It is given that the events $A$ and $B$ are such that $P(A)=\frac{1}{4}, P(A \mid B)=\frac{1}{2} \quad$ and $\quad P(B \mid A)=\frac{2}{3}$. Then $P(B)$ is

Question

It is given that the events $A$ and $B$ are such that $P(A)=\frac{1}{4}, P(A \mid B)=\frac{1}{2} \quad$ and $\quad P(B \mid A)=\frac{2}{3}$. Then $P(B)$ is, $\frac{1}{6}$, $\frac{1}{3}$, $\frac{2}{3}$, $\frac{1}{2}$

MARKS App · Accepted Answer

$P(A)=1 / 4, P(A / B)=\frac{1}{2}, P(B / A)=2 / 3$ By conditional probability, $\begin{aligned} & P(A \cap B)=P(A) P(B / A)=P(B) P(A / B) \ & \Rightarrow \frac{1}{4} \times \frac{2}{3}=P(B) \times \frac{1}{2} \Rightarrow P(B)=\frac{1}{3} \end{aligned}$

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