Introduction to Probability Models 12th edition by Sheldon M. Ross Solution manual

P(L)
=
1/4
3/4
=
1
3
19. E = event at least 1 six P(E)
=
number of ways to get E
number of samples pts
=
11
36
D = event two faces are different P(D)
= 1 − Prob(two faces the same)
= 1 −
6
36
=
5
6
P(E|D) =
P(ED)
P(D)
=
10/36
5/6
=
1
3
20. Let E = event same number on exactly two of the dice; S = event all three numbers
are the same; D = event all three numbers are different. These three events are
mutually exclusive and define the whole sample space. Thus, 1 = P(D)+ P(S)+
P(E), P(S) = 6/216 = 1/36; for D have six possible values for first die, five for
second, and four for third.
∴ Number of ways to get D = 6 · 5 · 4 = 120.
P(D) = 120/216 = 20/36
∴ P(E) = 1 − P(D) − P(S)
= 1 −
20
36
−
1
36
=
5
12
21. Let C = event person is color blind.
P(Male|C) =
P(C|Male)P(Male)
P(C|MaleP(Male) + P(C|Female)P(Female)
=
.05 × .5
.05 × .5 + .0025 × .5
=
2500
2625
=
20
21