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

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1 Probability Homework Solution #3 2.48 (a) The results follow directly from the definition of conditional probability. [] [|] PA B PAB PB = If , then [ ] 0 by Corollary 3 and thus [ | ] 0. If then , and thus [ | ] . If then , and thus [ | ] 1. AB P P PA ABA P ABB P A B ∩= ∩ = = ⊂∩ = = ⊃∩ = = = (b) If [ | ] [ ], then multiplying both sides by [ ], we have: [ ] [ ] [ ] [][] We then also have that [ | ] [ ]. We conclude that if [ | ] [ ], then and tend t PA B PAPB PAPB PB A PAB PA B A => ∩> = > o occur jointly. 2.49 (i) [ | ] for [ ] 0 0[ ] , ] | ] ,[] [ ] [ | ] 1 Therefore, | ] 1 P AB P B P ≤∩ < ∩⊂ ∩ ≤ ≤≤ (ii) [ ] 1 PS B BS P SB ⊂⇒ = = = (iii) () [( ) ] | [( ) ( )] (Axiom III) [|] [|] PA C B PA CB PA B C B PA B PC B
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This note was uploaded on 11/29/2009 for the course EE 131A taught by Professor Lorenzelli during the Fall '08 term at UCLA.

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

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