# P a b p ab p b but since p ab 0 the result follows b

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P ( A | B ) = P ( AB ) /P ( B ). But since P ( AB ) 0, the result follows. (b) P | B ) = 1 P | B ) = P B ) /P ( B ) = P ( B ) /P ( B ) = 1. (c) For A 1 , A 2 , . . . ∈ F with A i A j = for i 6 = j , P ( i =1 A i | B ) = i =1 P ( A i | B ) P ( i =1 A i | B ) = P (( i =1 A i ) B ) /P ( B ) = P ( i =1 A i B ) /P ( B ) = i =1 P ( A i B ) /P ( B ) = i =1 P ( A i | B ), since the A i B sets are disjoint. 2

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(d) AB = ∅ ⇒ P ( A | B ) = 0 . P ( A | B ) = P ( AB ) /P ( B ) = P ( ) /P ( B ) = 0 . (e) P ( B | B ) = 1 P ( B | B ) = P ( BB ) /P ( B ) = P ( B ) /P ( B ) = 1 (f) A B P ( A | B ) P ( A ) Since A B , AB = A . Then P ( A | B ) = P ( AB ) /P ( B ) = P ( A ) /P ( B ) P ( A ) since P ( B ) 1. (g) B A P ( A | B ) = 1. Since A B , AB = B . Then P ( A | B ) = P ( AB ) /P ( B ) = P ( B ) /P ( B ) = 1 . 5. Prove the law of total probability. Let { A i } be a partition of A . Note that A = n i =1 A i A . Note also that P ( AA i ) = P ( A | A i ) P ( A i ). P ( A ) = P ( n i =1 AA i ) = n X i =1 P ( AA i ) = n X i =1 P ( A | A i ) P ( A i ) . 6. Prove Bayes rule P ( A | B ) = P ( AB ) /P ( B ) = P ( B | A ) P ( A ) /P ( B ) 7. Suppose A and B are independent events. Show that A and B c are also independent. By independence, P ( AB ) = P ( A ) P ( B ). P ( A ) = P ( A ( B B c )) = P ( AB AB c ) = P ( AB ) + P ( AB c ) = P ( A ) P ( B ) + P ( AB c ) so P ( AB c ) = P ( A ) - P ( A ) P ( B ) = P ( A )(1 - P ( B )) = P ( A ) P ( B c ) .
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