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Introduction to Probability 2nd Edition Problem Solutions
(last updated: 7/31/08)
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Dimitri P. Bertsekas and John N. Tsitsiklis
Massachusetts Institute of Technology
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Athena Scienti?c, Belmont, Massachusetts
1
CHAPTER 1
Solution to Problem 1.1. We have A = {2, 4, 6}, so A ∪ B = {2, 4, 5, 6}, and (A ∪ B)c = {1, 3}. On the other hand, Ac ∩ B c = {1, 3, 5} ∩ {1, 2, 3} = {1, 3}. Similarly, we have A ∩ B = {4, 6}, and (A ∩ B)c = {1, 2, 3, 5}. On the other hand, Ac ∪ 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 , we have 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 = Ac and T = B, to obtain the ?rst relation Ac = (Ac ∩ B) ∪ (Ac ∩ B c ). Interchange the rol