 # Bayes`s Rule, Section 5.7 as PDF file

```Section 5.7 Bayes’s Rule 5 - 1
5.7 BAYES’S RULE
Objectives
1 Use the Rule of Total Probability
2 Use Bayes’s Rule to compute probabilities
Now let’s look at probability experiments with sample spaces that will be divided into
two or more mutually exclusive events. This is going to be a continuation of the study
of conditional probability presented in Section 5.4. We will use the following notation:
The symbol x means "and" and
the symbol h means "or".
In Other Words
E ¨ F represents event E and F
E ´ F represents event E or F
Two rules for computing probabilities that will be used in this section are the Addition
Rule for Disjoint Events and the General Multiplication Rule. We restate these rules
using this notation.
If E and F are disjoint (mutually exclusive) events, then
P(E ´ F) = P(E) + P(F)
In general, if E, F, G, . . . are disjoint (mutually exclusive) events, then
P(E ´ F ´ G ´ g ) = P(E) + P(F) + P(G) + g
General Multiplication Rule
The probability that two events E and F both occur is
P(E ¨ F) = P(E) # P(F E)
1 Use the Rule of Total Probability
Let’s begin by looking at the following example.
Example 1
Introduction to the Rule of Total Probability
Problem Suppose that there are two urns. Urn I contains four black and three white
balls. Urn II contains four black, seven white, and two red balls. Suppose that we roll
a fair die. If the die is a one or a two, we randomly select a ball from urn I. However,
if the die is a three, four, five, or six, we randomly select a ball from urn II. What is the
probability that the ball is black?
Approach We define the events as follows:
U1: Urn I is chosen
U2: Urn II is chosen
B: Black ball is chosen
W: White ball is chosen
R: Red ball is chosen
We answer the question using a tree diagram.
Solution Figure 18 shows a tree diagram along with the probabilities corresponding to
each branch.
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5 - 2
Chapter 5 Probability
Figure 18
1
3
2
3
P(B and U1) P(B U1) P(U1) P(B|U1)
13 47
4
7
B
3
7
W
4
13
B
U1
U2
7
13
2
13
4
21
P(B and U2) P(B U2) P(U2) P(B|U2)
W
R
4
2
3 13
8
39
So
P(B) = P[(B ¨ U1) ´ (B ¨ U2)]
= P(B ¨ U1) + P(B ¨ U2)
=
Now Work Problem 29(a)
4
8
36
+
=
= 0.396
21 39 91
There is a 39.6% probability that the ball is black.
Suppose that A1 and A2 are two nonempty sets. In addition, A1 and A2 are disjoint.
Let S be the sample space such that the union of sets A1 and A2 is S. That is,
A1 A2 A1 ¨ A2 = A1 ´ A2 = S
Under these circumstances, we say that A1 and A2 form a partition of S. See Figure 19(a).
For example, urn I and urn II partitioned the sample space into two disjoint events in
Example 1.
Figure 19
A1
A2
S
A1
A2
S
E A1
E A2
E
(a)
(b)
If we define E to be any event in the sample space S, then we can write event E as
E = (E ¨ A1) ´ (E ¨ A2)
Figure 19(b) illustrates this. We can see that the events (E ¨ A1) and (E ¨ A2) are
disjoint, so P((E ¨ A1) ¨ (E ¨ A2)) = 0. Therefore, we can express the probability of
event E using the Addition Rule for Disjoint Events:
P(E) = P(E ¨ A1) + P(E ¨ A2)
Using the General Multiplication Rule, we obtain
P(E) = P(A1) # P(E A1) + P(A2) # P(E A2)
(1)
We can use Formula (1) to find the probability of an event E when the sample space is
partitioned into two sets, A1 and A2. Figure 20 shows a tree diagram that leads directly
Figure 20
P(E|A1)
P(A1)
A1
EC
P(E|A2)
P(A2)
E
P(E) P(A1) P(E|A1) P(A2) P(E|A2)
E
A2
EC
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Section 5.7 Bayes’s Rule 5 - 3
to the result in Formula (1). It is probably better to construct a tree diagram like the one
in Figure 20, rather than memorize Formula (1).
Now let’s redo Example 1 using the results of Formula (1).
Example 2
Using Formula (1)
Problem Redo Example 1 using Formula (1).
