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英文版应用随机过程第九版习题答案.pdf
Instructor’s Manual to Accompany
Introduction to
Probability Models
Ninth Edition
Sheldon M. Ross
University of California
Berkeley, California
AMSTERDAM • BOSTON • HEIDELBERG • LONDON
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ISBN 13: 978-0-12-373875-2
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For information on all Academic Press publications
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Printed in the United States of America
06 07 08 09
10
7
9
8
6
5 4 3 2 1
Contents
Chapter 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1
Chapter 2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7
Chapter 3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17
Chapter 4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 33
Chapter 5 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 43
Chapter 6 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 59
Chapter 7 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 71
Chapter 8 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 81
Chapter 9 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 95
Chapter 10 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 99
Chapter 11 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 105
Chapter 1
1. S = {(R, R), (R, G), (R, B), (G, R), (G, G), (G, B),
(B, R), (B, G), (B, B)}.
The probability of each point in S is 1/9.
7.
2. S = {(R, G), (R, B), (G, R), (G, B), (B, R), (B, G)}.
3. S = {(e1, e2, . . . , en), n ≥ 2} where ei ∈ (heads,
In addition, en = en−1 = heads and for
tails}.
i = 1, . . . , n − 2 if ei = heads, then ei+1 = tails.
P{4 tosses} = P{(t, t, h, h)} + P{(h, t, h, h)}
If (E ∪ F)c occurs, then E ∪ F does not occur, and
so E does not occur (and so Ec does); F does not
occur (and so Fc does) and thus Ec and Fc both
occur. Hence,
(E ∪ F)c ⊂ EcFc.
If EcFc occurs, then Ec occurs (and so E does not),
and Fc occurs (and so F does not). Hence, neither
E or F occur and thus (E ∪ F)c does. Thus,
EcFc ⊂ (E ∪ F)c
.
and the result follows.
8. 1 ≥ P(E ∪ F) = P(E) + P(F) − P(EF).
9. F = E ∪ FEc, implying since E and FEc are disjoint
that P(F) = P(E) + P(FE)c.
10. Either by induction or use
4
= 2
1
2
= 1
8
(a) F(E ∪ G)c = FEcGc.
(b) EFGc.
(c) E ∪ F ∪ G.
(d) EF ∪ EG ∪ FG.
(e) EFG.
(f)
(g) (EF)c(EG)c(FG)c.
(h) (EFG)c.
(E ∪ F ∪ G)c = EcFcGc.
4.
5.
6.
. If he wins, he only wins $1, while if he loses, he
3
4
loses $3.
If E(F ∪ G) occurs, then E occurs and either F or G
occur; therefore, either EF or EG occurs and so
E(F ∪ G) ⊂ EF ∪ EG.
Similarly, if EF ∪ EG occurs, then either EF or EG
occur. Thus, E occurs and either F or G occurs; and
so E(F ∪ G) occurs. Hence,
EF ∪ EG ⊂ E(F ∪ G),
which together with the reverse inequality proves
the result.
· · · Ec
1
E2 ∪ Ec
1
Ec
2
1
n−1
En,
Ei = E1 ∪ Ec
E3 ∪ · · · ∪ Ec
n∪
1
and as each of the terms on the right side are
mutually exclusive:
P(∪
i
Ei) = P(E1) + P(Ec
E2) + P(Ec
1
1
· · · Ec
En)
n−1
+P(Ec
1
≤ P(E1) + P(E2) + · · · + P(En).
E3) + · · ·
(why?)
Ec
2
11. P{sum is i} =
i − 1
,
36
13 − i
36
i = 2, . . . , 7
,
i = 8, . . . , 12.
12. Either use hint or condition on initial outcome as:
P{E before F}
= P{E before F | initial outcome is E}P(E)
+P{E before F | initial outcome is F}P(F)
+P{E before F | initial outcome neither E
or F}[1 − P(E) − P(F)]
1
2
= 1 · P(E) + 0 · P(F) + P{E before F}
= [1 − P(E) − P(F)].
Therefore, P{E before F} =
P(E)
P(E) + P(F) .
13. Condition an initial toss
P{win} =
12∑
i=2
P{win | throw i}P{throw i}.
Now,
P{win| throw i} = P{i before 7}
=
0
i − 1
5 + 1
1
13 − i
19 − 1
i = 2, 12
i = 3, . . . , 6
i = 7, 11
i = 8, . . . , 10,
where above is obtained by using Problems 11
and 12.
P{win} ≈ .49.
= P
P{A wins on (2n + 1)st toss}
14. P{A wins} = ∞∑
n=0
(1 − P)2nP
= ∞∑
n=0
[(1 − P)2]n
∞∑
n=0
1
1 − (1 − P)2
P
2P − P2
2 − P .
= 1 − P
2 − P .
P{B wins} = 1 − P{A wins}
=
= 1
= P
16. P(E ∪ F) = P(E ∪ FEc)
= P(E) + P(FEc)
since E and FEc are disjoint. Also,
P(F) = P(FE ∪ FEc)
= P(FE) + P(FEc) by disjointness.
Hence,
P(E ∪ F) = P(E) + P(F) − P(EF).
Answers and Solutions
17. Prob{end} = 1 − Prob{continue}
= 1 − P({H, H, H} ∪ {T, T, T})
= 1 − [Prob(H, H, H) + Prob(T, T, T)].
