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4 Simulation, Expected Value, Odds, and Conditional Probability

4 Simulation, Expected Value, Odds, and Conditional Probability

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Section 11.4 Simulation, Expected Value, Odds, and Conditional Probability 529

A cereal company has put six types of toy cars in its cereal

boxes, one car per box. If the cars are distributed uniformly,

what is the experimental probability that you will get all six types of cars if you buy

10 boxes?

Problem-Solving Strategy

Do a Simulation



S O L U T I O N Simulate the experiment by using the whole numbers from 1 through

6 to represent the different cars. Use a table of random digits as the random-number

generator (Figure 11.25). (Six numbered slips of paper or chips, drawn at random

from a hat, or a six-sided die would also work. In this example we disregard 0, 7, 8, 9,

since there are six types of toy cars.) Start anywhere in the table. (We started at the

upper left.) Read until 10 numbers from 1 through 6 occur, ignoring 0, 7, 8, and 9.

Record the sequence of numbers [Figure 11.25(b)]. Each such sequence of 10 numbers

is a simulated outcome. (Simulated outcomes are separated by a vertical bar.) Six of

these sequences appear in Figure 11.25(b). Successful outcomes contain 1, 2, 3, 4, 5,

and 6 (corresponding to the six cars) and are marked “yes.” Based on the simulation,

our estimate of the probability is 62 = 0.3. Using a computer to simulate the experiment

yields an estimate of 0.257.





Figure 11.25



The previous example used a simulation to find the probability that you would get

all six prizes if 10 boxes were purchased. For many collectors, a more useful question

would be “How many boxes should I buy in order to get all six prizes?” This question

is addressed in the following example.

A cereal company has put six types of toy cars in its cereal

boxes, one car per box. If the cars are distributed uniformly,

how many boxes should I buy in order to get all six types of cars?

S O L U T I O N To simulate this experiment, let the numbers on a die represent each

of the six different types of cars. We will roll the die until all six numbers have turned

up and then record how many rolls it took to obtain all six numbers. By repeating

this 25 times, we will have the data necessary to determine a reasonable estimate of

the number of boxes that should be purchased. The numbers below represent five of

the 25 trials.



2, 2, 6, 5, 6, 4, 4, 2, 4, 4, 5, 1, 6, 2, 3



→ 15 rolls (boxes)



5, 5, 4, 2, 3, 1, 1, 5, 1, 6



→ 10 rolls (boxes)



2, 5, 5, 4, 4, 5, 6, 3, 2, 2, 6, 3, 4, 4, 2, 1



→ 16 rolls (boxes)



4, 2, 2, 6, 3, 2, 3, 5, 6, 5, 6, 6, 3, 3, 6, 6, 3, 4, 4, 6, 1 → 21 rolls (boxes)

4, 5, 3, 6, 5, 2, 1



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→ 7  rolls (boxes)



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530



Chapter 11 Probability

After repeating this process 20 more times, we computed the average of the rolls

needed to get all six toy cars. Based on this simulation, the average was 15.12 rolls. On

average, a collector should buy 15 to 16 boxes of cereal to get all six types of cars. ■

Another way to phrase the question in the previous example would be to say,

“How many boxes would a collector expect to have to buy in order to get all six types

of toy cars?” This idea of expected outcomes leads us to the next idea.







Check for Understanding: Exercise/Problem Set A #1–6



For Sterling’s weekly lawn-mowing job, his parents have given him

two choices for his method of payment.

Choice 1: He receives $10.

Choice 2: His parents place four $1 bills, one $5 bill, and two $10 bills in a bag and Sterling

draws two bills from the bag.

In the long run, which is the better deal for Sterling? Justify your answer with mathematical

reasoning.



Expected Value

Probability can be used to determine values such as admission to games (with payoffs) and insurance premiums, using the idea of expected value.

A cube has three red faces, two green faces, and a blue face.

A game consists of rolling the cube twice. You pay $2 to

play. If both faces are the same color, you are paid $5 (you win $3). If not, you lose

the $2 it costs to play. Will you win money in the long run?