Approach We will define the events as we did in Example 1. We want the probability of
selecting a black ball. This can occur by selecting urn I and then selecting a black ball or
by selecting urn II and then selecting a black ball. With B representing the event “black
ball selected,” we have
P(B) = P(B ¨ U1) + P(B ¨ U2)
= P(U1) # P(B U1) + P(U2) # P(B U2)
where
1
3
4
P(B U1) =
7
P(U1) =
2
3
4
P(B U2) =
13
P(U2) =
Solution Substituting, we obtain
P(B) = P(U1) # P(B U1) + P(U2) # P(B U2)
=
Now Work Problem 29(b)
1#4 2# 4
4
8
36
+
=
+
=
= 0.396
3 7 3 13 21 39 91
This result agrees with the result in Example 1.
What if we partition the sample space into three sets, A1, A2, and A3? Then we have
S = A1 ´ A2 ´ A3
A1 ¨ A2 = A1 ¨ A3 = A2 ¨ A3 = A1 A2 A3 Now, some event E can be written
E = (E ¨ A1) ´ (E ¨ A2) ´ (E ¨ A3)
Figure 21 shows the event E.
Figure 21
A1
A2
A3
S
E A1
E A2
E A3
E
Because the events E ¨ A1, E ¨ A2, and E ¨ A3 are all disjoint, the probability of
event E is
P(E) = P(E ¨ A1) + P(E ¨ A2) + P(E ¨ A3)
= P(A1) # P(E A1) + P(A2) # P(E A2) + P(A3) # P(E A3)
(2)
The tree diagram in Figure 22 shows the results of Formula (2).
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5 - 4
Chapter 5 Probability
Figure 22
P(A1)
P(A2)
P(A3)
Example 3
P(E|A1)
E
P(E|A2)
EC
E
A1
A2
A3
P(E|A3)
E
E
P(E) P(A1) P(E|A1) P(A2) P(E|A2)
P(A3) P(E|A3)
C
EC
Using Formula (2)
Problem Suppose that a calculator manufacturer buys integrated circuits from three
different suppliers: supplier I, supplier II, and supplier III. Based on past experience, it
is known that 2% of circuits from supplier I are defective, 1% of circuits from supplier
II are defective, and 1.6% of circuits from supplier III are defective. The calculator
manufacturer buys 20% of its circuits from supplier I, 35% of its circuits from supplier II,
and 45% of its circuits from supplier III. What is the probability that a randomly selected
integrated circuit is found to be defective?
Approach Let D represent the event that the integrated circuit is defective. Let S 1 , S 2 ,
and S 3 represent the events that the circuit came from suppliers I, II, or III, respectively.
We have the following probabilities:
P(S 1) = 0.2
P(S 2) = 0.35
P(S 3) = 0.45
P(D S 1) = 0.02 P(D S 2) = 0.01 P(D S 3) = 0.016
We wish to know the probability that the integrated circuit is defective. Since the sample space
is partitioned into three disjoint events corresponding to suppliers I, II, and III, the probability
of obtaining a defective circuit is the probability that the circuit is defective and comes from
supplier I or defective and comes from supplier II or defective and comes from supplier III.
P(D) = P((D ¨ S1) ´ (D ¨ S2) ´ (D ¨ S3))
These three events are disjoint, so we have
P(D) = P(D ¨ S1) + P(D ¨ S2) + P(D ¨ S3)
Using the General Multiplication Rule, this simplifies to
P(D) = P(S 1) # P(D S 1) + P(S 2) # P(D S 2) + P(S 3) # P(D S 3)
(3)
Solution Substituting into (3), we obtain
P(D) = 0.2(0.02) + 0.35(0.01) + 0.45(0.016)
= 0.0147
Now Work Problems 31(a)–(b)
There is a 1.47% probability that the integrated circuit is defective.
To generalize the results of Formulas (1) and (2), we use the following definition:
Definition
A sample space S is partitioned into n subsets A1 ,A2 , p ,An provided that
1. The n subsets are pairwise disjoint. That is, Ai ¨ Aj = for all i j.
2. A1 ´ A2 ´ g ´ An = S
3. Ai for all i. That is, each subset is nonempty.
Now let S be a sample space and A1 , A2 , p , An be a partition of the sample space.
