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(\bar{A\cap B}\right) = \frac{1}{6} $ (Since $\bar A \cap B = B - A$)