A first course in probability PDF
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A F i r s t C o u r s A First Course in Probability e i Sheldon Ross n P Ninth Edition r o b a b i l i t y R o s s N i n t h E d i t i o n ISBN 978-1-29202-492-9 9 781292 024929 A First Course in Probability Sheldon Ross Ninth Edition ISBN 10: 1-292-02492-5 ISBN 13: 978-1-292-02492-9 Pearson Education Limited Edinburgh Gate Harlow Essex CM20 2JE England and Associated Companies throughout the world Visit us on the World Wide Web at: www.pearsoned.co.uk © Pearson Education Limited 2014 All rights reserved. No part of this publication may be reproduced, stored in a retrieval system, or transmitted in any form or by any means, electronic, mechanical, photocopying, recording or otherwise, without either the prior written permission of the publisher or a licence permitting restricted copying in the United Kingdom issued by the Copyright Licensing Agency Ltd, Saffron House, 6–10 Kirby Street, London EC1N 8TS. All trademarks used herein are the property of their respective owners. The use of any trademark in this text does not vest in the author or publisher any trademark ownership rights in such trademarks, nor does the use of such trademarks imply any affi liation with or endorsement of this book by such owners. ISBN 10: 1-292-02492-5 ISBN 13: 978-1-292-02492-9 British Library Cataloguing-in-Publication Data A catalogue record for this book is available from the British Library Printed in the United States of America 11233444262930924513193797131 P E A R S O N C U S T O M L I B R AR Y Table of Contents 1. Combinatorial Analysis Sheldon Ross 1 2. Axioms of Probability Sheldon Ross 23 3. Conditional Probability and Independence Sheldon Ross 61 4. Random Variables Sheldon Ross 123 5. Continuous Random Variables Sheldon Ross 191 6. Jointly Distributed Random Variables Sheldon Ross 239 7. Properties of Expectation Sheldon Ross 303 8. Limit Theorems Sheldon Ross 397 9. Simulation Sheldon Ross 429 Common Distributions Sheldon Ross 447 Index 451 I II Combinatorial Analysis Contents 1 Introduction 5 MultinomialCoefficients 2 TheBasicPrincipleofCounting 6 TheNumberofIntegerSolutionsof 3 Permutations Equations 4 Combinations 1 Introduction Hereisatypicalproblemofinterestinvolvingprobability:Acommunicationsystem istoconsistofnseeminglyidenticalantennasthataretobelinedupinalinearorder. The resulting system will then be able to receive all incoming signals—and will be called functional—as long as no two consecutive antennas are defective. If it turns out that exactly m of the n antennas are defective, what is the probability that the resultingsystemwillbefunctional?Forinstance,inthespecialcasewheren=4and m=2,thereare6possiblesystemconfigurations,namely, 0 1 1 0 0 1 0 1 1 0 1 0 0 0 1 1 1 0 0 1 1 1 0 0 where 1 means that the antenna is working and 0 that it is defective. Because the resultingsystemwillbefunctionalinthefirst3arrangementsandnotfunctionalin the remaining 3, it seems reasonable to take 3 = 1 as the desired probability. In 6 2 the case of general n and m, we could compute the probability that the system is functionalinasimilarfashion.Thatis,wecouldcountthenumberofconfigurations thatresultinthesystem’sbeingfunctionalandthendividebythetotalnumberofall possibleconfigurations. Fromtheprecedingdiscussion,weseethatitwouldbeusefultohaveaneffec- tive method for counting the number of ways that things can occur. In fact, many problemsinprobabilitytheorycanbesolvedsimplybycountingthenumberofdif- ferent ways that a certain event can occur. The mathematical theory of counting is formallyknownascombinatorialanalysis. FromChapter1ofAFirstCourseinProbability,NinthEdition.SheldonRoss. Copyright©2014byPearsonEducation,Inc.Allrightsreserved. 