S O L U T I O N Use a probability tree diagram (Figure 11.26). Let W be the event

7 . Hence 7

that you win. Then W = {RR, GG, BB}, and P (W ) = 12 ¥ 12 + 13 ¥ 13 + 61 ¥ 61 = 18

18

11 (about 61%) of the time you will lose. If

(about 39%) of the time you will win, and 18

you play the game 18 times, you can expect to win 7 times and lose 11 times on average. Hence, your winnings, in dollars, will be 3 × 7 + ( −2 ) × 11 = −1. That is, you can

expect to lose $1 if you play the game 18 times. On the average, you will lose $1/18

per game (about 6¢).





Problem-Solving Strategy

Draw a Diagram



Figure 11.26



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Section 11.4 Simulation, Expected Value, Odds, and Conditional Probability 531

In Example 11.33, the amount in dollars that we expect to “win” on each play

7 + ( −2 ) × 11 = − 1 , called the expected value. Expected value is

of the game is 3 × 18

18

18

defined as follows.



DEFINITION 11.2

Expected Value

Suppose that the outcomes of an experiment are real numbers (values) called

v1, v2 , . . . , vn , and suppose that the outcomes have probabilities p1, p2 , . . . , pn respectively. The FYQFDUFEWBMVF, E , of the experiment is the sum

E = v1 ¥ p1 + v2 ¥ p2 + ⋅ ⋅ ⋅ + vn ¥ pn .



The expected value of an experiment is the average value of the outcomes over

many repetitions. The next example shows how insurance companies use expected

values.

TABLE 11.4

AMOUNT OF CLAIM

(NEAREST $2000)



PROBABILITY



0

$2000

4000

6000

8000

10000



0.80

0.10

0.05

0.03

0.01

0.01



Suppose that an insurance company has broken down yearly

automobile claims for drivers from age 16 through 21, as

shown in Table 11.4. How much should the company charge as its average premium

in order to break even on its costs for claims?

S O L U T I O N Use the notation from the definition for expected value. Let n = 6



(the number of claim categories), and let the values v1, v2 , . . . , vn and the probabilities

p1, p2 , . . . , pn be as listed in Table 11.5.

Thus the expected value, E = 0( 0.80 ) + 2000( 0.10 ) + 4000( 0.05) + 6000( 0.03) +

8000( 0.01) + 10, 000( 0.01) = 760. Since the average claim value is $760, the average

automobile insurance premium should be set at $760 per year for the insurance

company to break even on its claims costs.





TABLE 11.5

v

v1 = 0

v 2 = 2000

v 3 = 4000

v 4 = 6000

v 5 = 8000

v 6 = 10, 000



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p

p1 = 0.80

p2 = 0.10

p3 = 0.05

p 4 = 0.03

p5 = 0.01

p6 = 0.01







Check for Understanding: Exercise/Problem Set A #7–10



Odds

The term odds is used often in the English language in situations ranging from

horse racing to medical research. For example, hepatitis C is a disease of the liver.

If a person is a chronic carrier of the virus, the odds of his developing cirrhosis of

the liver are 1 : 4. Under certain treatments for hepatitis C, the odds of achieving a

substantial decrease in the presence of the virus are 2 : 3. The use of the term odds

may sound familiar, but what do these odds really mean and how are they related

to probability?

When computing the probability of an event occurring, we examine the ratio of

the favorable outcomes compared to the total number of possible outcomes. When

people speak about odds in favor of an event, they are comparing the number of

favorable outcomes of an event to the number of unfavorable outcomes of the event.

This comparison assumes, as we will in this section, that outcomes are equally likely.

According to this description, a person who is a chronic carrier of hepatitis C has

one chance of developing cirrhosis and four chances of not developing it. Similarly,

a person who undergoes a certain treatment has two chances of a substantial benefit

and three chances of not having a substantial benefit.

In general, let E be an event in the sample space S and E be the event complementary to E . Then odds are defined formally as follows.



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532



Chapter 11 Probability



DEFINITION 11.3

Odds for Events with Equally Likely Outcomes

The PEETJOGBWPS of event E are n ( E ) : n ( E ).

