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part8 - Probability Homework Solution#3 2.48(a The results...

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1 Probability Homework Solution #3 2.48 (a) The results follow directly from the definition of conditional probability. [ ] [ | ] [ ] P A B P A B P B = If , then [ ] 0 by Corollary 3 and thus [ | ] 0. [ ] If , then , and thus [ | ] . [ ] [ ] If , then , and thus [ | ] 1. [ ] A B P A B P A B P A A B A B A P A B P B P B A B A B B P A B P B = ∅ = = = = = = = (b) [ ] If [ | ] [ ], then multiplying both sides by [ ], we have: [ ] [ ] [ ] [ ] [ ] [ ] [ ] We then also have that [ | ] [ ]. [ ] [ ] We conclude that if [ | ] [ ], then and tend t P A B P A B P A P B P B P A B P A P B P A B P A P B P B A P B P A P A P A B P A B A = > > = > = > o occur jointly. 2.49 (i) [ ] [ | ] for [ ] 0 [ ] 0 [ ], 0 [ ] 0 [ | ] , [ ] [ ] [ | ] 1 Therefore, 0 [ | ] 1 P A B P A B P B P B P A B P B P A B A B B P A B P B P A B P A B = > < (ii) [ ] [ ] [ | ] 1 [ ] [ ] P S
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