HW%232_f08_sol

# HW%232_f08_sol - CE 3102 Fall 2008 Homework#2 Solutions 1...

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CE 3102 Fall 2008 Homework #2 Solutions 1. One way to make our reasoning clear is lay two columns, with claims and justifications. (a) Claim Justification Suppose A and B are independent Supposition If A and B are independent, then P(A|B)=P(A) (2.13) p. 52, A&T P(A|B) P(A) Since P(A|B)>P(A) A and B are not independent proof by contradiction ( a k a modus tollens ) (b) Claim Justification Suppose A and B are mutually exclusive A B= Definition of mutually exclusive events P( ) = 0 P ( )=1-P(S)=1-1=0 P(A B ) = 0 P(A|B)=P(A B)/P(B) Definition of conditional probability P(A|B)=0 Since P(A B)=0 P(A|B) P(A) Since P(A) 0 P ( A | B ) > P ( A ) G i v e n A and B are not mutually exclusive Proof by contraction

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(c ) Claim Justification P(A B)=P(A|B)P(B) Definition of conditional probability P(A|B) > P(A) Given P(A B)>P(A)P(B) Substitution of P(A) for P(A|B) (d) Claim Justification ) ( ) ( ) ( B A P B A P A P + = Total probability theorem ) ( ) ( ) ( B
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HW%232_f08_sol - CE 3102 Fall 2008 Homework#2 Solutions 1...

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