Let E be any event. Then
E = (E ¨ A1) ´ (E ¨ A2) ´ g ´ (E ¨ An)
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Section 5.7 Bayes’s Rule 5 - 5
Because E ¨ A1 , E ¨ A2 , p , E ¨ An are disjoint, we have
P(E) = P(E ¨ A1) + P(E ¨ A2) + g + P(E ¨ An)
Finally, we replace P(E ¨ A1) with P(A1) # P(E A1) using the General Multiplication Rule.
We do the same with the remaining probabilities. This gives us the Rule of Total Probability.
In Other Words
Rule of Total Probability
The Rule of Total Probability can
be thought of as a weighted
average of the conditional
probabilities, where the P(Ai)
are the weights.
Let E be an event that is a subset of a sample space S. Let A1 , A2 , p , An be a
partition of the sample space S. Then
P(E) = P(A1) # P(E A1) + P(A2) # P(E A2) + g + P(An) # P(E An)
Example 4
Using the Rule of Total Probability
Problem According to the U.S. Census Bureau, 27.3% of U.S. adult women have never
married (single), 49.8% of U.S. adult women are married, and 22.8% of U.S. adult women
are widowed, divorced, or separated (other). Of the single women, 6.7% are unemployed;
of the married women, 2.6% are unemployed; of the “other” women, 4.9% are unemployed.
What is the probability that a randomly selected U.S. adult woman is unemployed?
Approach Define the following events:
U: unemployed
S: single
M: married
O: other
We have the following probabilities:
P(S) = 0.273
P(M) = 0.498
P(O) = 0.228
P(U S) = 0.067 P(U M) = 0.026 P(U O) = 0.049
Then, according to the Rule of Total Probability,
P(U) = P(S) # P(U S) + P(M) # P(U M) + P(O) # P(U O)
Solution
P(U) = 0.273 # 0.067 + 0.498 # 0.026 + 0.228 # 0.049
= 0.042
Now Work Problem 33(a)
The probability that a randomly selected U.S. adult woman is unemployed is
0.042 or 4.2%.
2 Use Bayes’s Rule to Compute Probabilities
Let’s begin with an example.
Example 5
Introduction to Bayes’s Rule
Problem Refer to Examples 1 and 2. Suppose that a black ball was selected. What is
the probability that the black ball came from urn I?
Approach From Example 2, we know that
1
2
P(U2) =
3
3
4
4
36
P(B U1) =
P(B U2) =
P(B) =
7
13
91
We wish to know P(U1 B). Using the Conditional Probability Rule, we write P(U1 B) as
P(U1) =
P(U1 B) =
P(U1 ¨ B)
P(B)
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5 - 6
Chapter 5 Probability
But,
P(U1 ¨ B) = P(U1) # P(B U1) and P(B) = P(U1) # P(B U1) + P(U2) # P(B U2)
So we have
P(U1 B) =
P(U1 ¨ B)
P(U1) # P(B U1)
=
#
P(B)
P(U1) P(B U1) + P(U2) # P(B U2)
(4)
Solution Substituting, we obtain
P(U1 B) =
Now Work Problem 29(c)
P(U1) # P(B U1)
P(B)
1#4
3 7
13
=
=
= 0.481
36
27
91
There is a 48.1% probability that a randomly selected black ball came from urn I.
Equation (4) is a special case of Bayes’s Rule when the sample space is partitioned
into two subsets. The general formula is given next.
Bayes’s Rule
Let A1 , A2 , p , An be a partition of a sample space S. Then, for any event E for
which P(E) 7 0, the probability of event Ai for i = 1, 2, p , n, given the event E, is
P(Ai E) =
P(Ai ¨ E)
P(E)
=
Example 6
P(Ai) # P(E Ai)
P(A1) # P(E A1) + P(A2) # P(E A2) + g + P(An) # P(E An)
Unemployed Women
Problem According to the U.S. Census Bureau, 27.3% of U.S. adult women have
never married (single), 49.8% of U.S. adult women are married, and 22.8% of U.S. adult
women are widowed, divorced, or separated (other). Of the single women, 6.7% are
unemployed; of the married women, 2.6% are unemployed; of the “other” women, 4.9%
are unemployed. Suppose that a randomly selected U.S. adult woman is found to be
unemployed. What is the probability that she is single?