The PEETBHBJOTU event E are n ( E ) : n ( E ).



If a six-sided die is tossed, what are the odds in favor of the

following events?

B Getting a 4

CGetting a prime

DGetting a number greater than 0

EGetting a number greater than 6

SOLUTION



B 1 : 5 , since there is one 4 and five other numbers

C3 : 3 = 1 : 1, since there are three primes (2, 3, and 5) and three nonprimes

D6 : 0 since all numbers are favorable to this event

E0 : 6 since no numbers are favorable to this event







Notice that in Example 11.35(c) it is reasonable to allow the second number in the

odds ratio to be zero.

Just like the sets E and E combine to make the entire sample space, the number

of favorable outcomes combines with the number of unfavorable outcomes to yield

the total number of possible outcomes. This connection allows a smooth transition

between odds and probability.

It is possible to determine the odds in favor of an event E directly from its probability. For example, if P ( E ) = 57 , we would expect that, in the long run, E would occur

five out of seven times and not occur two of the seven times. Thus the odds in favor

of E would be 5 : 2. When determining the odds in favor of E , we compare n ( E ) and

n ( E ). Now consider P ( E ) = 57 . and P ( E ) = 72 . If we compare these two probabilities

in the same order, we have P ( E ) : P ( E ) = 57 : 72 = 57 ÷ 72 = 52 = 5 : 2, the odds in favor of

E . The following discussion justifies the latter method of calculating odds. The odds

in favor of E are

n (E )

P (E )

n ( E ) n (S ) P ( E )

=

=

=

n (E ) n (E ) P (E ) 1 − P (E )

n (S )

Thus we can find the odds in favor of an event directly from the probability of an event.



THEOREM 11.4

The odds in favor of the event E are

P ( E ) : [1 − P ( E )]    or    P ( E ) : P ( E ).

The odds against E are

[1 − P ( E )] : P ( E )    or    P ( E ) : P ( E ).



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Section 11.4 Simulation, Expected Value, Odds, and Conditional Probability 533

In fact, this result is used to define odds using probabilities in the case of unequally

likely outcomes as well as equally likely outcomes.

Find the odds in favor of event E , where E has the following

probabilities.

BP ( E ) =



1

2



5

C P ( E ) = 34      DP ( E ) = 13



SOLUTION



( 12 ) = 12 : 12 = 1 : 1

COdds in favor of E = 34 : (1 − 34 ) = 34 : 14 = 3 : 1

5 : 1− 5 = 5 : 8 = 5:8

DOdds in favor of E = 13

( 13 ) 13 13

BOdds in favor of E = 12 : 1 −







Now suppose that you know the odds in favor of an event E . Can the probability

of E be found? The answer is “yes!” For example, if the odds in favor of E are 2 : 3,

this means that the ratio of favorable outcomes to unfavorable outcomes is 2 : 3. Thus

in a sample space with five elements with two outcomes favorable to E and three

unfavorable, P ( E ) = 52 = 2 2+ 3 . In general, we have the following.



THEOREM 11.5

If the odds in favor of E are a : b, then

P (E ) =



a

.

a+b



Find P ( E ) given that the odds in favor of (or against) E are

as follows.

B Odds in favor of E are 3 : 4.

DOdds against E are 7 : 3.



C Odds in favor of E are 9 : 2 .

E Odds against E are 2 : 13.



SOLUTION



3

3

=

3+4 7

3

3

=

DP ( E ) =

7 + 3 10

BP ( E ) =







9

9

=

9 + 2 11

13

13

=

E P ( E ) =

2 + 13 15

C P ( E ) =







Check for Understanding: Exercise/Problem Set A #11–17



Conditional Probability

When drawing cards from a standard deck of 52 cards, we know that the probability

4 = 1 , since there are four aces in the deck. If a second draw is

of drawing an ace is 52

13

made, what is the probability of drawing an ace given that one ace has already been

3 .

drawn? Our sample space is now 51 cards with three aces, so the probability is 51

Even though both probabilities deal with drawing an ace, the results are different

because the second example has an extra condition that reduces the size of the sample

space. When such conditions are added that change the size of the sample space, it is

referred to as conditional probability.



