Question

If $A$ and $B$ are events such that
$P(A\cup B)=\frac{3}{4},P(A\cap B)=\frac{1}{4},P\left(\overline{A}\right)=\frac{2}{3}$, then $P(\overline{A}\cap B)$ is

• A. $\frac{5}{12}$
• B. $\frac{3}{8}$
• C. $\frac{5}{8}$
• D. $\frac{1}{4}$

Answer 1

Answer: (A)

$\frac{5}{12}$

Given, $P(A\cup B)=\frac{3}{4},P(A\cap B)=\frac{1}{4},P\left(\overline{A}\right)=\frac{2}{3}$

We need to find $P(\overline{A}\cap B)$.

Therefore, $P\left(B\right)=\frac{2}{3}$

We know $P\left(B\right)=P(A\cup B)+P(A\cap B)-P\left(A\right)$

Therefore, $P(\overline{A}\cap B)=\frac{2}{3}-\frac{1}{4}=\frac{5}{12}$