Q106 / 180 · COMEDK 2024 (Memory Based) · Mathematics
If $A$ and $B$ are events, such that $P(A)=\frac{1}{4}$ $P\left(\frac{A}{B}\right)=\frac{1}{2}$ and $P\left(\frac{B}{A}\right)=\frac{2}{3}$ then $P(B)$ is
- A$\frac{1}{3}$
- Correct answerB$\frac{2}{3}$
- C$\frac{1}{2}$
- D$\frac{1}{6}$
Solution
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, $P\left(\frac{E_1}{E_2}\right)=\frac{P\left(E_1 \cap E_2\right)}{P\left(E_2\right)}$
$\Rightarrow \quad P\left(\frac{B}{A}\right)=\frac{P(A \cap B)}{P(A)}$
$\begin{aligned} \Rightarrow \quad \frac{2}{3} \times \frac{1}{4} & =P(A \cap B) \Rightarrow \frac{1}{6}=P(A \cap B) \\ \Rightarrow \quad P\left(\frac{A}{B}\right) & =\frac{P(A \cap B)}{P(B)} \\ P(B) & =\frac{1}{6} \times \frac{4}{1} \\ P(B) & =\frac{2}{3}\end{aligned}$
We know that, $P\left(\frac{E_1}{E_2}\right)=\frac{P\left(E_1 \cap E_2\right)}{P\left(E_2\right)}$
$\Rightarrow \quad P\left(\frac{B}{A}\right)=\frac{P(A \cap B)}{P(A)}$
$\begin{aligned} \Rightarrow \quad \frac{2}{3} \times \frac{1}{4} & =P(A \cap B) \Rightarrow \frac{1}{6}=P(A \cap B) \\ \Rightarrow \quad P\left(\frac{A}{B}\right) & =\frac{P(A \cap B)}{P(B)} \\ P(B) & =\frac{1}{6} \times \frac{4}{1} \\ P(B) & =\frac{2}{3}\end{aligned}$
Hence, the correct answer is option: 'B'.