· 1
2
+ 1
2
· 1
2
· 1
2
· 1
2
1
2
1
4
· 1
4
· 1
4
+ 3
4
· 3
4
· 3
4
Fair coin: Prob{end} = 1 −
= 3
4
Biased coin: P{end} = 1 −
= 9
16
.
.
18. Let B = event both are girls; E = event oldest is
=
P(B)
P(E)
P(BE)
P(E)
girl; L = event at least one is a girl.
(a) P(B|E) =
= 1/4
1/2
(b) P(L) = 1 − P(no girls) = 1 − 1
4
= 1/4
3/4
P(B|L) =
P(BL)
P(L)
P(B)
P(L)
=
= 1
2
.
,
= 3
4
= 1
3
.
19. E = event at least 1 six P(E)
= number of ways to get E
number of sample pts
= 11
36
D = event two faces are different P(D)
= 1 − Prob(two faces the same)
P(ED)
= 1 − 6
P(D)
36
P(E|D) =
= 5
6
= 10/36
5/6
= 1
3
.
20. Let E = event same number on exactly two of
the dice; S = event all 3 numbers are the same;
D = event all 3 numbers are different. These
3 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 6 possible
values for first die, 5 for second, and 4 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)
= 5
12
= 1 − 20
36
− 1
36
.
21. Let C = event person is color blind.
Answers and Solutions
3
P(Male|C)
=
=
P(C|Male) P(Male)
P(C|Male P(Male) + P(C|Female) P(Female)
.05 × .5
.05 × .5 + .0025 × .5
= 2500
2625
= 20
21
.
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 inde-
pendence, we obtain that
P{2n points are needed}
= (2p(1 − p))n−1(p2 + (1 − p)2),
n ≥ 1.
The probability that A wins on trial n is
(2p(1 − p))n−1 p2 and so
P{A wins} = p2
(2p(1 − p))n−1
∞∑
n=1
=
p2
1 − 2p(1 − p) .
25.
23. P(E
1
)P(E
2
|E1)P(E
3
E2)
P(E
)
1
P(E
)
1
· · · En).
= P(E
= P(E
1
1
|E1E2) · · · P(En|E1 · · · En−1)
· · · P(E1 · · · En)
P(E
1
· · · En−1)
P(E
E2E3)
E2)
1
P(E
1
24. Let a signify a vote for A and b one for B.
(a) P2,1 = P{a, a, b} = 1/3.
(b) P3,1 = P{a, a} = (3/4)(2/3) = 1/2.
(c) P3,2 = P{a, a, a} + P{a, a, b, a}
= (3/5)(2/4)[1/3 + (2/3)(1/2)] = 1/5.
(d) P4,1 = P{a, a} = (4/5)(3/4) = 3/5.
(e) P4,2 = P{a, a, a} + P{a, a, b, a}
= (4/6)(3/5)[2/4 + (2/4)(2/3)] = 1/3.
(f) P4,3 = 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) P5,1 = P{a, a} = (5/6)(4/5) = 2/3.
(h) P5,2 = 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 that
(i) P5,3 = 1/4,
(j) P5,4 = 1/9.
(k) In all the cases above, Pn,m =
(a) P{pair} = P{second card is same
n − n
n + n .
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.
52
13
39
13
26. P(E
P(E
P(E
P(E
P(E
27. P(E
1
2
3
4
1
48
12
36
3
12
1
4
) =
1
|E1) =
|E1E2) =
2
1
|E1E2E3) = 1.
E2E3E4) = 39 · 26 · 13
51 · 50 · 49
) = 1
24
12
.
.
= 39 · 38 · 37
51 · 50 · 49
= 26 · 25
38 · 37
.
= 13/25.
26
13
1
P(E2|E1) = 39/51, since 12 cards are in the ace of
spades pile and 39 are not.
|E1E2) = 26/50, since 24 cards are in the piles
P(E
of the two aces and 26 are in the other two piles.
|E1E2E3) = 13/49.
P(E
3
4
So
P{ each pile has an ace} = (39/51)(26/50)(13/49).
Answers and Solutions
4
28. Yes. P(A|B) > P(A) is equivalent to P(AB) >
P(E) = .6.
P(A)P(B) which is equivalent to P(B|A) > P(B).
(a) P(E|F) = 0.
(b) P(E|F) = P(EF)/P(F) = P(E)/P(F) ≥
(c) P(E|F) = P(EF)/P(F) = P(F)/P(F) = 1.
(a) P{George|exactly 1 hit}
P{George, not Bill}
P{G, not B}
P{exactly 1}
=
P{G, not B} + P{B, not G)}
29.
30.
=
=
(.4)(.3)
(.4)(.3) + (.7)(.6)
= 2/9.
(b) P{G|hit}
= P{G, hit}/P{hit}
= P{G}/P{hit} = .4/[1 − (.3)(.6)]
= 20/41.
31. Let S = event sum of dice is 7; F = event first
die is 6.
P(S) = 1
6
P(FS) = 1
36
= 1/36
1/6
= 1
6
.
P(F|S) =
P(F|S)
P(S)
32. Let Ei = event person i selects own hat.
Thus,
∑
P(Ei1Ei2 · · · Eik)
(n − k)!
i1<···
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