Question 1198614
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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.



<pre>
From  P(B)=0.30  and  P(A│B)=0.60,  we find  P(A &#8745 B)  = 0.30*0.60 = 0.18.

From  P(B')=0.7  and  P(A│B')=0.5,  we find  P(A &#8745 B') = 0.50*0.70 = 0.35.



From  P(A &#8745 B)  = 0.18  and  P(A &#8745 B') = 0.35  we find

      P(A) = P(A &#8745 B) + P(A &#8745 B') = 0.18 + 0.35 = 0.53.



From  P(A &#8745 B) = 0.18  and  P(A) = 0.53  we find

      P(B|A) = P(A &#8745 B) / P(A) = {{{0.18/0.53}}} = {{{18/53}}} = 0.3396  (rounded).    <U>ANSWER</U>
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Solved.