1 - Introduction to Probability 2nd Edition Problem...

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Introduction to Probability 2nd Edition Problem Solutions (last updated: 7/31/08) c ± Dimitri P. Bertsekas and John N. Tsitsiklis Massachusetts Institute of Technology WWW site for book information and orders http://www.athenasc.com Athena Scientifc, Belmont, Massachusetts 1

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CHAPTER 1 Solution to Problem 1.1. We have A = { 2 , 4 , 6 } ,B = { 4 , 5 , 6 } , so A B = { 2 , 4 , 5 , 6 } ,and ( A B ) c = { 1 , 3 } . On the other hand, A c B c = { 1 , 3 , 5 }∩{ 1 , 2 , 3 } = { 1 , 3 } . Similarly, we have A B = { 4 , 6 } ( A B ) c = { 1 , 2 , 3 , 5 } . On the other hand, A c B c = { 1 , 3 , 5 }∪{ 1 , 2 , 3 } = { 1 , 2 , 3 , 5 } . Solution to Problem 1.2. (a) By using a Venn diagram it can be seen that for any sets S and T ,wehave S =( S T ) ( S T c ) . (Alternatively, argue that any x must belong to either T or to T c ,so x belongs to S if and only if it belongs to S T or to S T c .) Apply this equality with S = A c and T = B , to obtain the Frst relation A c A c B ) ( A c B c ) . Interchange the roles of A and B to obtain the second relation. (b) By De Morgan’s law, we have ( A B ) c = A c B c , and by using the equalities of part (a), we obtain ( A B ) c = ( ( A c B ) ( A c B c ) ) ( ( A B c ) ( A c B c ) ) A c B ) ( A c B c ) ( A B c ) . (c) We have A = { 1 , 3 , 5 } and B = { 1 , 2 , 3 } A B = { 1 , 3 } . Therefore, ( A B ) c = { 2 , 4 , 5 , 6 } , 2
and A c B = { 2 } ,A c B c = { 4 , 6 } B c = { 5 } . Thus, the equality of part (b) is veriFed. Solution to Problem 1.5. Let G and C be the events that the chosen student is a genius and a chocolate lover, respectively. We have P ( G )=0 . 6, P ( C . 7, and P ( G C . 4. We are interested in P ( G c C c ), which is obtained with the following calculation: P ( G c C c )=1 P ( G C ( P ( G )+ P ( C ) P ( G C ) ) =1 (0 . 6+0 . 7 0 . 4) = 0 . 1 . Solution to Problem 1.6. We Frst determine the probabilities of the six possible outcomes. Let a = P ( { 1 } )= P ( { 3 } P ( { 5 } )and b = P ( { 2 } P ( { 4 } P ( { 6 } ). We are given that b =2 a . By the additivity and normalization axioms, 1 = 3 a +3 b = 3 a +6 a =9 a .Thu s , a / 9, b / 9, and P ( { 1 , 2 , 3 } )=4 / 9. Solution to Problem 1.7. The outcome of this experiment can be any Fnite sequence of the form ( a 1 ,a 2 ,...,a n ), where n is an arbitrary positive integer, a 1 2 n 1 belong to { 1 , 3 } ,and a n belongs to { 2 , 4 } . In addition, there are possible outcomes in which an even number is never obtained. Such outcomes are inFnite sequences ( a 1 2 ,... ), with each element in the sequence belonging to { 1 , 3 } . The sample space consists of all possible outcomes of the above two types. Solution to Problem 1.8. Let p i be the probability of winning against the opponent played in the i th turn. Then, you will win the tournament if you win against the 2nd player (probability p 2 ) and also you win against at least one of the two other players [probability p 1 +(1 p 1 ) p 3 = p 1 + p 3 p 1 p 3 ]. Thus, the probability of winning the tournament is p 2 ( p 1 + p 3 p 1 p 3 ) . The order (1 , 2 , 3) is optimal if and only if the above probability is no less than the probabilities corresponding to the two alternative orders, i.e., p 2 ( p 1 + p 3 p 1 p 3 ) p 1 ( p 2 + p 3 p 2 p 3 ) , p 2 ( p 1 + p 3 p 1 p 3 ) p 3 ( p 2 + p 1 p 2 p 1 ) .

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1 - Introduction to Probability 2nd Edition Problem...

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