983,429

were audited.

461,729

122,158,336-

Of the 122,158,336taxpayers, 461,729

earned less than $25,000and were audited.

= P1less than $25,0002 + P1audited2 - P1less than $25,000 and audited2

P1less than $25,000 or audited2


� Find the probability of one eventand a second event occurring.

Table 10.5 (repeated)

< $25,000$25,000–$49,999

$50,000–$99,999 » $100,000 Total

Audit 461,729 191,150 163,711 166,839 983,429

No audit 51,509,900 30,637,782 26,300,262 12,726,963 121,174,907

Total 51,971,629 30,828,932 26,463,973 12,893,802 122,158,336

Source: Internal Revenue Service

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If two events are independent, we can calculate the probability of the firstoccurring and the second occurring by multiplying their probabilities.

And Probabilities with Independent EventsIf and are independent events, then

P1A and B2 = P1A2 # P1B2.

BA

Independent Events on a Roulette Wheel

Figure 10.17 shows a U.S. roulette wheel that has 38 numbered slots (1 through 36, 0,and 00). Of the 38 compartments, 18 are black, 18 are red, and two are green. A playhas the dealer spin the wheel and a small ball in opposite directions. As the ballslows to a stop, it can land with equal probability on any one of the 38 numberedslots. Find the probability of red occurring on two consecutive plays.

Solution The wheel has 38 equally likely outcomes and 18 are red. Thus, theprobability of red occurring on a play is or The result that occurs on each playis independent of all previous results. Thus,

The probability of red occurring on two consecutive plays is

Some roulette players incorrectly believe that if red occurs on two consecutiveplays, then another color is “due.” Because the events are independent, the outcomesof previous spins have no effect on any other spins.

Check Point 10 Find the probability of green occurring on two consecutiveplays on a roulette wheel.

The and rule for independent events can be extended to cover three or moreevents. Thus, if and are independent events, then

P1A and B and C2 = P1A2 # P1B2 # P1C2.

CA, B,

81361 .

P1red and red2 = P1red2 # P1red2 =

919

#919

=

81361

L 0.224.

919 .18

38 ,

EXAMPLE 10

Figure 10.17 A U.S. roulette wheel

Independent Events in a Family

The picture in the margin shows a family that has had nine girls in a row. Find theprobability of this occurrence.

Solution If two or more events are independent, we can find the probability ofthem all occurring by multiplying their probabilities.The probability of a baby girl is

so the probability of nine girls in a row is used as a factor nine times.

The probability of a run of nine girls in a row is (If another child is born intothe family, this event is independent of the other nine, and the probability of a girlis still )

Check Point 11 Find the probability of a family having four boys in a row.

12 .

1512 .

= a12b

9

=

1512

P1nine girls in a row2 =

12

#12

#12

#12

#12

#12

#12

#12

#12

12

12 ,

EXAMPLE 11

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Exercise Set 10.7Practice and Application ExercisesThe table shows the distribution, by marital status and gender, of the 212.5 million Americans ages 18 or older. Use the data in the table tosolve Exercises 1–10.

Never Married

Male

Female

Total

23.3

51.9

28.6

Widowed

11.3

14.0

2.7

Divorced

12.7

21.7

9.0

Married

62.8

124.9

62.1

Total

110.1

212.5

102.4

Total never married:28.6 + 23.3 = 51.9

Total widowed:2.7 + 11.3 = 14.0

Total adult population:102.4 + 110.1 = 212.5

Total male:28.6 + 62.1 + 2.7 + 9.0 = 102.4

Total female:23.3 + 62.8 + 11.3 + 12.7 = 110.1

Total married:62.1 + 62.8 = 124.9

Total divorced:9.0 + 12.7 = 21.7

Marital Status of the United States Population, Ages 18 or Older, in Millions

If one person is randomly selected from the population described inthe table, find the probability, to the nearest hundredth, that the person

1. is divorced.2. has never been married.3. is female.4. is male.5. is a widowed male.6. is a widowed female.7. Among those who are divorced, find the probability of

selecting a woman.8. Among those who are divorced, find the probability of

selecting a man.9. Among adult men, find the probability of selecting a married

person.10. Among adult women, find the probability of selecting a

married person.

In Exercises 11–16, a die is rolled. Find the probability of getting

11. a 4. 12. a 5.13. an odd number. 14. a number greater than 3.15. a number greater than 4. 16. a number greater than 7.

In Exercises 17–20, you are dealt one card from a standard 52-carddeck. Find the probability of being dealt

17. a queen. 18. a diamond.19. a picture card.20. a card greater than 3 and less than 7.

In Exercises 21–22, a fair coin is tossed two times in succession. Thesample space of equally likely outcomes is Find the probability of getting

21. two heads.22. the same outcome on each toss.

In Exercises 23–24, you select a family with three children. If represents a male child and a female child, the sample space ofequally likely outcomes is

Find the probability of selecting a familywith

23. at least one male child.

