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If $A$ and $B$ are events, such that $P(A)=\frac{1}{4}$ $P(A / B)=\frac{1}{2}$ and $P(B / A)=\frac{2}{3}$, then $P(B)$ is
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Verified Answer
The correct answer is:
$\frac{1}{3}$
Given,
$P(A)=\frac{1}{4}, P\left(\frac{A}{B}\right)=\frac{1}{2}$ and $P\left(\frac{B}{A}\right)=\frac{2}{3}$
We know that, $P\left(\frac{E_1}{E_2}\right)=\frac{P\left(E_1 \cap E_2\right)}{P\left(E_2\right)}$
$\begin{aligned} & \Rightarrow \quad \frac{2}{3}=\frac{P(A \cap B)}{1 / 4} \quad[\because A \cap B=B \cap A] \\ & \Rightarrow \quad P(A \cap B)=\frac{2}{3} \times \frac{1}{4}=\frac{1}{6}\end{aligned}$
$\because \quad P\left(\frac{A}{B}\right)=\frac{1}{2}$
$\therefore \quad \frac{P(A \cap B)}{P(B)}=\frac{1}{2}$
$\Rightarrow \quad P(B)=2 P(A \cap B)=2 \times \frac{1}{6}=\frac{1}{3}$
$P(A)=\frac{1}{4}, P\left(\frac{A}{B}\right)=\frac{1}{2}$ and $P\left(\frac{B}{A}\right)=\frac{2}{3}$
We know that, $P\left(\frac{E_1}{E_2}\right)=\frac{P\left(E_1 \cap E_2\right)}{P\left(E_2\right)}$
$\begin{aligned} & \Rightarrow \quad \frac{2}{3}=\frac{P(A \cap B)}{1 / 4} \quad[\because A \cap B=B \cap A] \\ & \Rightarrow \quad P(A \cap B)=\frac{2}{3} \times \frac{1}{4}=\frac{1}{6}\end{aligned}$
$\because \quad P\left(\frac{A}{B}\right)=\frac{1}{2}$
$\therefore \quad \frac{P(A \cap B)}{P(B)}=\frac{1}{2}$
$\Rightarrow \quad P(B)=2 P(A \cap B)=2 \times \frac{1}{6}=\frac{1}{3}$
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