Introduction to Probability Models 11th Edition by Sheldon M. Ross Solution manual
P(B)
P(E)
=
1/4
1/2
=
1
2
(b) P(L) = 1 − P(no girls) = 1 −
1
4
=
3
4 ,
P(B|L) =
P(BL)
P(L)
=
P(B)
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
Instructor’s Manual to Accompany 5
22. Let trial 1 consist of the first two points; trial 2 the next two points, and so on. The
probability that each player wins one point in a trial is 2p(1 − p). Now a total of
2n points are played if the first (a − 1) trials all result in each player winning one
of the points in that trial and the nth trial results in one of the players winning both
points. By independence, we obtain
P{2n points are needed}
= (2p(1 − p)) n−1 (p 2 + (1 − p) 2 ), n ≥ 1
The probability that A wins on trial n is (2p(1 − p)) n−1 p 2 and so
P{A wins} = p 2
∞
?
n=1
(2p(1 − p)) n−1
=
p 2
1 − 2p(1 − p)
23. P(E 1 )P(E 2 |E 1 )P(E 3 |E 1 E 2 )... P(E n |E 1
... E n−1 )
= P(E 1 )
P(E 1 E 2 )
P(E 1 )
P(E 1 E 2 E 3 )
P(E 1 E 2 )
...
P(E 1 ... E n )
P(E 1 ... E n−1 )
= P(E 1 ... E n )
24. Let a signify a vote for A and b one for B.
(a) P 2,1 = P{a,a,b} = 1/3
(b) P 3,1 = P{a,a} = (3/4)(2/3) = 1/2
(c) P
3,2 = P{a,a,a} + P{a,a,b,a}
= (3/5)(2/4)[1/3 + (2/3)(1/2)] = 1/5
(d) P 4,1 = P{a,a} = (4/5)(3/4) = 3/5
(e) P
4,2 = P{a,a,a} + P{a,a,b,a}
= (4/6)(3/5)[2/4 + (2/4)(2/3)] = 1/3
(f) P
4,3 = P{always ahead|a,a}(4/7)(3/6)
= (2/7)[1 − P{a,a,a,b,b,b|a,a}
− P{a,a,b,b|a,a} − P{a,a,b,a,b,b|a,a}]
= (2/7)[1 − (2/5)(3/4)(2/3)(1/2)
−(3/5)(2/4) − (3/5)(2/4)(2/3)(1/2)]
= 1/7
(g) P 5,1 = P{a,a} = (5/6)(4/5) = 2/3
(h) P
5,2 = P{a,a,a} + P{a,a,b,a}
= (5/7)(4/6)[(3/5) + (2/5)(3/4)] = 3/7
By the same reasoning we have
6 Introduction to Probability Models
(i) P 5,3 = 1/4
(j) P 5,4 = 1/9
(k) In all the cases above, P n,m =
n−n
n+n
25. (a) P{pair} = P{second card is same denomination as first}
= 3/51
(b)
P{pair|different suits}
=
P{pair,different suits}
P{different suits}
= P{pair}/P{different suits}
=
3/51
39/51
= 1/13
26. P(E 1 ) =
? 4
1
?? 48
12
??? 52
13
?
=
39.38.37
51.50.49
P(E 2 |E 1 ) =
? 3
1
? ? 36
12
??? 39
13
?
=
26.25
38.37
P(E 3 |E 1 E 2 ) =
? 2
1
? ? 24
12
??? 26
13
?
= 13/25
P(E 4 |E 1 E 2 E 3 ) = 1
P(E 1 E 2 E 3 E 4 ) =
39.26.13
51.50.49
27. P(E 1 ) = 1
P(E 2 |E 1 ) = 39/51, since 12 cards are in the ace of spades pile and 39 are not.
P(E 3 |E 1 E 2 ) = 26/50, since 24 cards are in the piles of the two aces and 26 are in
the other two piles.
P(E 4 |E 1 E 2 E 3 ) = 13/49
So
P{each pile has an ace} = (39/51)(26/50)(13/49)
28. Yes. P(A|B) > P(A) is equivalent to P(AB) > P(A)P(B), which is equivalent
to P(B|A) > P(B).
29. (a) P(E|F) = 0
(b) P(E|F) = P(EF)/P(F) = P(E)/P(F) ≥ P(E) = .6
(c) P(E|F) = P(EF)/P(F) = P(F)/P(F) = 1
30. (a) P{George|exactly 1 hit} =
P{George, not Bill}
P{exactly 1}
=
P{G, not B}
P{G, not B} + P{B, not G)}
=
(.4)(.3)
(.4)(.3) + (.7)(.6)