Question

If A and B are any two events having $P(A\cup B)=\frac{1}{2}andP\left(\overline{A}\right)=\frac{2}{3}$ then probability of $\overline{A}\cap B$ is
(a) $\frac{1}{2}$

(b) $\frac{2}{3}$

(c) $\frac{1}{6}$

(d) $\frac{1}{3}$

Answer 1

For any 2 events A and B
$$\begin{gathered} P(A\cup B)=\frac{1}{2} \\ P\left(\overline{A}\right)=\frac{2}{3}=\frac{2}{3} \\ \mathrm{i}.\mathrm{e}.P\left(A\right)=1-\frac{2}{3}=\frac{1}{3} \\ P\left(\overline{A}\cap B\right)=P(A\cup B)-P\left(A\right) \\ =\frac{1}{2}-\frac{1}{3} \\ =\frac{3-2}{6}=\frac{1}{6} \\ \mathrm{i}.\mathrm{e}.P\left(\overline{A}\cap B\right)=\frac{1}{6} \end{gathered}$$
Hence, the correct answer is option C.