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If P(B)=0.30, P(A│B)=0.60, P(B’)=0.70 and P(A│B’)=0.50, find P(B│A).
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The condition P(B') = 0.7 is excessive in formulation of this problem.
It follows automatically from the given condition P(B) = 0.30.
From P(B)=0.30 and P(A│B)=0.60, we find P(A ∩ B) = 0.30*0.60 = 0.18.
From P(B')=0.7 and P(A│B')=0.5, we find P(A ∩ B') = 0.50*0.70 = 0.35.
From P(A ∩ B) = 0.18 and P(A ∩ B') = 0.35 we find
P(A) = P(A ∩ B) + P(A ∩ B') = 0.18 + 0.35 = 0.53.
From P(A ∩ B) = 0.18 and P(A) = 0.53 we find
P(B|A) = P(A ∩ B) / P(A) = = = 0.3396 (rounded). ANSWER