Approach Define the following events:
U:
S:
M:
O:
unemployed
single
married
other
We have the following probabilities:
P(S) = 0.273
P(M) = 0.498
P(O) = 0.228
P(U S) = 0.067 P(U M) = 0.026 P(U O) = 0.049
and, from Example 4, we know that P(U) = 0.042.
We wish to determine the probability that a woman is single, given the knowledge
that she is unemployed. That is, we wish to determine P(S U). We will use Bayes’s Rule
as follows:
P(S U) =
P(S ¨ U) P(S) # P(U S)
=
P(U)
P(U)
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Section 5.7 Bayes’s Rule 5 - 7
Solution
0.273(0.067)
= 0.436
0.042
There is a 43.6% probability that a randomly selected unemployed woman is single.
P(S U) =
Now Work Problem 33(b)
In Other Words
A priori probabilities are
probabilities computed prior
to any knowledge. A posteriori
probabilities are computed after
gaining some knowledge.
Example 7
Historical Note
Thomas Bayes was
born in 1702 in
London, England.
His father was a
Nonconformist
minister. Thomas
followed in his father’s footsteps
and was ordained as well. Bayes’s
rule was presented in his paper
“Essay towards solving a problem
in the doctrine of chances”
published posthumously in the
Philosophical Transactions of the
Royal Society of London in 1764.
Today, Bayesian theory is used
by many Internet search engines
such as Google to help retrieve
information. Bayesian theory is also
heavily used by companies such as
Microsoft and Intel.
We say that all the probabilities P(Ai) are a priori probabilities. These are
probabilities of events prior to any knowledge regarding the event. However, the
probabilities P(Ai E) are a posteriori probabilities because they are probabilities
computed after some knowledge regarding the event. In Example 6, the a priori
probability of a randomly selected woman being single is 0.273. The a posteriori
probability of a woman being single, knowing that she is unemployed, is 0.436. Notice
the information that Bayes’s Rule gives us. Without any knowledge of the employment
status of the woman, there is a 27.3% probability that she is single. But with the
knowledge that the woman is unemployed, the likelihood of her being single increases
to 43.6%.
Let’s do one more example to illustrate Bayes’s Rule.
Work Disability
Problem Persons are classified as work disabled if they have a health problem that
prevents them from working in the type of work they can do. Table 12 contains the
proportion, by age, of Americans who are 16 years of age or older who are work
disabled.
Table 12
Age
Event
Proportion Work Disabled
16–24
A1
0.064
25–34
A2
0.097
35–44
A3
0.151
45–54
A4
0.229
55 and older
A5
0.459
Source: U.S. Census Bureau
If we let M represent the event that a randomly selected American who is 16 years of
age or older is male, the Census Bureau also gives us the following probabilities:
P(male 16924) = P(M A1) = 0.515 P(male 25934) = P(M A2) = 0.509
P(male 35944) = P(M A3) = 0.500 P(male 45954) = P(M A4) = 0.492
P(male 55 and older) = P(M A5) = 0.449
(a) If a work-disabled American aged 16 years of age or older is randomly selected,
what is the probability that the American is male?
(b) If a work-disabled American randomly selected is male, what is the probability that
he is 25 to 34 years of age?
Approach
(a) We will use the Rule of Total Probability to compute P(M) as follows:
P(M) = P(A1) # P(M A1) + P(A2) # P(M A2) + P(A3) # P(M A3)
+ P(A4) # P(M A4) + P(A5) # P(M A5)
(b) We use Bayes’s Rule to compute P(25934 male) as follows:
P(A2 M) =
P(A2) # P(M A2)
P(M)
where P(M) is found from part (a).
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5 - 8
Chapter 5 Probability
Solution
(a) P(M) = P(A1) # P(M A1) + P(A2) # P(M A2) + P(A3) # P(M A3)
+ P(A4) # P(M A4) + P(A5) # P(M A5)
= (0.064)(0.515) + (0.097)(0.509) + (0.151)(0.500)
+ (0.229)(0.492) + (0.459)(0.449)
= 0.477
There is a 47.7% probability that a randomly selected work-disabled American is male.
