Introduction to Probability Models (12th Edition) – Sheldon M. Ross – Complete Solution Manual Chapters 1–11 | Probability Models A+ Study Guide

Solution Manual For
Introduction to Probability Models,12th Edition by Sheldon M. Ross
Chapter 1-11
In IntroductionExams serve as a fundamental tool in evaluating a student's understanding of a subject, particularly in fields as diverse as business, law,
and mathematics. These disciplines troductionExams serve as a fundamental tool in evaluating a student's understanding of a subject, particularly in
fields as diverse as business, law, and mathematics. These disciplines


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.
2. S = {(R, G),(R, B), (G, R), (G, B), (B, R), (B, G)}
3. S = {(e1, e 2 ,. .., en), n ≥ 2} where ei ∈(heads, tails}. In addition, en = en—1 =
heads and for i = 1 , . . . , n — 2 if ei = heads, then ei+1 = tails.
P{4 tosses}= P{(t, t, h, h)}+ P{(h, t, h, h)}
1 4 1
=2 =
2 8
4. (a) F(E ∪ G)c = FE c G c
(b) EFG c
(c) E F G
(d) EF EG FG
(e) EFG
(f) (E ∪ F ∪ G)c = E c F c G c
(g) (EF) c (EG) c (FG) c
(h) (EFG) c
5. 34 . If he wins, he only wins $1, while if he loses, he loses $3.
6. 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

,2


Similarly, if EF ∪ EG occurs, then either EF or EG occurs. 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.
7. 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 ⊂ Ec Fc
If Ec Fc occurs, then Ec occurs (and so E does not), and Fc occurs (and so F does
not). Hence, neither E or F occurs and thus (E ∪ F)c does. Thus,
Ec Fc ⊂ (E ∪ F)c
IntroductionExams serve as a fundamental tool in evaluating a student's understanding of a subject, particularly in fields as
diverse as business, law, and mathematics. These disciplines

and the result follows.
8. 1 ≥ P(E ∪ F) = P(E) + P(F) — P(EF)
9. F = E ∪ FE c, implying since E and FE are disjoint that P(F) = P(E) +
c

P(FE) c .
10. Either by induction or use
n
c c c c c
∪ Ei = E1 ∪ E1 E2 ∪ E1 E2 E3 ∪ · · · ∪ E1 · · · En—1 En
1
and as each of the terms on the right side are mutually exclusive:
P( ∪ Ei ) = P(E 1 ) + P(E c E2) + P(E c Ec E3) + · · ·
i 1 1 2
c c
+ P(E 1 · · · En—1 En)
≤ P(E 1 ) + P(E 2 ) + · · · + P(E n ) (why?)

36
11. P{sum is i } =
i = 8 , . . . , 12
13—i
36 ,
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 · P(E) + 0 · P(F) + P{E before F }
= [1 — P(E) — P(F)]
P(E)
Therefore, P{E before F } = P (E)+ P(F)
13. Condition an initial toss
12
P{win}= P{win|throw i }P{throw i }
i =2

, 3


Now,
P{win|throw i } = P{i before 7}
,
10 i = 2, 12
, i—
⎨ 5 1 i = 3,..., 6
,
=, +
13 i
1 i = 7, 11
19 — 1
where above is obtained by using Problems 11 and 12.

P{win}≈ .49.
∞
14. P{ A wins}= P{ A wins on (2n + 1)st toss}
n=0
∞
= (1 — P) P
n=0
∞
=P [(1 — P) ]
n=0
1
=P
1 — (1 — P)2
P
=
2 P P2
1
=
2—P
P{B wins}= 1 — P{ A wins}
1— P
=
2— P
IntroductionExams serve as a fundamental tool in evaluating a student's understanding of a subject, particularly

in fields as diverse as business, law, and mathematics. These disciplines


16. P(E ∪ F) = P(E ∪ FE c )
= P(E) + P(FE )
c


since E and FE c are disjoint. Also,

P(E) = P(FE ∪ FE c )
= P(FE) + P(FE ) by disjointness
c


Hence,

P(E ∪ F) = P(E) + P(F) — P(EF)
17. Prob{end}= 1 — Prob{continue}
= 1 — P({H, H, H } ∪ {T, T, T })

, 4
1 1 1 1 1 1
Fair coin: Prob{end}= 1 —
· · + · ·
2 2 2 2 2 2
3
=
4 1 1 1 3 3 3
Biased coin: P{end}= 1 —
· · + · ·
4 4 4 4 4 4
9
=
16
18. Let B = event both are girls; E = event oldest is girl; L = event at least one is a girl.
P(BE) P(B) 1/4 1
(a) P(B|E) = = = =
P(E) P(E) 2
(b) P(L) = 1 — P(no girls) = 1 1 3,
— =
4 4
P(BL) P(B) 1/4 1
P(B|L) = = = =
P(L) P(L) 3
19. E = event at least 1 six P(E)
number of ways to get E 11
= =
number of samples pts 36
D = event two faces are different P(D)
= 1 — Prob(two faces the same)
6 5 P(E D) 10/36 1
=1— = P(E |D) = = =
36 6 P(D) 5/6 3
IntroductionExams serve as a fundamental tool in evaluating a student's understanding of a subject, particularly in fields as diverse as
business, law, and mathematics. These disciplines

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) = = for D have six possible values for first die, five for
6/216 1/36;
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)
20 1 5
=1— — =
36 36 12
21. Let C = event person is color blind.
P(C|Male)P(Male)
P(Male|C) =
P(C |Male P(Male) + P(C |Female)P(Female)
.05 × .5
=
.05 × .5 + .0025 × .5
2500 20
= =
2625 21