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534



Chapter 11 Probability

When tossing three fair coins, what is the probability of getting two tails given that the first coin came up heads?

S O L U T I O N When tossing three coins, there are eight outcomes: HHH, HHT,

HTH, THH, HTT, THT, TTH, TTT. In this case, however, the condition of the first

coin being a head has been added. This changes the possible outcomes to be {HHH,

HHT, HTH, HTT}. There are only four possible outcomes in this reduced sample

space. The only outcome fitting the description of having two tails in this sample

space is HTT. Thus the conditional probability of getting two tails given that the first

of the three coins is a head is 14 .





Problem-Solving Strategy

Draw a Diagram



In Example 11.38 the original sample space is reduced to those outcomes having the given condition, namely H on the first coin. Let A be the event that exactly

two tails appear among the three coins, and let B be the event the first coin comes

up heads. Then A = {HTT, THT, TTH} and B = {HHH, HHT, HTH, HTT}.

We see that there is only one way for A to occur given that B occurs, namely

1

HTT. (Note that A ∩ B = {HTT}.) Thus the probability of A given B is . The

4

1

notation P ( A | B ) means “the probability of A given B.” So, P ( A | B ) = . Notice that

4

1 1/ 8 P ( A ∩ B )

P( A | B ) = =

=

. That is, P ( A | B ) is the relative frequency of event A

4 4 /8

P (B )

within event B. This suggests the following.



DEFINITION 11.4

Conditional Probability

Let A and B be events in a sample space S where P ( B ) ≠ 0. The DPOEJUJPOBMQSPC

BCJMJUZ that event A occurs, given that event B occurs, denoted P ( A | B ), is

P (A | B ) =



P (A ∩ B )

.

P (B )



A Venn diagram can be used to illustrate the definition of conditional probability. A sample space S of equally likely outcomes is shown in Figure 11.27(a). The

reduced sample space, given that event B occurs, appears in Figure 11.27(b). From

P ( A ∩ B ) 2 /12 2

=

= . From Figure 11.27(b) we see that

Figure 11.27(a) we see that

P (B )

7 /12 7

P (A | B ) =



2

P (A ∩ B )

. Thus P ( A | B ) =

.

7

P (B )



Figure 11.27



The next example illustrates conditional probability in the case of unequally likely

outcomes.



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Section 11.4 Simulation, Expected Value, Odds, and Conditional Probability 535

Suppose a 20-sided die has the following numerals on its

faces: 1, 1, 2, 2, 2, 3, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16.

The die is rolled once and the number on the top face is recorded. Let A be the

event the number is prime, and B be the event the number is odd. Find P ( A | B ) and

P ( B | A).

S O L U T I O N Assuming that the die is balanced, P ( A) =



9

20 , since there are 9 ways

that a prime can appear. Similarly, P ( B ) = 10

,

since

there

are 10 ways that an odd

20

6

number can occur. Also, P ( A ∩ B ) = P ( B ∩ A) = 20 , since an odd prime can appear

in six ways. Thus



P (A | B ) =



P ( A ∩ B ) 6 / 20 3

=

=

P (B )

10 / 20 5



P ( B | A) =



P ( B ∩ A) 6 / 20 2

=

= .

P ( A)

9 / 20 3



and



These results can be checked by reducing the sample space to reflect the given information, then assigning probabilities to the events as they occur as subsets of the

reduced sample space.









Check for Understanding: Exercise/Problem Set A #18–21



©Ron Bagwell



The French naturalist Buffon devised his famous needle problem from

which p may be determined using probability methods. The method

is as follows. Draw a number of parallel lines at a distance of 2 inches

apart. Then drop a needle, whose length is one inch, at random onto

the parallel lines. Buffon showed that the probability that the needle

will touch one of the lines is 1/p . Thus, p can be approximated by

repeatedly tossing the needle onto the parallel lines and dividing the

total number of times that a needle is tossed by the total number of

times that it lands crossing one of the parallel lines.



