Suggested Homework
Section 6.4
September 24, 2007
The answers to the odd questions are in the back of the book.
6.4 #6: Given the probability tree on p.306, find:
(a) Pr(I|B).
(b) Pr(II |B)
(c) Pr(III|B)
Solution. This is an application of Bayes’ Theorem. First, notice that:
Pr(B) =Pr(I)Pr(B|I)+P r(II)P r(B—II)+Pr(III)Pr(B|III)
= (.6)(.8) + (.3)(.3) + (.1)(.4) = .48 + .09 + .04 = .61.
(a) So, now we can apply Bayes’ to get:
Pr(I|B) = Pr(I)Pr(B|I)
Pr(B)=(.6)(.8)
.61 =.48
.61 =.787.
(b) Again, we can apply Bayes’ to get:
Pr(II |B) = Pr(II)Pr(B|I I)
Pr(B)=(.3)(.3)
.61 =.09
.61 =.148.
(c) Again we can apply Bayes’ to get:
Pr(III|B) = Pr(III)Pr(B|III)
Pr(B)=(.1)(.4)
.61 =.04
.61 =.065.
Or, we could have noticed that Pr(III|B) = 1−Pr(II|B)−Pr(I|B).
6.4 #8: Given the probability tree on p.306, find:
(a) Pr(Y|a).
(b) Pr(Y|b)
(c) Pr(Y|c)
Solution. This is an application of Bayes’ Theorem. First, notice that:
Pr(a) =Pr(X)Pr(a|X)+Pr(Y)Pr(a|Y) = (.7)(.5) + (.3)(.3) = .35 + .09 = .44.
Pr(b) =Pr(X)Pr(b|X)+P r(Y)P r(b—Y) =(.7)(.1)+(.3)(.6)=.07+.18=.25.
P r(c)=P r(X)P r(c—X)+Pr(Y)Pr(c|Y) = (.7)(.4) + (.3)(.1) = .28 + .03 = .31.
(a) So, now we can apply Bayes’ to get:
Pr(Y|a) = Pr(Y)Pr(Y|a)
Pr(a)=(.3)(.5)
.44 =.15
.44 =.340.
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