Question

If $P\left(A\right)=\frac{1}{4}$, $P\left(B\right)=\frac{1}{2}$, $P(A\cup B)=\frac{5}{8}$ then $P(A\cap B)$ is

• A. $\frac{1}{2}$
• B. $\frac{1}{8}$
• C. $\frac{3}{8}$
• D. None of these

Answer 1

Answer: (B)

$\frac{1}{8}$

Given $P\left(A\right)=\frac{1}{4},P\left(B\right)=\frac{1}{2},P(A\cup B)=\frac{5}{8}$.

We have to find $P(A\cap B)$.

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

Substitute the given values, we get

$\frac{5}{8}=\frac{1}{4}+\frac{1}{2}-P(A\cap B)$

$\Rightarrow P(A\cap B)=\frac{1}{4}+\frac{1}{2}-\frac{5}{8}$

$=\frac{2+4-5}{8}=\frac{1}{8}$

$\therefore P(A\cap B)=\frac{1}{8}$