EXERCISE/PROBLEM SET A

EXERCISES

 A penny gumball machine contains gumballs in eight different colors. Assume that there are a large number of

gumballs equally divided among the eight colors.

  B Estimate how many pennies you will have to use to get

one of each color.

  CCut out eight identical pieces of paper and mark them

with the digits 1–8. Put the pieces of paper in a container. Without looking, draw one piece and record

its number. Replace the piece, mix the pieces up, and

draw again. Repeat this process until all digits have

appeared. Record how many draws it took. Repeat this



c11.indd 535



experiment a total of 10 times and average the number

of draws needed.

 Use the Chapter 11 eManipulative activity Simulation on

our Web site to simulate Problem 1 by doing the following.

 J Click on one each of the numbers 1 through 8.

 JJ Press START and watch until all 8 numbers have drawn at

least once.

 JJJ Press PAUSE and record the number of draws.

 JW Clear the draws and repeat. (Perform at least 20

repetitions.)

 W Average the number of draws needed.



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536



Chapter 11 Probability



 A cloakroom attendant receives five coats from five women

and gets them mixed up. She returns the coats at random.

Follow the following steps to find the experimental probability that at least one woman receives her own coat.

 J Cut out five pieces of paper, all the same size, and label

them A, B, C, D, and E.

 JJ Put the pieces in a container and mix them up.

JJJ Draw the pieces out, one at a time, without replacing

them, and record the order.

 JW Repeat steps 2 and 3 a total of 25 times.

 W Count the number of times at least one letter is in the

appropriate place (A in first place, B in second place, etc.).

From this simulation, what is the approximate probability

that at least one woman receives her own coat?

 You are going to bake a batch of 100 oatmeal cookies.

Because raisins are expensive, you will only put 150 raisins

in the batter and mix the batter well. Follow the steps given

to find the experimental probability that a cookie will end

up without a raisin.

  J Draw a 10 × 10 grid as illustrated. Each cell is represented

by a two-digit number. The first digit is the horizontal

scale and the second digit is the vertical scale. For example, 06 and 73 are shown.



  C If your calculator can generate random numbers,

generate another set of 150 numbers and repeat this

experiment.

  A young couple is planning their family and would like to

have one child of each sex. On average, how many children should they plan for in order to have at least one boy

and one girl.

  B Describe a simulation that could be used to answer the

above question.

  C Perform at least 30 trials of the simulation and record

your results.

  D Repeat parts a and b using the Chapter 11 eManipulative activity Simulation on our Web site.

  Explain how to simulate tossing two coins using the random number table in Exercise 4. Simulate tossing two

coins 25 times and record your results.

  From the data given, compute the expected value of the

outcome.

OUTCOME

PROBABILITY



−2000



0



1000



3000



1

4



1

6



1

4



1

3



  A study of attendance at a football game shows the following pattern. What is the expected value of the attendance?

WEATHER

Extremely cold

Cold

Moderate

Warm



  JJ Given is a portion of a table of random digits. For each

two-digit number in the table, place an × in the appropriate cell of your grid.

15

85

47

13

10

60

57

16

74

59

14

42

73

75

49



77

40

69

26

55

20

62

70

99

34

85

07

47

16

25



01

51

35

87

33

00

94

48

16

71

40

50

16

00

36



64

40

90

40

20

84

04

02

92

55

52

15

49

21

12



69

10

95

20

47

22

99

00

99

84

68

69

79

11

07



69

15

16

40

54

05

65

59

31

91

60

86

69

42

25



58

33

17

81

16

06

50

68

31

59

41

97

80

44

90



40

94

45

46

86

67

89

53

05

46

94

40

76

84

89



81

11

86

08

11

26

18

31

36

44

98

25

16

46

55



16

65

29

09

16

77

74

55

48

45

18

88

60

84

25



  B Tally the number of squares that have no raisin indicated. What is the probability of selecting a cookie without

a raisin?



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ATTENDANCE



WEATHER PROBABILITY



30,000

40,000

52,000

65,000



0.06

0.44

0.35

0.15



  A player rolls a fair die and receives a number of dollars

equal to the number of dots showing on the face of the

die.