(b)
P(A2) # P(M A2)
(0.097)(0.509)
=
= 0.104
P(M)
0.477
There is a 10.4% probability that a randomly selected work-disabled American who is
male is 25 to 34 years of age.
P(A2 M) =
Notice that the a priori probability (0.097) and the a posteriori probability (0.104) do not
differ much. This means that the knowledge that the individual is male does not yield
much information regarding the age of the work-disabled individual.
Now Work Problem 35
Vocabulary and skill building
1. If E and F are two events that are not disjoint, draw a Venn
diagram that represents E ¨ F.
2. If E and F are two events that are not disjoint, draw a Venn
diagram that represents E ´ F.
3. Suppose that S is a sample space. What would it mean to
partition S into three disjoint subsets A1, A2, and A3? Draw a
figure to help support your explanation.
4. What must be true regarding the sum of the probability of
events that make up the partitions of the sample space?
5. Describe the Rule of Total Probability in your own words.
6. What is the difference between a priori and a posteriori
probabilities? Which is Bayes’s Rule used for?
In Problems 7–20, find the indicated probabilities by referring to
the given tree diagram and by using Bayes’s Rule.
P(E |A) 0.6
A
P(A) 0.2
P (B) 0.55
P(C) 0.25
B
C
EC
P(EC | A) 0.4
P(E |B) 0.7 E
EC
P(EC | B) 0.3
P(E |C) 0.4 E
P(EC | C) 0.6
7. P(E A)
10. P(E c B)
13. P(E)
16. P(A E c)
19. P(B E)
E
8. P(E B)
11. P(E C)
14. P(E c)
17. P(C E)
20. P(C E c)
EC
9. P(E c A)
12. P(E c C)
15. P(A E)
18. P(B E c)
21. Suppose that events A1 and A2 form a partition of the sample
space S with P(A1) = 0.55 and P(A2) = 0.45. If E is an event
that is a subset of S and P(E A1) = 0.06 and P(E A2) = 0.08,
find P(E).
22. Suppose that events A1 and A2 form a partition of the sample
space S with P(A1) = 0.35 and P(A2) = 0.65. If E is an event
that is a subset of S and P(E A1) = 0.12 and P(E A2) = 0.09,
find P(E).
23. Suppose that events A1, A2, and A3 form a partition
of the sample space S with P(A1) = 0.35, P(A2) = 0.45,
and P(A3) = 0.2. If E is an event that is a subset of S and
P(E A1) = 0.25, P(E A2) = 0.18, and P(E A3) = 0.14, find
P(E).
24. Suppose that events A1, A2, and A3 form a partition
of the sample space S with P(A1) = 0.3, P(A2) = 0.65,
and P(A3) = 0.05. If E is an event that is a subset of S and
P(E A1) = 0.05, P(E A2) = 0.25, and P(E A3) = 0.5, find
P(E).
25. Use the information given in Problem 21 to find:
(a) P(A1 E)
(b) P(A2 E)
26. Use the information given in Problem 22 to find:
(a) P(A1 E)
(b) P(A2 E)
27. Use the information given in Problem 23 to find:
(a) P(A1 E) (b) P(A2 E) (c) P(A3 E)
28. Use the information given in Problem 24 to find:
(a) P(A1 E) (b) P(A2 E)
(c) P(A3 E)
29. Urn I contains five black and seven white balls. Urn II
contains six black, five white, and three red balls. We roll a fair
die. If the die is a one, three or five, we randomly select a ball from
urn I. However, if the die is a two, four, or six, we randomly select
a ball from urn II.
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Section 5.7 Bayes’s Rule 5 - 9
(a)Determine the probability that the ball is black by drawing a
tree diagram.
(b)Determine the probability that the ball is black using the
Rule of Total Probability.
(c)Suppose that a randomly chosen ball is black. What is the
probability that the ball came from urn I?
(d)Suppose that a randomly chosen ball is red. What is the
probability that the ball came from urn I? urn II?
30. Urn I contains eleven black and nine white balls and urn II
contains three black, eight white, and six red balls. We roll a fair
die. If the die is a one, three, or five, we randomly select a ball
from urn I. However, if the die is a two, four, or six, we randomly
select a ball from urn II.
(a)Determine the probability that the ball is black by drawing a
tree diagram.