1 CombinatorialAnalysis 2 The Basic Principle of Counting Thebasicprincipleofcountingwillbefundamentaltoallourwork.Looselyput,it statesthatifoneexperimentcanresultinanyofmpossibleoutcomesandifanother experiment can result in any of n possible outcomes, then there are mn possible outcomesofthetwoexperiments. Thebasicprincipleofcounting Suppose that two experiments are to be performed. Then if experiment 1 can resultinanyoneofmpossibleoutcomesandif,foreachoutcomeofexperiment 1, there are n possible outcomes of experiment 2, then together there are mn possibleoutcomesofthetwoexperiments. Proof of the Basic Principle: The basic principle may be proven by enumerating allthepossibleoutcomesofthetwoexperiments;thatis, (1,1), (1,2), ..., (1,n) (2,1), (2,2), ..., (2,n) # # # (m,1), (m,2), ..., (m,n) where we say that the outcome is (i, j) if experiment 1 results in its ith possible outcomeandexperiment2thenresultsinitsjthpossibleoutcome.Hence,thesetof possible outcomes consists of m rows, each containing n elements. This proves the result. Example A small community consists of 10 women, each of whom has 3 children. If one 2a woman and one of her children are to be chosen as mother and child of the year, howmanydifferentchoicesarepossible? Solution Byregardingthechoiceofthewomanastheoutcomeofthefirstexperi- mentandthesubsequentchoiceofoneofherchildrenastheoutcomeofthesecond experiment, we see from the basic principle that there are 10 * 3 = 30 possible choices. . Whentherearemorethantwoexperimentstobeperformed,thebasicprinciple canbegeneralized. Thegeneralizedbasicprincipleofcounting Ifrexperimentsthataretobeperformedaresuchthatthefirstonemayresult in any of n possible outcomes; and if, for each of these n possible outcomes, 1 1 therearen possibleoutcomesofthesecondexperiment;andif,foreachofthe 2 possibleoutcomesofthefirsttwoexperiments,therearen possibleoutcomes 3 ofthethirdexperiment;andif...,thenthereisatotalofn · n ···n possible 1 2 r outcomesoftherexperiments. Example Acollegeplanningcommitteeconsistsof3freshmen,4sophomores,5juniors,and 2b 2seniors.Asubcommitteeof4,consistingof1personfromeachclass,istobecho- sen.Howmanydifferentsubcommitteesarepossible? 2 CombinatorialAnalysis Solution Wemayregardthechoiceofasubcommitteeasthecombinedoutcomeof thefourseparateexperimentsofchoosingasinglerepresentativefromeachofthe classes.Itthenfollowsfromthegeneralizedversionofthebasicprinciplethatthere are3 * 4 * 5 * 2=120possiblesubcommittees. . Example Howmanydifferent7-placelicenseplatesarepossibleifthefirst3placesaretobe 2c occupiedbylettersandthefinal4bynumbers? Solution By the generalized version of the basic principle, the answer is 26 · 26 · 26 · 10 · 10 · 10 · 10=175,760,000. . Example How many functions defined on n points are possible if each functional value is 2d either0or1? Solution Let the points be 1,2,...,n. Since f(i) must be either 0 or 1 for each i=1,2,...,n,itfollowsthatthereare2npossiblefunctions. . Example InExample2c,howmanylicenseplateswouldbepossibleifrepetitionamongletters 2e ornumberswereprohibited? Solution In this case, there would be 26 · 25 · 24 · 10 · 9 · 8 · 7 = 78,624,000 possiblelicenseplates. . 3 Permutations Howmanydifferentorderedarrangementsofthelettersa,b,andcarepossible? By direct enumeration we see that there are 6, namely, abc, acb, bac, bca, cab, andcba.Eacharrangementisknownasapermutation.Thus,thereare6possible permutations of a set of 3 objects. This result could also have been obtained fromthebasicprinciple,sincethefirstobjectinthepermutationcanbeanyof the3,thesecondobjectinthepermutationcanthenbechosenfromanyofthe remaining 2, and the third object in the permutation is then the remaining 1. Thus,thereare3 · 2 · 1=6possiblepermutations. Supposenowthatwehavenobjects.Reasoningsimilartothatwehavejustused forthe3lettersthenshowsthatthereare n(n − 1)(n − 2)···3 · 2 · 1=n! differentpermutationsofthenobjects. Whereas n! (read as “n factorial”) is defined to equal 1 · 2···n when n is a positiveinteger,itisconvenienttodefine0!toequal1. Example How many different batting orders are possible for a baseball team consisting of 9 3a players? Solution Thereare9!=362,880possiblebattingorders. . Example A class in probability theory consists of 6 men and 4 women. An examination is 3b given,andthestudentsarerankedaccordingtotheirperformance.Assumethatno twostudentsobtainthesamescore. (a) Howmanydifferentrankingsarepossible? 3 CombinatorialAnalysis (b) Ifthemenarerankedjustamongthemselvesandthewomenjustamongthem- selves,howmanydifferentrankingsarepossible? Solution (a)Becauseeachrankingcorrespondstoaparticularorderedarrangement ofthe10people,theanswertothispartis10!