Question

If $A$ and $B$ are any two events having $P(A\cup B)=\frac{1}{2}$ and $P(\bar{A})=\frac{2}{3}$, then the probability of $\bar{A}\cap B$ is

• A. $\frac{1}{2}$
• B. $\frac{2}{3}$
• C. $\frac{1}{6}$
• D. $\frac{1}{3}$

Answer 1

Answer: (C)

We have $P\left(A\cup B\right)=\frac{1}{2}$
$\Rightarrow P\left(A\cup \left(B-A\right)\right)=\frac{1}{2}$
$\Rightarrow P\left(A\right)+P\left(B-A\right)=\frac{1}{2}$
(Since $A$ and $B-A$ are mutually exclusive)
$\Rightarrow 1-P\left(\bar{A}\right)+P\left(B-A\right)=\frac{1}{2}$
$\Rightarrow 1-\frac{2}{3}+P\left(B-A\right)=\frac{1}{2}$
$\Rightarrow P\left(B-A\right)=\frac{1}{6}$
$\Rightarrow P\left(\overline{A\cap B}\right)=\frac{1}{6}$ (Since $\bar{A}\cap B=B-A$)