  B If the game costs $1 to play, how much should the

player expect to win for each play?

  C If the game costs $2 to play, how much should the

player expect to win per play?

  D What is the most the player should be willing to pay to

play the game and not lose money in the long run?

 For visiting a resort, you will receive one gift. The probabilities and manufacturer’s suggested retail values of each

gift are as follows: gift A, 1 in 52,000 ($9272.00); gift B,

25,736 in 52,000 ($44.95); gift C, 1 in 52,000 ($2500.00);

gift D, 3 in 52,000 ($729.95); gift E, 25,736 in 52,000

($26.99); gift F, 3 in 52,000 ($1000.00); gift G, 180 in

52,000 ($44.99); gift H, 180 in 52,000 ($63.98); gift I, 160

in 52,000 ($25.00). Find the expected value of your gift.

 Which, if either, are more favorable odds, 50 : 50 or 100 : 100?

Explain.

  A die is thrown once.

  B If each face is equally likely to turn up, what is the probability of getting a 5?

  C What are the odds in favor of getting a 5?

  D What are the odds against getting a 5?



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Section 11.4 Simulation, Expected Value, Odds, and Conditional Probability 537

 A card is drawn at random from a standard 52-card deck.

Find the following odds.

  B In favor of drawing the ace of spades

  C Against drawing a 2, 3, or 4



 The spinner is spun once. (All central angles

equal 60°.)



 The spinner is spun once. Find the following odds.



  B In favor of getting a primary color (blue, red, or yellow)

  C Against getting red or green

 In each part, you are given the probability of event E . Find

the odds in favor of event E and the odds against event E .

  B 53

C 14

D 56

 In each part, you are given the following odds in favor of

event E . Find P ( E ).

  B 9 : 1

C 2 : 5

D 12 : 5

 Two fair dice are rolled, and the sum of the dots is

recorded. In each part, give an example of an event

having the given odds in its favor.

  B 1 : 1

C 1 : 5

D 1 : 3

 The diagram shows a sample space S of equally likely

outcomes and events A and B. Find the following probabilities.

  B P ( A)

C P ( B )

D P ( A | B )

E P ( B | A)



  B What is the probability that it lands on 4?

  C If you are told it has landed on an even number, what is

the probability that it landed on 4?

  D If you are told it has landed on an odd number, what is

the probability that it landed on 4?

 A container holds three red balls and five blue balls. One

ball will be drawn and discarded. Then a second ball is

drawn.

  B What is the probability that the second ball drawn is red

if you drew a red ball the first time?

  C What is the probability of drawing a blue ball second if

the first ball was red?

  D What is the probability of drawing a blue ball second if

the first ball was blue?

 A standard six-sided die is tossed. What is the probability

that it shows 2 if you know the following?

  B It shows an even number.

  C It shows a number less than 5.

  D It does not show a 6.

  E It shows 1 or 2.

  F It shows an even number less than 4.

  G It shows a number greater than 3.



PROBLEMS

 Given is the probability tree diagram for an experiment.

The sample space S = {a, b, c, d }. Also, event A = {a, b, c}

and event B = {b, c, d }. Find the following probabilities.

  B P ( A)

C P ( B )

D P ( A ∩ B )

  E P ( A ∪ B )

F P ( A | B )

G P ( B | A)



 Given is a tabulation of academic award winners in a school.



Class 1

Class 2

Class 3

Class 4

Class 5

Class 6

Boys

Girls



NUMBER OF STUDENTS

RECEIVING AWARDS



NUMBER OF

MATH AWARDS



15

16

14

20

19

21

52

53



7

8

9

11

12

14

29

32



A student is chosen at random from the award winners.

Find the probabilities of the following events.

  B The student is in class 1.

  C The student is in class 4, 5, or 6.



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538



Chapter 11 Probability



D The student won a math award.

E The student is a girl.

F The student is a boy who won a math award.

G The student won a math award given that he or she is in

class 1.

  H The student won a math award given that he or she is in

class 1, 2, or 3.

  I The student is a girl, given that he or she won a math

award.

