It is given that in a random experiment events A and B are such that $\mathrm{P}(\mathrm{A})=\frac{1}{4}, P(A \mid B)=\frac{1}{2}$ and $P(B \mid A)=\frac{2}{3}$ then $P(B)=$ $\frac{1}{6}$

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Aicharya · Accepted Answer

We know that, $P\left(\frac{B}{A}ight)=\frac{P(A \cap B)}{P(A)}$ $\Rightarrow P\left(\frac{B}{A}ight) \cdot P(A)=P(A \cap B) \Rightarrow \frac{2}{3} 	imes \frac{1}{4}=\frac{1}{6}=P(A \cap B)$ And $P\left(\frac{A}{B}ight)=\frac{P(A \cap B)}{P(B)} \Rightarrow P(B)=\frac{P(A \cap B)}{P(A / B)}$ $\Rightarrow P(B)=\frac{1 / 6}{1 / 2}=\frac{1}{6} 	imes \frac{2}{1}=\frac{1}{3} .$

It is given that in a random experiment events A and B are such that $\mathrm{P}(\mathrm{A})=\frac{1}{4},…

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