24. at least two female children.

FMF, FFM, FFF6.5MMM, MMF, MFM, MFF, FMM,

FM

TT6.5HH, HT, TH,

In Exercises 25–26, a single die is rolled twice. The 36 equally likelyoutcomes are shown as follows:

Second Roll

Fir

st R

oll

(6, 1) (6, 6)(6, 5)(6, 4)(6, 3)(6, 2)

(5, 1) (5, 6)(5, 5)(5, 4)(5, 3)(5, 2)

(4, 1) (4, 6)(4, 5)(4, 4)(4, 3)(4, 2)

(3, 1) (3, 6)(3, 5)(3, 4)(3, 3)(3, 2)

(2, 1) (2, 6)(2, 5)(2, 4)(2, 3)(2, 2)

(1, 1) (1, 6)(1, 5)(1, 4)(1, 3)(1, 2)

Find the probability of getting25. two numbers whose sum is 4.

26. two numbers whose sum is 6.

27. To play the California lottery, a person has to select 6 out of51 numbers, paying $1 for each six-number selection. If youpick six numbers that are the same as the ones drawn by thelottery, you win mountains of money. What is the probabilitythat a person with one combination of six numbers will win?What is the probability of winning if 100 different lotterytickets are purchased?

28. A state lottery is designed so that a player chooses six numbersfrom 1 to 30 on one lottery ticket.What is the probability that aplayer with one lottery ticket will win? What is the probabilityof winning if 100 different lottery tickets are purchased?

Exercises 29–30 involve a deck of 52 cards. If necessary, refer to thepicture of a deck of cards, Figure 10.11 on page 1019.

29. A poker hand consists of five cards.a. Find the total number of possible five-card poker hands.b. A diamond flush is a five-card hand consisting of all

diamonds. Find the number of possible diamond flushes.c. Find the probability of being dealt a diamond flush.

30. If you are dealt 3 cards from a shuffled deck of 52 cards, findthe probability that all 3 cards are picture cards.

Source: U.S. Census Bureau

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Educational Attainment, in Millions, of the UnitedStates Population, Ages 25 and Over

LessThan

4 YearsHigh

School

4 YearsHigh

SchoolOnly

SomeCollege[Lessthan

4 years]

4 YearsCollege

[orMore] Total

Male 14 25 20 23 82

Female 15 31 24 22 92

Total 29 56 44 45 174

Source: U.S. Census Bureau

The table shows the educational attainment of the U.S. population,ages 25 and over. Use the data in the table, expressed in millions, tosolve Exercises 31–36.

1

2

3

45

6

7

8

Find the probability that the pointer will stop on43. an odd number or a number less than 6.

44. an odd number or a number greater than 3.

Use this information to solve Exercises 45–46. The mathematicsdepartment of a college has 8 male professors, 11 female professors,14 male teaching assistants, and 7 female teaching assistants. If aperson is selected at random from the group, find the probabilitythat the selected person is45. a professor or a male.46. a professor or a female.

In Exercises 47–50, a single die is rolled twice. Find the probabilityof rolling47. a 2 the first time and a 3 the second time.

48. a 5 the first time and a 1 the second time.

49. an even number the first time and a number greater than2 the second time.

50. an odd number the first time and a number less than 3 thesecond time.

51. If you toss a fair coin six times, what is the probability ofgetting all heads?

52. If you toss a fair coin seven times, what is the probability ofgetting all tails?

53. The probability that South Florida will be hit by a majorhurricane (category 4 or 5) in any single year is (Source: National Hurricane Center)

a. What is the probability that South Florida will be hit by amajor hurricane two years in a row?

b. What is the probability that South Florida will be hit by amajor hurricane in three consecutive years?

c. What is the probability that South Florida will not be hitby a major hurricane in the next ten years?

d. What is the probability that South Florida will be hit by amajor hurricane at least once in the next ten years?

Writing in Mathematics54. Describe the difference between theoretical probability and

empirical probability.

55. Give an example of an event whose probability must bedetermined empirically rather than theoretically.

56. Write a probability word problem whose answer is one of thefollowing fractions: or or

57. Explain how to find the probability of an event notoccurring. Give an example.

58. What are mutually exclusive events? Give an example of twoevents that are mutually exclusive.

59. Explain how to find or probabilities with mutually exclusiveevents. Give an example.

60. Give an example of two events that are not mutually exclusive.

61. Explain how to find or probabilities with events that are notmutually exclusive. Give an example.

62. Explain how to find and probabilities with independentevents. Give an example.

63. The president of a large company with 10,000 employees isconsidering mandatory cocaine testing for every employee.The test that would be used is 90% accurate, meaning thatit will detect 90% of the cocaine users who are tested,and that 90% of the nonusers will test negative. This also

13 .1

416

116 .

Find the probability, expressed as a simplified fraction, that arandomly selected American, aged 25 or over,

31. has not completed four years (or more) of college.

32. has not completed four years of high school.

