Q62 / 180 · KCET 2022 · Mathematics
If $A$ and $B$ are two events such that $P(A)=\frac{1}{2}$, $P(B)=\frac{1}{2}$ and $P(A \mid B)=\frac{1}{4}$, then $P\left(A^{\prime} \cap B^{\prime}\right)$ is
- A$\frac{1}{4}$
- B$\frac{3}{16}$
- C$\frac{1}{12}$
- Correct answerD$\frac{1}{8}$
Solution
Given, $P(A)=\frac{1}{2}, P(B)=\frac{1}{2}$ and $P(A \mid B)=\frac{1}{4}$
$$
\begin{gathered}
P(A \mid B)=\frac{P(A \cap B)}{P(B)} \Rightarrow \frac{1}{4}=\frac{P(A \cap B)}{1 / 2} \\
\Rightarrow P(A \cap B)=\frac{1}{4} \times \frac{1}{2}=\frac{1}{8}
\end{gathered}
$$
$$
\begin{aligned}
& \text { Now, } P\left(A^{\prime} \cap B^{\prime}\right)=P(A \cup B)^{\prime}=1-P(A \cup B) \\
& =1-[P(A)+P(B)-P(A \cap B)]=1-\left[\frac{1}{2}+\frac{1}{2}-\frac{1}{8}\right] \\
& =1-1+\frac{1}{8}=\frac{1}{8}
\end{aligned}
$$
Hence, the correct answer is option: 'D'.