(b)Determine the probability that the ball is black using the
Rule of Total Probability.
(c)If a randomly chosen ball is black, what is the probability that
the ball came from urn I?
(d)If a randomly chosen ball is red, what is the probability that
the ball came from urn I? urn II?
31. Urn I contains six black and nine white balls; urn II contains
five black, twelve white, and nine red balls; and urn III contains
eleven black, six white, and twelve red balls. We roll a fair die. If
the die is a one, three, or five, we randomly select a ball from urn
I. If the die is a two or four, we randomly select a ball from urn II.
If the die is a six, we randomly select a ball from urn III.
(a)Determine the probability that the ball is black by drawing a
tree diagram.
(b)Determine the probability that the ball is black using the
Rule of Total Probability.
(c)If a randomly chosen ball is black, what is the probability that
the ball came from urn I?
(d)If a randomly chosen ball is red, what is the probability that
the ball came from urn I? urn II? urn III?
32. Urn I contains seven black and three white balls; urn II
contains twelve black, five white, and six red balls; and urn III
contains one black, eight white, and four red balls. We roll a fair
die. If the die is a one, we randomly select a ball from urn I. If the
die is a two, three, or four, we randomly select a ball from urn II.
If the die is a five or six, we randomly select a ball from urn III.
(a)Determine the probability that the ball is black by drawing a
tree diagram.
(b)Determine the probability that the ball is black using the
Rule of Total Probability.
(c)If a randomly chosen ball is black, what is the probability that
the ball came from urn I?
(d)If a randomly chosen ball is red, what is the probability that
the ball came from urn I? urn II? urn III?
34. Elisa Test The standard test for the HIV virus is the Elisa test,
which tests for the presence of HIV antibodies. If an individual
does not have the HIV virus, the test will come back negative
for the presence of HIV antibodies 99.8% of the time and will
come back positive for the presence of HIV antibodies 0.2% of
the time (a false positive). If an individual has the HIV virus, the
test will come back positive 99.8% of the time and will come back
negative 0.2% of the time (a false negative). Approximately 0.7%
of the world population has the HIV virus.
(a)What is the probability that a randomly selected individual
has a test that comes back positive?
(b)What is the probability that a randomly selected individual
has the HIV virus if the test comes back positive?
35. Educational Attainment The data in the following table
represent the proportion of Americans 25 years of age or older at
various levels of educational attainment in 2008.
Level
Event
Proportion
A1
0.134
A2
0.312
Some college, no degree
A3
0.172
Associate’s degree
A4
0.088
Bachelor’s degree
A5
0.191
A6
0.103
Source: U.S. Census Bureau
If we let M represent the event that a randomly selected American
who is 25 years of age or older is male, we can also obtain the
following probabilities from the Census Bureau data.
P(M A1) = 0.505
P(M A2) = 0.482
P(M A3) = 0.467
P(M A4) = 0.433
P(M A5) = 0.480
P(M A6) = 0.513
(a)What is the probability that a randomly selected American
25 years of age or older is male?
(b)What is the probability that an American male 25 years of
age or older has an advanced degree?
(c)What is the probability that an American male 25 years of
age or older is not a high school graduate?
36. Educational Attainment Refer to Problem 35. If we let
E represent the event that a randomly selected American who
is 25 years of age or older is employed, we can also obtain the
following probabilities from the Census Bureau data.
Applying the Concepts
P(E A1) = 0.412
P(E A2) = 0.586
P(E A3) = 0.659
33. Color Blindness The most common form of color blindness is
red–green color blindness. People with this type of color blindness
cannot distinguish between green and red. Approximately 8% of all
males have red–green color blindness, while only about 0.64% of
women have red–green color blindness. According to the U.S. Census
Bureau, 49.1% of all Americans are male and 50.9% are female.
(a)What is the probability that a randomly selected American is
color blind?
(b)What is the probability that a randomly selected American
who is color blind is female?
P(E A4) = 0.737
P(E A5) = 0.757
P(E A6) = 0.776
(a)What is the probability that a randomly selected American
25 years of age or older is employed?
(b)What is the probability that an employed American 25 years
of age or older has a bachelor’s degree?
(c)What is the probability that an employed American 25 years
of age or older is not a high school graduate?