  J The student won a math award, given that she is a girl.



 B There are ten ways the American League can win a

six-game World Series. (There are 10 branches that

contain four A’s and two N’s, where the last one is A.)

If P ( A) = p and P ( N ) = q, what is the probability of the

American League winning the World Series in six games?

(NOTE: q = 1 − p.)

  C There are also ten ways the National League team can

win a six-game series. What is the probability of that

event?

  D What is the probability that the World Series will end at

six games?



 In the World Series, the team that wins four out of seven

games is the winner.

  B Would you agree or disagree with the following statement? The prospects for a long series decrease when the

teams are closely matched.

  C If the probability that the American League team wins

any game is p, what is the probability that it wins the

series in four games?

  D If the probability that the National League team wins

any game is q, what is the probability that it wins the

series in four games? (NOTE: q = 1 − p.)

  E What is the probability that the series ends at four games?

  F Complete the following table for the given odds.



B There are 20 ways each for the American League team

or National League team to win a seven-game World

Series. If P ( A) = p and P ( N ) = q, what is the probability

of the American League winning? (NOTE: q = 1 − p.)

  C What is the probability of the National League winning?

  D What is the probability of the World Series going all

seven games?























ODDS FAVORING AMERICAN LEAGUE



1:1



2:1



3:1



  Summarize the results from Problems 24 to 27. Here

  B P ( A) = p and P ( N ) = q, where p + q = 1.

X = NUMBER OF GAMES



4



5



6



7



P (AMERICAN WINS)

P (NATIONAL WINS)



3:2



P (X GAMES IN SERIES)



p



  C If the odds in favor of the American League are 1 : 1,

complete the following table.



q

P (AMERICAN IN 4 GAMES)

P (NATIONAL IN 4 GAMES)



X = NUMBER OF GAMES



P (4-GAME SERIES)



P (X )



  G What conclusion can you state from this evidence about

the statement in part a?

B In a five-game World Series, there are four ways the

American League could win (NAAAA, ANAAA,

AANAA and AAANA). Here, event A is an American

League win, event N a National League victory. If

P ( A) = p and P ( N ) = q, what is the probability of each

sequence? What is the probability of the American

League winning the series in five games?

  C Similarly, there are four ways the National League

could win (verify this). What is the probability of the

National League winning in five games?

  D What is the probability the series will end at five games?



4



5



6



7



  D Find the expected value for the length of the series.

 A snack company has put five different prizes in its snack

boxes, one per box. Assuming that the same number of each

toy has been used, follow the directions to conduct a simulation that will answer the question “How many boxes of

snacks should you expect to buy in order to get all five toys?”

  B Describe how to use the Chapter 11 eManipulative

activity Simulation on our Web site to perform a simulation of this problem.

  C Perform a simulation of at least 30 trials and record

your results.

 Eight points are evenly spaced around a circle. How many

segments can be formed by joining these points?



EXERCISE/PROBLEM SET B

EXERCISES

  A candy bar company is having a contest. On the inside

of each package, N, U, or T is printed in ratios 3 : 2 : 1. To

determine how many packages you should buy to spell

NUT, perform the following simulation.

  J Using a die, let 1, 2, 3 represent N; let 4, 5 represent U;

and let 6 represent T.



c11.indd 538



  JJ Roll the die and record the corresponding letter.

Repeat rolling the die until each letter has been

obtained.

  JJJ Repeat step 2 a total of 20 times.

Average the number of packages purchased in each case.



7/31/2013 9:06:02 AM



Section 11.4 Simulation, Expected Value, Odds, and Conditional Probability 539

  Use the Chapter 11 eManipulative activity Simulation

on our Web site to simulate Exercise 1 by doing the

following.

  J Click on one 1, two 2s, and three 3s.

  JJ Press START and watch until all three numbers 1, 2, and 3

appear at least once.

  JJJ Press PAUSE and record the number of draws.

  JW Clear the draws and repeat. (Perform at least 20

repetitions.)

  W Average the number of candy bars purchased.

  A family wants to have five children. To determine the

probability that they will have at least four of the same

sex, perform the following simulation.