33. has completed four years of high school only or less than fouryears of college.

34. has completed less than four years of high school or fouryears of high school only.

35. has completed four years of high school only or is a man.

36. has completed four years of high school only or is a woman.

In Exercises 37–42, you are dealt one card from a 52-card deck.Find the probability that37. you are not dealt a king.

38. you are not dealt a picture card.

39. you are dealt a 2 or a 3.

40. you are dealt a red 7 or a black 8.

41. you are dealt a 7 or a red card.

42. you are dealt a 5 or a black card.

In Exercises 43–44, it is equally probable that the pointer onthe spinner shown will land on any one of the eight regions, num-bered 1 through 8. If the pointer lands on a borderline, spin again.

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means that the test gives 10% false positive. Suppose that1% of the employees actually use cocaine. Find the proba-bility that someone who tests positive for cocaine use is,indeed, a user.

Hint: Find the following probability fraction:

This fraction is given by

What does this probability indicate in terms of the percent-age of employees who test positive who are not actuallyusers? Discuss these numbers in terms of the issue ofmandatory drug testing. Write a paper either in favor of oragainst mandatory drug testing, incorporating the actualpercentage accuracy for such tests.

Critical Thinking ExercisesMake Sense? In Exercises 64–67, determine whether eachstatement makes sense or does not make sense, and explainyour reasoning.

64. The probability that Jill will win the election is 0.7 and theprobability that she will not win is 0.4.

65. Assuming the next U.S. president will be a Democrat or aRepublican, the probability of a Republican president is 0.5.

66. The probability that I will go to graduate school is 1.5.67. When I toss a coin, the probability of getting heads or tails is

1, but the probability of getting heads and tails is 0.68. The target in the figure shown contains four squares. If a dart

thrown at random hits the target, find the probability that itwill land in a yellow region.

90% of 1% of 10,000the number who test positive who actually use

.

cocaine plus the number who test positivewho do not use cocaine

the number of employees who test positiveand are cocaine usersthe number of employees who test positive

.

12 in.

9 in.

6 in.

3 in.

69. Suppose that it is a week in which the cash prize in Florida’sLOTTO is promised to exceed $50 million. If a personpurchases 22,957,480 tickets in LOTTO at $1 per ticket (allpossible combinations), isn’t this a guarantee of winning thelottery? Because the probability in this situation is 1, what’swrong with doing this?

70. Some three-digit numbers, such as 101 and 313, read thesame forward and backward. If you select a number from allthree-digit numbers, find the probability that it will read thesame forward and backward.

71. In a class of 50 students, 29 are Democrats, 11 are businessmajors, and 5 of the business majors are Democrats. If onestudent is randomly selected from the class, find the proba-bility of choosing

Preview ExercisesExercises 75–77 will help you prepare for the material covered inthe first section of the next chapter.

75. Use the table to complete each statement.

a. is undefined because _________________________.

b. If is less than 4 and approaches 4 from the left, thevalues of are getting closer to the integer _____.

c. If is greater than 4 and approaches 4 from the right, thevalues of are getting closer to the integer _____.

76. Graph the function in Exercise 75.

Begin by simplifying the function’s equation. How does yourgraph illustrate the statement in Exercise 75(b) and thestatement in Exercise 75(c)?

77. Graph the compound function:

f1x2 = b2x - 4 if x Z 3-5 if x = 3 .

f1x2 =

x2- 6x + 8x - 4

,

f1x2x

f1x2x

f142

a. a Democrat who is not a business major.b. a student who is neither a Democrat nor a business major.

72. On New Year’s Eve, the probability of a person driving whileintoxicated or having a driving accident is 0.35. If the proba-bility of driving while intoxicated is 0.32 and the probabilityof having a driving accident is 0.09, find the probability of aperson having a driving accident while intoxicated.

73. a. If two people are selected at random, the probability thatthey do not have the same birthday (day and month) is

Explain why this is so. (Ignore leap years and assume 365 days in a year.)

b. If three people are selected at random, find the probabilitythat they all have different birthdays.

c. If three people are selected at random, find the probabilitythat at least two of them have the same birthday.

d. If 20 people are selected at random, find the probabilitythat at least 2 of them have the same birthday.

e. How large a group is needed to give a 0.5 chance of atleast two people having the same birthday?

Group Exercise74. Research and present a group report on state lotteries. Include

answers to some or all of the following questions:Which statesdo not have lotteries? Why not? How much is spent per capitaon lotteries? What are some of the lottery games? What is theprobability of winning top prize in these games? What incomegroups spend the greatest amount of money on lotteries? Ifyour state has a lottery, what does it do with the money itmakes? Is the way the money is spent what was promisedwhen the lottery first began?

365365

# 364365 .

x 3.9 3.99 3.999 4 4.001 4.01 4.1

1.9 1.99 1.999 Undefined 2.001 2.01 2.1f(x) = x2-6x+8x-4

x approaches 4 from the left. x approaches 4 from the right.

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