# 180asolns2 - Math 180A Homework 2 Solutions October 8 2006...

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Math 180A Homework 2 Solutions October 8, 2006 Problem 1 a) P (( A B ) C ) = 1 - P ( A B ) = 1 - ( P ( A ) + P ( B ) - P ( A C )) = 1 - ( . 65 + . 55 - . 25) = . 05 b) P ( A B C ) + P ( B A C ) = P ( A ) - P ( A B ) + P ( B ) - P ( A B ) = . 65 - . 25 + . 55 - . 25 = . 70 c) P ( A B ) = P ( A ) + P ( B ) - P ( A B ) = . 95 d) P (( A B ) C ) = 1 - P ( A B ) = . 75 e) P (( A B C ) ( B A C ) | A B ) = P (( A B C ) ( B A C ) ( A B )) P ( A B ) = P ( A B C ) ( B A C ) P ( A B ) = . 70 . 95 0 . 737 f) P ( A B | A B ) = P (( A B ) ( A B )) P ( A B ) = P ( A B ) P ( A B ) = . 25 . 95 0 . 263 g) P ( B C | A ) = P ( A B C ) P ( A ) = P ( A ) - P ( A B ) P ( A ) = . 4 . 65 0 . 615 1

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Problem 2 Let A be the event that the first dice is as large as four, and B be the event that the sum is 8. There are five outcomes in B = { (2 , 6) , (3 , 5) , (4 , 4) , (5 , 3) , (6 , 2) } . Also clearly, A B = { (4 , 4) , (5 , 3) , (6 , 2) } . Hence P ( A | B ) = 3 5 . Problem 3 Here is the tree: RR RB BB BR RRR RRB RBR RBB BRR BRB BBR BBB 2/3 3/7 4/7 2/7 5/7 1/2 3/8 5/8 1/2 3/8 5/8 1/4 3/4 (1/7) (4/21) (4/21) (10/21) 1/3 R (1/3) B (2/3) (1/14) (1/14) (5/42) (1/14) (5/42) (5/42) (5/14) (1/14) Note how the probabilities for the nodes at a given level sum to one.
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