  J Use five coins, where H = girl and T = boy.

  JJ Toss the five coins and record how they land.

  JJJ Repeat step 2 a total of 30 times.

  JW Count the outcomes that have at least four of the

same sex.

What is the approximate probability of having at least

four of the same sex?

  A bus company overbooks the 22 seats on its bus to the

coast. It regularly sells 25 tickets. Assuming that there is

a 0.1 chance of any passenger not showing up, complete

the following steps to find the probability that at least one

passenger will not have a seat.

  J Let the digit 0 represent not showing up and the digits

1–9 represent showing up. Is P( 0 ) = 0.1?

  JJ Given here is a portion of a random number table. Each

row of 25 numbers represents the 25 tickets sold on a

given day. In the first row, how many passengers did not

show up (how many zeros appear)?



c11.indd 539



07018

52444

72161

17918

13623

27426

96039

68282

54262

66920

53348

34482

99268

95342

38556

39159



31172

65625

57299

75071

76165

97534

21338

98888

21477

27544

39044

42758

98715

97178

60373

04795



12572

97918

87521

91057

43195

89707

88169

25545

33097

72780

04072

40128

07545

10401

77935

51163



23968

46794

44351

46829

50205

97453

69530

69406

48125

91384

62210

48136

27317

31615

64608

84475



55216

62370

99981

47992

75736

90836

53300

29470

92982

47296

01209

30254

52459

95784

28949

60722



41786

95627

98738

75214

73904

33329

66364

68349

19193



18169

30768

15548

61575

89123

08896

94799

16984

99621



96649

30607

42263

27805

19271

94662

62211

86532

66899



92406

89023

79489

21930

15792

05781

37539

96186

12351



42733

60730

85118

94726

72675

59187

80172

53893

72438



  JJJ Count the number of rows that have two or fewer zeros.

They represent days in which someone will not have a seat.



 B From this simulation, what is the probability that at

least one passenger will not have a seat?



 C If your calculator can generate random numbers,

generate another five sets of 25 numbers and repeat

the experiment.

  A bag of chocolate candies has the following colors: 5 yellow, 7 green, 8 red, 5 orange, and 4 brown. What is the

probability of getting two of the same color when pouring

two out of the bag?

  B Describe a simulation that could be used to answer the

preceding question.

  C Perform at least 30 trials of the simulation and record

your results.

 D Repeats parts a and b using the Chapter 11 eManipulative activity Simulation on our Web site.

  Explain how to simulate the rolling of a die using the

random-number table in Exercise 4. Simulate 20 rolls of a

die and record your results.

  From the following data, compute the expected value of

the payoff.

PAYOFF



−2



0



2



3



PROBABILITY



0.3



0.1



0.4



0.2



  A laboratory contains ten electronic microscopes, of

which two are defective. Four microscopes are to be

tested. All microscopes are equally likely to be chosen. A

sample of four microscopes can have zero, one, or two

defective ones, with the probabilities given. What is the

expected number of defective microscopes in the sample?

NUMBER OF DEFECTIVES



0



1



2



PROBABILITY



1

3



8

15



2

15



  A student is considering applying for two scholarships.

Scholarship A is worth $1000 and scholarship B is worth

$5000. Costs involved in applying are $10 for scholarship

A and $25 for scholarship B. The probability of receiving

scholarship A is 0.05 and of scholarship B is 0.01.

  B What is the student’s expected value for applying for

scholarship A?

  C What is the student’s expected value for applying for

scholarship B?

  D If the student can apply for only one scholarship, which

should she apply for?

 According to a publisher’s records, 20% of the books published break even, 30% lose $1000, 25% lose $10,000, and

25% earn $20,000. When a book is published, what is the

expected income for the book?

 Which, if either, are more favorable odds, 60 : 40 or

120 : 80 ? Explain.

  Two dice are thrown.

  B Find the odds in favor of the following events.

 J Getting a sum of 7

JJ Getting a sum greater than 3

JJJ Getting a sum that is an even number

  C Find the odds against each of the events in part a.



7/31/2013 9:06:03 AM



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