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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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