Q61 / 120 · VITEEE 2017 · Mathematics
$A$ and $B$ are events such that $\mathrm{P}(\mathrm{A} \cup \mathrm{B})=3 / 4$, $P(A \cap B)=1 / 4, P(\bar{A})=2 / 3$ then $P(\bar{A} \cap B)$ is
- Correct answerA$5 / 12$
- B$3 / 8$
- C$5 / 8$
- D$1 / 4$
Solution
$$
\begin{array}{l}
P(A \cup B)=P(A)+P(B)-P(A \cap B) \\
\Rightarrow \frac{3}{4}=1-P(\bar{A})+P(B)-\frac{1}{4} \\
\Rightarrow 1=1-\frac{2}{3}+P(B) \Rightarrow P(B)=\frac{2}{3} \\
\text { Now, } P(\bar{A} \cap B)=P(B)-P(A \cap B)=\frac{2}{3}- \\
\frac{1}{4}=\frac{5}{12}
\end{array}
$$
Hence, the correct answer is option: 'A'.