CSE21 HW #5 Solution
1. (5 Points) P(A) = 0.5, P(B) = 0.6, P(B|A) = 0.8. Find P(A|B) =?
Solution P(A|B) = P(A∩B)
P(B)=P(B|A)P(A)
P(B)=0.8·0.5
0.6=2
3.
2. (10 Points)
(a) P(A) = 1
4,P(B|A) = 1
3,P(A|B) = 1
2.P(A∪B) =?.
Solution(4 pts) P(A∩B) = P(B|A)P(A) = 1
3·1
4=1
12 .
P(B) = P(A∩B)
P(A|B)=
1
12
1
2
=1
6.
P(A∪B) = P(A) + P(B)−P(A∩B) = 1
4+1
6−1
12 =1
3.
(b) Given that P(A) = p,P(B) = 1 −ε. Prove
p−ε
1−ε≤P(A|B)≤p
1−ε.
Solution(6 pts)
Proof P(A|B) = P(A∩B)
P(B)≤(P(A)
(P(B)=p
1−ε.
P(A|B) = P(A∩B)
P(B)=P(A)+P(B)−P(A∪B)
P(B)=p+1−ε−P(A∩B)
1−ε≥p+1−ε−1
1−ε=p−ε
1−ε.
3. (10 Points) Let Adenote the event that the marble selected is yellow, and let Bdenote the
event that the marble is not black. We want to find P(A|B). By definition P(A|B) = P(A∪B)
P(B).
A∪Bis the event that the marble selected is yellow and not black, therefore A∪B=A.
P(A|B) = P(A∪B)
P(B)=P(A)
P(B).P(A) = 10
15+10+15 =1
4.P(B) = 15+10
15+10+15 =5
8.P(A|B) = 2
5.
4. (10 Points) A fair die is thrown twice, independently. Let xbe the first score and ybe the
second score. Let A={x+y= 10}and B={x > y}. Find: P(A|B) and P(B|A).
Solution P(A) = 3
6×6=1
12 . [There are 3 combinations of xand ywhich satisfy x+y= 10:
(x, y) = (4,6) or (5,5) or (6,4).]
P(B) = 15
36 =5
12 . [There are 15 combinations of xand ywhich satisfy x > y:(x, y) = (6,1) or
(6,2), or (6,3) or (6,4) or (6,5) or (5,1) or (5,2) or (5,3) or (5,4) or (4,1) or (4,2) or (4,3)
or (3,1) or (3,2) or (2,1).]
P(A∩B) = 1
36 . [(x, y) = (6,4).]
P(A|B) = P(A∩B)
P(B)=1
15 .
P(B|A) = P(A∩B)
P(A)=1
3.
5. (15 Points) Two boxes are given as follows:
Box Acontains 8 red balls, 2 white balls, and 2 blue balls.
Box Bcontains 2 red balls and 6 white balls.
A fair die is tossed: if a 3 or 6 appears, a ball is randomly chosen from A; otherwise a ball is
randomly chosen from B.
(a) Find the probability that the ball is:
(i) red, (ii) white, (iii) blue.
Solution(9 pts) Construct the corresponding stochastic tree diagram as in Fig 1. We
seek P(R) (the probability that the ball is red), P(W) (the probability that the ball is
white), P(B) (the probability that the ball is blue).
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