=3,628,800. (b)Since there are 6! possible rankings of the men among themselves and 4! possiblerankingsofthewomenamongthemselves,itfollowsfromthebasicprinciple thatthereare(6!)(4!)=(720)(24)=17,280possiblerankingsinthiscase. . Example Ms.Joneshas10booksthatsheisgoingtoputonherbookshelf.Ofthese,4aremath- 3c ematicsbooks,3arechemistrybooks,2arehistorybooks,and1isalanguagebook. Ms. Jones wants to arrange her books so that all the books dealing with the same subjectaretogetherontheshelf.Howmanydifferentarrangementsarepossible? Solution There are 4! 3! 2! 1! arrangements such that the mathematics books are firstinline,thenthechemistrybooks,thenthehistorybooks,andthenthelanguage book.Similarly,foreachpossibleorderingofthesubjects,thereare4!3!2!1!pos- sible arrangements. Hence, as there are 4! possible orderings of the subjects, the desiredansweris4!4!3!2!1!=6912. . Weshallnowdeterminethenumberofpermutationsofasetofnobjectswhen certain of the objects are indistinguishable from one another. To set this situation straightinourminds,considerthefollowingexample. Example HowmanydifferentletterarrangementscanbeformedfromthelettersPEPPER? 3d Solution We first note that there are 6! permutations of the letters P E P P E R 1 1 2 3 2 whenthe3P’sandthe2E’saredistinguishedfromoneanother.However,consider anyoneofthesepermutations—forinstance,P P E P E R.Ifwenowpermutethe 1 2 1 3 2 P’s among themselves and the E’s among themselves, then the resultant arrange- mentwouldstillbeoftheformPPEPER.Thatis,all3!2!permutations P P E P E R P P E P E R 1 2 1 3 2 1 2 2 3 1 P P E P E R P P E P E R 1 3 1 2 2 1 3 2 2 1 P P E P E R P P E P E R 2 1 1 3 2 2 1 2 3 1 P P E P E R P P E P E R 2 3 1 1 2 2 3 2 1 1 P P E P E R P P E P E R 3 1 1 2 2 3 1 2 2 1 P P E P E R P P E P E R 3 2 1 1 2 3 2 2 1 1 areoftheformPPEPER.Hence,thereare6!/(3!2!) = 60possibleletterarrange- mentsofthelettersPEPPER. . Ingeneral,thesamereasoningasthatusedinExample3dshowsthatthereare n! n !n ! ··· n ! 1 2 r differentpermutationsofnobjects,ofwhichn arealike,n arealike,...,n are 1 2 r alike. Example A chess tournament has 10 competitors, of which 4 are Russian, 3 are from the 3e United States, 2 are from Great Britain, and 1 is from Brazil. If the tournament resultlistsjustthenationalitiesoftheplayersintheorderinwhichtheyplaced,how manyoutcomesarepossible? 4 CombinatorialAnalysis Solution Thereare 10! =12,600 4!3!2!1! possibleoutcomes. . Example How many different signals, each consisting of 9 flags hung in a line, can be made 3f froma setof 4 white flags, 3red flags, and 2 blue flags ifallflags of the same color areidentical? Solution Thereare 9! =1260 4!3!2! differentsignals. . 4 Combinations Weareofteninterestedindeterminingthenumberofdifferentgroupsofr objects that could be formed from a total of n objects. For instance, how many different groups of 3 could be selected from the 5 items A, B, C, D, and E? To answer this question,reasonasfollows:Sincethereare5waystoselecttheinitialitem,4waysto thenselectthenextitem,and3waystoselectthefinalitem,therearethus5 · 4 · 3 ways of selecting the group of 3 when the order in which the items are selected is relevant.However,sinceeverygroupof3—say,thegroupconsistingofitemsA,B, and C—will be counted 6 times (that is, all of the permutations ABC, ACB, BAC, BCA, CAB, and CBA will be counted when the order of selection is relevant), it followsthatthetotalnumberofgroupsthatcanbeformedis 5 · 4 · 3 =10 3 · 2 · 1 Ingeneral,asn(n − 1)···(n − r + 1)representsthenumberofdifferentwaysthat a group of r items could be selected from n items when the order of selection is relevant,andaseachgroupofritemswillbecountedr!timesinthiscount,itfollows thatthenumberofdifferentgroupsofritemsthatcouldbeformedfromasetofn itemsis n(n − 1)···(n − r + 1) n! = r! (n − r)!r! Notationandterminology (cid:2) (cid:3) n Wedefine ,forr … n,by r (cid:2) (cid:3) n n! = r (n − r)!r! (cid:2) (cid:3) n and say that (read as “n choose r”) represents the number of possible r combinationsofnobjectstakenratatime.† (cid:2) (cid:3) (cid:2) (cid:3) (cid:2) (cid:3) †Byconvention,0!isdefinedtobe1.Thus, n = n = 1.Wealsotake n tobeequalto0when eitheri < 0ori > n. 0 n i 5
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