Introduction to Probability Models 11th Edition by Sheldon M. Ross Solution manual
Instructor’s Manual
to Accompany
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 = {(e 1 ,e 2 ,...,e n ), n ≥ 2} where e i ∈(heads,tails}. In addition, e n = e n−1 =
heads and for i = 1,...,n − 2 if e i = heads, then e i+1 = tails.
P{4 tosses} = P{(t,t,h,h)} + P{(h,t,h,h)}
= 2
? 1
2
? 4
=
1
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.
3
4 . 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
Introduction to Probability Models, Eleventh Edition. http://dx.doi.org/10.1016/B978-0-12-407948-9.00019-0
© 2014 Elsevier Inc. All rights reserved.
2 Introduction to Probability Models
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 E c
does); F does not occur (and so F c does) and thus E c and F c both occur. Hence,
(E ∪ F) c ⊂ E c F c
If E c F c occurs, then E c occurs (and so E does not), and F c occurs (and so F does
not). Hence, neither E or F occurs and thus (E ∪ F) c does. Thus,
E c F c ⊂ (E ∪ F) c
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 c are disjoint that P(F) = P(E) +
P(FE) c .
10. Either by induction or use
n
∪
1
E i = E 1 ∪ E c
1 E 2 ∪ E
c
1 E
c
2 E 3 ∪ ··· ∪ E
c
1 ··· E
c
n−1 E n
and as each of the terms on the right side are mutually exclusive:
P(∪
i
E i ) = P(E 1 ) + P(E c
1 E 2 ) + P(E
c
1 E
c
2 E 3 ) + ···
+ P(E c
1 ··· E
c
n−1 E n )
≤ P(E 1 ) + P(E 2 ) + ··· + P(E n ) (why?)
11. P{sum is i} =
? i−1
36
, i = 2,...,7
13−i
36
, 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 · 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}
Instructor’s Manual to Accompany 3
Now,
P{win|throw i} = P{i before 7}
=
⎧
⎪
⎨
⎪
⎩
0 i = 2,12
i − 1
5 + 1
i = 3,...,6
1 i = 7,11
13 − i
19 − 1
i = 8,...,10
where above is obtained by using Problems 11 and 12.
P{win} ≈ .49.
14. P{A wins} =
∞
?
n=0
P{A wins on (2n + 1)st toss}
=
∞
?
n=0
(1 − P) 2n P
= P
∞
?
n=0
[(1 − P) 2 ] n
= P
1
1 − (1 − P) 2
=
P
2P − P 2
=
1
2 − P
P{B wins} = 1 − P{A wins}
=
1 − P
2 − P
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 c ) by disjointness
Hence,
P(E ∪ F) = P(E) + P(F) − P(EF)
17. Prob{end} = 1 − Prob{continue}
= 1 − P({H, H, H} ∪ {T,T,T})
= 1 − [Prob(H, H, H) + Prob(T,T,T)].
4 Introduction to Probability Models
Fair coin: Prob{end} = 1 −
? 1
2
·
1
2
·
1
2
+
1
2
·
1
2
·
1
2
?
=
3
4
Biased coin:P{end} = 1 −
? 1
4
·
1
4
·
1
4
+
3
4
·
3
4
·
3
4
?
=
9
16
18. Let B = event both are girls; E = event oldest is girl; L = event at least one is a girl.
(a) P(B|E) =
P(BE)
P(E)
=
to Accompany
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 = {(e 1 ,e 2 ,...,e n ), n ≥ 2} where e i ∈(heads,tails}. In addition, e n = e n−1 =
heads and for i = 1,...,n − 2 if e i = heads, then e i+1 = tails.
P{4 tosses} = P{(t,t,h,h)} + P{(h,t,h,h)}
= 2
? 1
2
? 4
=
1
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.
3
4 . 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
Introduction to Probability Models, Eleventh Edition. http://dx.doi.org/10.1016/B978-0-12-407948-9.00019-0
© 2014 Elsevier Inc. All rights reserved.
2 Introduction to Probability Models
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 E c
does); F does not occur (and so F c does) and thus E c and F c both occur. Hence,
(E ∪ F) c ⊂ E c F c
If E c F c occurs, then E c occurs (and so E does not), and F c occurs (and so F does
not). Hence, neither E or F occurs and thus (E ∪ F) c does. Thus,
E c F c ⊂ (E ∪ F) c
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 c are disjoint that P(F) = P(E) +
P(FE) c .
10. Either by induction or use
n
∪
1
E i = E 1 ∪ E c
1 E 2 ∪ E
c
1 E
c
2 E 3 ∪ ··· ∪ E
c
1 ··· E
c
n−1 E n
and as each of the terms on the right side are mutually exclusive:
P(∪
i
E i ) = P(E 1 ) + P(E c
1 E 2 ) + P(E
c
1 E
c
2 E 3 ) + ···
+ P(E c
1 ··· E
c
n−1 E n )
≤ P(E 1 ) + P(E 2 ) + ··· + P(E n ) (why?)
11. P{sum is i} =
? i−1
36
, i = 2,...,7
13−i
36
, 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 · 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}
Instructor’s Manual to Accompany 3
Now,
P{win|throw i} = P{i before 7}
=
⎧
⎪
⎨
⎪
⎩
0 i = 2,12
i − 1
5 + 1
i = 3,...,6
1 i = 7,11
13 − i
19 − 1
i = 8,...,10
where above is obtained by using Problems 11 and 12.
P{win} ≈ .49.
14. P{A wins} =
∞
?
n=0
P{A wins on (2n + 1)st toss}
=
∞
?
n=0
(1 − P) 2n P
= P
∞
?
n=0
[(1 − P) 2 ] n
= P
1
1 − (1 − P) 2
=
P
2P − P 2
=
1
2 − P
P{B wins} = 1 − P{A wins}
=
1 − P
2 − P
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 c ) by disjointness
Hence,
P(E ∪ F) = P(E) + P(F) − P(EF)
17. Prob{end} = 1 − Prob{continue}
= 1 − P({H, H, H} ∪ {T,T,T})
= 1 − [Prob(H, H, H) + Prob(T,T,T)].
4 Introduction to Probability Models
Fair coin: Prob{end} = 1 −
? 1
2
·
1
2
·
1
2
+
1
2
·
1
2
·
1
2
?
=
3
4
Biased coin:P{end} = 1 −
? 1
4
·
1
4
·
1
4
+
3
4
·
3
4
·
3
4
?
=
9
16
18. Let B = event both are girls; E = event oldest is girl; L = event at least one is a girl.
(a) P(B|E) =
P(BE)
P(E)
=