Transcription of Introduction to Probability 2nd Edition Problem Solutions
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Introduction to Probability 2nd Edition Problem Solutions
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).
Jul 31, 2008 · In addition, there are possible outcomes in which an even number is never obtained. Such outcomes are infinite sequences (a 1,a 2,...), with each element in the sequence belonging to {1,3}. The sample space ... so the probability of doubles is 6/36 = 1/6. (b) The conditioning event (sum is 4 or less) consists of the 6 outcomes ...
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