Question

Probability $P\left(A\right)=\frac{4}{5},P\left(B^{\prime }\right)=\frac{2}{5}$ and $P(A\cap B)=\frac{1}{2}$, then $P(A\cap B^{\prime })$ is equal to

• A. $\frac{3}{10}$
• B. $\frac{5}{2}$
• C. $\frac{2}{5}$
• D. $\frac{5}{7}$

Answer 1

The correct option is A $\frac{3}{10}$

Given, $P\left(A\right)=\frac{4}{5},P\left(B^{\prime }\right)=\frac{2}{5}$
and $P(A\cap B)=\frac{1}{2}$
$\therefore P\left(B\right)=1-P\left(B^{\prime }\right)$
$=1-\frac{2}{5}=\frac{3}{5}$
We know that,
$P(A\cup B)=P\left(A\right)+P\left(B\right)-P(A\cap B)$
$=\frac{4}{5}+\frac{3}{5}-\frac{1}{2}=\frac{7}{5}-\frac{1}{2}$
$=\frac{14-5}{10}=\frac{9}{10}$
$\therefore P(A\cap B^{\prime })=P(A\cup B)-P\left(B\right)$
$=\frac{9}{10}-\frac{3}{5}$
$=\frac{9-6}{10}=\frac{3}{10}$