Introduction to Probability 2nd Edition Problem Solutions

1 Introduction to Probability2nd EditionProblem Solutions (last updated: 7/31/08)c Dimitri P. Bertsekas and John N. TsitsiklisMassachusetts Institute of TechnologyWWW site for book information and Scientific, Belmont, Massachusetts1C H A P T E R 1 Solution to Problem haveA={2,4,6}, B={4,5,6},soA B={2,4,5,6}, and(A B)c={1,3}.On the other hand,Ac Bc={1,3,5} {1,2,3}={1,3}.Similarly, we haveA B={4,6}, and(A B)c={1,2,3,5}.On the other hand,Ac Bc={1,3,5} {1,2,3}={1,2,3,5}.Solution to Problem (a) By using a Venn diagram it can be seen that for anysetsSandT, we haveS= (S T) (S Tc).(Alternatively, argue that anyxmust belong to eitherTor toTc, soxbelongs toSif and only if it belongs toS Tor toS Tc.) Apply this equality withS=AcandT=B, to obtain the first relationAc= (Ac B) (Ac Bc).Interchange the roles ofAandBto obtain the second relation.(b) By De Morgan s law, we have(A B)c=Ac Bc,and by using the equalities of part (a), we obtain(A B)c=((Ac B) (Ac Bc)) ((A Bc) (Ac Bc))= (Ac B) (Ac Bc) (A Bc).

2 (c) We haveA={1,3,5}andB={1,2,3}, soA B={1,3}. Therefore,(A B)c={2,4,5,6},2andAc B={2}, Ac Bc={4,6}, A Bc={5}.Thus, the equality of part (b) is to Problem the events that the chosen student isa genius and a chocolate lover, respectively. We haveP(G) = ,P(C) = , andP(G C) = We are interested inP(Gc Cc), which is obtained with the followingcalculation:P(Gc Cc) = 1 P(G C) = 1 (P(G)+P(C) P(G C))= 1 ( + ) = to Problem first determine the probabilities of the six possibleoutcomes. Leta=P({1}) =P({3}) =P({5}) andb=P({2}) =P({4}) =P({6}).We are given thatb= 2a. By the additivity and normalization axioms, 1 = 3a+ 3b=3a+ 6a= 9a. Thus,a= 1/9,b= 2/9, andP({1,2,3}) = 4 to Problem outcome of this experiment can be any finite sequenceof the form (a1,a2,..,an), wherenis an arbitrary positive integer,a1,a2,..,an 1belong to{1,3}, andanbelongs to{2,4}.

3 In addition, there are possible outcomesin which an even number is never obtained. Such outcomes are infinite sequences(a1,a2,..), with each element in the sequence belonging to{1,3}. The sample spaceconsists of all possible outcomes of the above two to Problem the Probability of winning against the opponentplayed in theith turn. Then, you will win the tournament if you win against the 2ndplayer (probabilityp2) and also you win against at least one of the two other players[probabilityp1+ (1 p1)p3=p1+p3 p1p3]. Thus, the Probability of winning thetournament isp2(p1+p3 p1p3).The order (1,2,3) is optimal if and only if the above Probability is no less than theprobabilities corresponding to the two alternative orders, ,p2(p1+p3 p1p3) p1(p2+p3 p2p3),p2(p1+p3 p1p3) p3(p2+p1 p2p1).It can be seen that the first inequality above is equivalent top2 p1, while the secondinequality above is equivalent top2 to Problem (a) Since = ni=1Si, we haveA=n i=1(A Si),while the setsA Siare disjoint.

4 The result follows by using the additivity axiom.(b) The eventsB Cc,Bc C,B C, andBc Ccform a partition of , so by part(a), we haveP(A) =P(A B Cc) +P(A Bc C) +P(A B C) +P(A Bc Cc).(1)3 The eventA Bcan be written as the union of two disjoint events as follows:A B= (A B C) (A B Cc),so thatP(A B) =P(A B C) +P(A B Cc).(2)Similarly,P(A C) =P(A B C) +P(A Bc C).(3)Combining Eqs. (1)-(3), we obtain the desired to Problem the eventsA BcandAc Bare disjoint, wehave using the additivity axiom repeatedly,P((A Bc) (Ac B))=P(A Bc)+P(Ac B) =P(A) P(A B)+P(B) P(A B).Solution to Problem (a) Each possible outcome has Probability 1/36. Thereare 6 possible outcomes that are doubles, so the Probability of doubles is 6/36 = 1/6.(b) The conditioning event (sum is 4 or less) consists of the 6 outcomes{(1,1),(1,2),(1,3),(2,1),(2,2),( 3,1)},2 of which are doubles, so the conditional Probability of doubles is 2/6 = 1/3.

5 (c) There are 11 possible outcomes with at least one 6, namely, (6,6), (6,i), and (i,6),fori= 1,2,..,5. Thus, the Probability that at least one die is a 6 is 11/36.(d) There are 30 possible outcomes where the dice land on different numbers. Out ofthese, there are 10 outcomes in which at least one of the rolls is a 6. Thus, the desiredconditional Probability is 10/30 = 1 to Problem the event that the first toss is a head andletBbe the event that the second toss is a head. We must compare the conditionalprobabilitiesP(A B|A) andP(A B|A B). We haveP(A B|A) =P((A B) A)P(A)=P(A B)P(A),andP(A B|A B) =P((A B) (A B))P(A B)=P(A B)P(A B).SinceP(A B) P(A), the first conditional Probability above is at least as large, soAlice is right, regardless of whether the coin is fair or not. In the case where the coinis fair, that is, if all four outcomesHH,HT,TH,TTare equally likely, we haveP(A B)P(A)=1/41/2=12,P(A B)P(A B)=1/43/4= generalization of Alice s reasoning is that ifA,B, andCare events such thatB CandA B=A C(for example, ifA B C), then the eventAis at least4as likely if we know thatBhas occurred than if we know thatChas occurred.

