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hw3solution

# hw3solution - Econ 3190 Fall 2011 Solution to HW3 1(a P(B A...

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Econ 3190 Fall 2011 Solution to HW3 1. (a) P ( B \ A ) = P ( B j A ) ° P ( A ) = 0 : 75 ° 0 : 3 = 0 : 225 P ( A \ B \ C ) = P ( C j A \ B ) ° P ( A \ B ) = 0 : 2 ° 0 : 225 = 0 : 045 (b) P ( B ) = P ( B \ A )+ P ( B \ A 0 ) = P ( B j A ) ° P ( A )+ P ( B j A 0 ) ° P ( A 0 ) = 0 : 75 ° 0 : 3 + 0 : 2 ° 0 : 7 = 0 : 365 P ( B 0 ) = 1 ± P ( B ) = 1 ± 0 : 365 = 0 : 635 Think B 0 \ C as D; P ( B 0 \ C ) = P ( B 0 \ C \ A )+ P ( B 0 \ C \ A 0 ) = P ( C j B 0 \ A ) ° P ( B 0 \ A ) + P ( C j B 0 \ A 0 ) ° P ( B 0 \ A 0 ) Now we need to °nd the value of P ( B 0 \ A ) and P ( B 0 \ A 0 ) P ( B 0 \ A ) = P ( B 0 j A ) ° P ( A ) = [1 ± P ( B j A )] ° P ( A ) = (1 ± 0 : 75) ° 0 : 3 = 0 : 075 P ( B 0 \ A 0 ) = P ( B 0 ) ± P ( B 0 \ A ) = 0 : 635 ± 0 : 075 = 0 : 56 Plug in numeric values P ( B 0 \ C ) = 0 : 8 ° 0 : 075 + 0 : 9 ° 0 : 56 = 0 : 564 (c) P ( C ) = P ( C \ A \ B )+ P ( C \ A 0 \ B )+ P ( C \ A \ B 0 )+ P ( C \ A 0 \ B 0 ) = P ( C j A \ B ) ° P ( A \ B ) + P ( C j A 0 \ B ) ° P ( A 0 \ B ) + + P ( C j A \ B 0 ) ° P ( A \ B 0 ) + P ( C j A 0 \ B 0 ) ° P ( A 0 \ B 0 ) we know ( A \ B 0 ) = 1 ± P ( A \ B ) ± P ( A 0 \ B ) ± P ( A 0 \ B 0 ) = 1 ± 0 : 225 ± 0 : 075 ± 0 : 56 = 0 : 14 Plug in numeric values P ( C ) = 0 : 20 ° 0 : 225 + 0 : 15 ° 0 : 14 + 0 : 8 ° 0 : 075 + 0 : 90 ° 0 : 56 = 0 : 63 (d) the question is asking for P ( A j C \ B 0 ) = P ( A \ C \ B 0 ) P ( C \ B 0 ) = P ( C j A \ B 0 ) ° P ( A \ B 0 ) P ( C \ B 0 ) = 0 : 80 ° 0 : 075 0 : 564 = 0 : 106 38 2. Assume P ( A ) > 0 and P ( B ) > 0 , we know A and B can±t be both mutually exclusive and independent. Intuitively, mutually exclusive implies that if A occurs, B will not occur. This is already a dependent relationship.

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hw3solution - Econ 3190 Fall 2011 Solution to HW3 1(a P(B A...

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