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Question 63 of 89

Q.It is given that the events A and B are such that P(A)=14P(A)=\dfrac{1}{4}, P(A/B)=12P(A/B)=\dfrac{1}{2} and P(B/A)=23P(B/A)=\dfrac{2}{3}. Then P(B)P(B) is:

(a) 23\dfrac{2}{3}
(b) 12\dfrac{1}{2}
(c) 16\dfrac{1}{6}
(d) 13\dfrac{1}{3}
Puducherry TnboardTamil Nadu HSC First Year (DGE) Board 2019MCQ· 1mImportance★★★★★
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First get P(A∩B)P(A\cap B) from P(B∣A)=P(A∩B)/P(A)P(B|A)=P(A\cap B)/P(A), then use P(A∣B)=P(A∩B)/P(B)P(A|B)=P(A\cap B)/P(B) to solve for P(B)=1/3P(B)=1/3.

Given P(A)=14P(A)=\dfrac{1}{4}, P(A∣B)=12P(A|B)=\dfrac{1}{2}, P(B∣A)=23P(B|A)=\dfrac{2}{3}.

From P(B∣A)=P(A∩B)P(A)P(B|A) = \dfrac{P(A\cap B)}{P(A)}: P(A∩B)=P(B∣A)⋅P(A)=23×14=16P(A\cap B) = P(B|A)\cdot P(A) = \dfrac{2}{3}\times\dfrac{1}{4} = \dfrac{1}{6}.

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