6 Alice sreasoning corresponds to the special case whereC=A to Problem this Problem , there is a tendency to reason that sincethe opposite face is either heads or tails, the desired Probability is 1/2. This is, however,wrong, because given that heads came up, it is more likely that the two-headed coinwas chosen. The correct reasoning is to calculate the conditional probabilityp=P(two-headed coin was chosen|heads came up)=P(two-headed coin was chosen and heads came up)P(heads came up).We haveP(two-headed coin was chosen and heads came up) =13,P(heads came up) =12,so by taking the ratio of the above two probabilities, we obtainp= 2/3. Thus, theprobability that the opposite face is tails is 1 p= 1 to Problem the event that the batch will be A2 A3 A4, whereAi,i= 1,..,4, is the event that theith item isnot defective. Using the multiplication rule, we haveP(A) =P(A1)P(A2|A1)P(A3|A1 A2)P(A4|A1 A2 A3) =95100 9499 9398 9297= to Problem the definition of conditional probabilities, wehaveP(A B|B) =P(A B B)P(B)=P(A B)P(B)=P(A|B).

7 Solution to Problem the event that Alice does not find her paperin draweri. Since the paper is in draweriwith probabilitypi, and her search issuccessful with probabilitydi, the multiplication rule yieldsP(Ac) =pidi, so thatP(A) = 1 pidi. LetBbe the event that the paper is in drawerj. Ifj6=i, thenA B=B,P(A B) =P(B), and we haveP(B|A) =P(A B)P(A)=P(B)P(A)=pj1 , ifi=j, we haveP(B|A) =P(A B)P(A)=P(B)P(A|B)P(A)=pi(1 di)1 to Problem (a) Figure provides a sequential description for thethree different strategies. Here we assume 1 point for a win, 0 for a loss, and 1/2 point50 - 0 Timid playpwBold play( a )( b )(c)Bold playBold playBold playBold playTimid playTimid playTimid play0 - 01 - 02 - 01 - 11 - 10 - 10 - 20 - 20 - 11 - - 01 - 01 - 11 - 10 - 10 - :Sequential descriptions of the chess match histories under strategies(i), (ii), and (iii).

8 For a draw. In the case of a tied 1-1 score, we go to sudden death in the next game,and Boris wins the match (probabilitypw), or loses the match ( Probability 1 pw).(i) Using the total Probability theorem and the sequential description of Fig. (a),we haveP(Boris wins) =p2w+ 2pw(1 pw) termp2wcorresponds to the win-win outcome, and the term 2pw(1 pw)pwcorre-sponds to the win-lose-win and the lose-win-win outcomes.(ii) Using Fig. (b), we haveP(Boris wins) =p2dpw,corresponding to the draw-draw-win outcome.(iii) Using Fig. (c), we haveP(Boris wins) =pwpd+pw(1 pd)pw+ (1 pw) termpwpdcorresponds to the win-draw outcome, the termpw(1 pd)pwcorre-sponds to the win-lose-win outcome, and the term (1 pw)p2wcorresponds to lose-win-win outcome.(b) Ifpw<1/2, Boris has a greater Probability of losing rather than winning any onegame, regardless of the type of play he uses.

9 Despite this, the Probability of winningthe match with strategy (iii) can be greater than 1/2, provided thatpwis close enoughto 1/2 andpdis close enough to 1. As an example, ifpw= andpd= , withstrategy (iii) we haveP(Boris wins) = + (1 ) + (1 ) strategies (i) and (ii), the corresponding probabilities of a win can be calculatedto be approximately and , respectively. What is happening here is that withstrategy (iii), Boris is allowed to select a playing styleafterseeing the result of the firstgame, while his opponent is not. Thus, by being able to dictate the playing style ineach game after receiving partial information about the match s outcome, Boris gainsan to Problem (m,k) be the Probability that the starting playerwins when the jar initially containsmwhite andkblack balls. We have, using thetotal Probability theorem,p(m,k) =mm+k+km+k(1 p(m,k 1))= 1 km+kp(m,k 1).

10 The probabilitiesp(m,1),p(m,2),..,p(m,n) can be calculated sequentially using thisformula, starting with the initial conditionp(m,0) = to Problem derive a recursion for the probabilitypithat a whiteball is chosen from theith jar. We have, using the total Probability theorem,pi+1=m+ 1m+n+ 1pi+mm+n+ 1(1 pi) =1m+n+ 1pi+mm+n+ 1,starting with the initial conditionp1=m/(m+n). Thus, we havep2=1m+n+ 1 mm+n+mm+n+ 1=mm+ generally, this calculation shows that ifpi 1=m/(m+n), thenpi=m/(m+n).Thus, we obtainpi=m/(m+n) for to Problem ,n i(k) denote the Probability that afterkex-changes, a jar will containiballs that started in that jar andn iballs that started inthe other jar. We want to findpn,0(4). We argue recursively, using the total probability7theorem. We havepn,0(4) =1n 1n pn 1,1(3),pn 1,1(3) =pn,0(2) + 2 n 1n 1n pn 1,1(2) +2n 2n pn 2,2(2),pn,0(2) =1n 1n pn 1,1(1),pn 1,1(2) = 2 n 1n 1n pn 1,1(1),pn 2,2(2) =n 1n n 1n pn 1,1(1),pn 1,1(1) = these equations, we obtainpn,0(4) =1n2(1n2+4(n 1)2n4+4(n 1)2n4)=1n2(1n2+8(n 1)2n4).