7804_Ch05CD_pp01-10.indd 9
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5 - 10
Chapter 5 Probability
37. Voting Pattern The following data represent the proportion
of Americans who voted in the 2008 presidential election at
various levels of educational attainment.
Level
Event
Proportion
A1
0.069
A2
0.590
A3
0.341
Source: Statistical Abstract of the United States, 2010
If we let D represent the event that a randomly selected American
who voted in the 2008 presidential election voted Democratic, we
can also obtain the following probabilities:
P(D A1) = 0.720
P(D A2) = 0.570
P(D A3) = 0.510
(a)What is the probability that a randomly selected American
who voted in the 2008 presidential election voted
Democratic? (b)What is the probability that an American who voted
(c)What is the probability that an American who voted
Democratic has a grade school education?
38. Murder Victims The following data represent the proportion
of murder victims at various age levels in 2007.
Level
Event
Proportion
Less than 17 years
A1
0.080
17–29
A2
0.431
30–44
A3
0.271
45–59
A4
0.142
At least 60 years
A5
0.076
If we let M represent the event that a randomly selected murder
victim was male, we can also obtain the following probabilities
from the FBI data.
P(M A1) = 0.638
P(M A4) = 0.734
P(M A2) = 0.860
P(M A5) = 0.630
P(M A3) = 0.771
(a)What is the probability that a randomly selected murder
victim was male? (b)What is the probability that a randomly selected male murder
victim was 17 to 29 years of age?
(c)What is the probability that a randomly selected male murder
victim was less than 17 years of age?
39. Espionage The CIA suspects that one of its operatives is a
double agent. Past experience indicates that 95% of all operatives
suspected of espionage are, in fact, guilty. The CIA decides to
administer a polygraph to the suspected spy. It is known that if
a person is guilty, the polygraph returns a result that indicates
the person is guilty 90% of the time. If a person is innocent, the
polygraph returns a result that indicates the person is innocent 99%
of the time. What is the probability that this particular suspect is
innocent, given that the polygraph indicates that he is guilty?
Source: Federal Bureau of Investigation
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31. (a)
1.
S
U1
1
2
EF
E
1
3
F
U3
A2
S
7. P(E A) = 0.6
11. P(E C) = 0.4
15. P(A E) = 0.198
19. P(B E) = 0.636
23. P(E) = 0.1965
25. (a) P(A1 E) = 0.4783
27. (a) P(A1 E) = 0.4453
(c) P(A3 E) = 0.1425
29. (a)
1
2
1
2
U1
9.
13.
17.
21.
P(Ec A) = 0.4
P(E) = 0.605
P(C E) = 0.165
P(E) = 0.069
W
B
12
26
9
26
11
29
W
R
B
6
29
W
R
617
= 0.3273
1885
(b) P(B) = P(U1) # P(B U1) + P(U2) # P(B U2) + P(U3) #
P(B) =
617
= 0.3273
1885
(c) P(U1 B) = 0.6110
(d) P(U1 R) = 0;P(U2 R) = 0.626;P(U3 R) = 0.374
33. (a) P(colorblind) = 0.0425
(b) P(female colorblind) = 0.0766
35. (a) P(M) = 0.4810
(b) P(A6 M) = 0.1099
(d) P(A1 M) = 0.1407
37. (a) P(D) = 0.560
(b) P(A3 D) = 0.311
(c) P(A1 D) = 0.089
39. P(innocent testispositive) = 0.00058
A3
9
15
5
26
12
29
A1
B
U2
1
6
3. To partition a sample space S into three subsets. A1, A2, and A3
means that S is separated into three nonempty subsets, where the subsets
are all pairwise disjoint and A1 ´ A2 ´ A3 = S.
6
15
P(B U3) =
(b) P(A2 E) = 0.5217
(b) P(A2 E) = 0.4122
5
12
Black
7
12
White
6
14
U2
Black
5
14
3
14
White
Red
71
= 0.4226
168
(b) P(B) = P(U1) # P(B U1) + P(U2) # P(B U2) = 0.4226
(c) P(U1 B) = 0.4930
(d) P(U1 R) = 0;P(U2 R) = 1
P(B) =
AN5-1
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``` # red is , and the probability that the second marble . Therefore, # Assignment 5 cover sheet Student name: Student number: Tutor name: 