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Instructor’s Manual to Accompany Introduction to Probability Models Tenth Edition Sheldon M. Ross

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Instructor’s Manual to Accompany Introduction to Probability Models Tenth Edition Sheldon M. Ross

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Instructor’s Manual to Accompany

Introduction to
Probability Models
Tenth Edition




Sheldon M. Ross
University of Southern California
Los Angeles, CA




AMSTERDAM • BOSTON • HEIDELBERG • LONDON
NEW YORK • OXFORD • PARIS • SAN DIEGO
SAN FRANCISCO • SINGAPORE • SYDNEY • TOKYO
Academic Press is an imprint of Elsevier

,Academic Press is an imprint of Elsevier
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Copyright 
c 2010 Elsevier Inc. All rights reserved.


No part of this publication may be reproduced or transmitted in any form or by any means, electronic or
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Center and the Copyright Licensing Agency, can be found at our website: www.elsevier.com/permissions.

This book and the individual contributions contained in it are protected under copyright by the Publisher
(other than as may be noted herein).

Notices
Knowledge and best practice in this field are constantly changing. As new research and experience broaden our
understanding, changes in research methods, professional practices, or medical treatment may become necessary.

Practitioners and researchers must always rely on their own experience and knowledge in evaluating and
using any information, methods, compounds, or experiments described herein. In using such information or
methods they should be mindful of their own safety and the safety of others, including parties for whom they
have a professional responsibility.

To the fullest extent of the law, neither the Publisher nor the authors, contributors, or editors, assume any liability
for any injury and/or damage to persons or property as a matter of products liability, negligence or otherwise, or
from any use or operation of any methods, products, instructions, or ideas contained in the material herein.

ISBN: 978-0-12-381445-6


For information on all Academic Press publications
visit our Web site at www.elsevierdirect.com


Typeset by: diacriTech, India

09 10 9 8 7 6 5 4 3 2 1

, Contents

Chapter 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4
Chapter 2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10
Chapter 3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20
Chapter 4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 36
Chapter 5 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 47
Chapter 6 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 62
Chapter 7 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 73
Chapter 8 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 83
Chapter 9 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 96
Chapter 10 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 100
Chapter 11 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 106

, Chapter 1

1. S = {(R, R), (R, G), (R, B), (G, R), (G, G), (G, B), 7. If (E ∪ F)c occurs, then E ∪ F does not occur, and so
(B, R), (B, G), (B, B)} E does not occur (and so Ec does); F does not occur
The probability of each point in S is 1/9. (and so Fc does) and thus Ec and Fc both occur.
Hence,
2. S = {(R, G), (R, B), (G, R), (G, B), (B, R), (B, G)} (E ∪ F)c ⊂ Ec Fc

3. S = {(e1 , e2 , …, en ), n ≥ 2} where ei ∈ (heads, tails}. If Ec Fc occurs, then Ec occurs (and so E does not),
In addition, en = en−1 = heads and for i = 1, …, n − and Fc occurs (and so F does not). Hence, neither E
2 if ei = heads, then ei+1 = tails. or F occurs and thus (E ∪ F)c does. Thus,

Ec Fc ⊂ (E ∪ F)c
P{4 tosses} = P{(t, t, h, h)} + P{(h, t, h, h)}
 4 and the result follows.
1 1
=2 =
2 8
8. 1 ≥ P(E ∪ F) = P(E) + P(F) − P(EF)

4. (a) F(E ∪ G)c = FEc Gc 9. F = E ∪ FEc , implying since E and FEc are disjoint
(b) EFG c that P(F) = P(E) + P(FE)c .
(c) E ∪ F ∪ G 10. Either by induction or use
(d) EF ∪ EG ∪ FG n
∪ Ei = E1 ∪ Ec1 E2 ∪ Ec1 Ec2 E3 ∪ · · · ∪ Ec1 · · · Ecn−1 En
(e) EFG 1
c c c c
(f) (E ∪ F ∪ G) = E F G and as each of the terms on the right side are
c c
(g) (EF) (EG) (FG) c mutually exclusive:
P(∪Ei ) = P(E1 ) + P(Ec1 E2 ) + P(Ec1 Ec2 E3 ) + · · ·
(h) (EFG)c i
+ P(Ec1 · · · Ecn−1 En )
3 ≤ P(E1 ) + P(E2 ) + · · · + P(En ) (why?)
5. . If he wins, he only wins $1, while if he loses, he
4
loses $3. ⎧
⎪ i − 1,

⎨ i = 2, …, 7
6. If E(F ∪ G) occurs, then E occurs and either F or G 11. P{sum is i} =
36
occur; therefore, either EF or EG occurs and so ⎪
⎪ − i,
13
⎩ i = 8, …, 12
36
E(F ∪ G) ⊂ EF ∪ EG
12. Either use hint or condition on initial outcome as:
Similarly, if EF ∪ EG occurs, then either EF or EG
occurs. Thus, E occurs and either F or G occurs; and P{E before F}
so E(F ∪ G) occurs. Hence, = P{E before F | initial outcome is E}P(E)
EF ∪ EG ⊂ E(F ∪ G) + P{E before F | initial outcome is F}P(F)

which together with the reverse inequality proves + P{E before F | initial outcome neither E
the result. or F}[1 − P(E) − P(F)]


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