Two events A and B are given. If P(A) = frac{ | vedclass

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(B) Given that $P(A) = \frac{1}{4}$,$P(A|B) = \frac{1}{2}$,and $P(B|A) = \frac{2}{3}$.We know the definition of conditional probability: $P(B|A) = \fr

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$1/3$

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(B) Given that $P(A) = \frac{1}{4}$,$P(A|B) = \frac{1}{2}$,and $P(B|A) = \frac{2}{3}$.We know the definition of conditional probability: $P(B|A) = \frac{P(A \cap B)}{P(A)}$.Substituting the known values: $\frac{2}{3} = \frac{P(A \cap B)}{1/4}$.Therefore,$P(A \cap B) = \frac{2}{3} 	imes \frac{1}{4} = \frac{2}{12} = \frac{1}{6}$.Now,use the definition of conditional probability for $P(A|B)$: $P(A|B) = \frac{P(A \cap B)}{P(B)}$.Substituting the values: $\frac{1}{2} = \frac{1/6}{P(B)}$.Solving for $P(B)$: $P(B) = \frac{1/6}{1/2} = \frac{1}{6} 	imes 2 = \frac{2}{6} = \frac{1}{3}$.Thus,the value of $P(B)$ is $\frac{1}{3}$.

Two events $A$ and $B$ are given. If $P(A) = \frac{1}{4}$,$P(A|B) = \frac{1}{2}$,and $P(B|A) = \frac{2}{3}$,then what is the value of $P(B)$?

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