Question

If $A$ and $B$ are events with $P(A\cup B)=\frac{3}{4},P\left(A^{\prime }\right)=\frac{2}{3}$ and $P(A\cap B)=\frac{1}{4}$ then $P\left(B\right)$ is

• A. $\frac{1}{3}$
• B. $\frac{2}{3}$
• C. $\frac{3}{4}$
• D. $\frac{1}{4}$

Answer 1

Answer: (B)

$\frac{2}{3}$

Given $P(A\cup B)=\frac{3}{4},P\left(A^{\prime }\right)=\frac{2}{3},P(A\cap B)=\frac{1}{4}$.

We have to find $P\left(B\right)$.

Since $P\left(A^{\prime }\right)=\frac{2}{3},P\left(A\right)=1-P\left(A^{\prime }\right)=1-\frac{2}{3}=\frac{1}{3}$

Hence $P\left(A\right)=\frac{1}{3}$

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

Substituting the values we get,

$\frac{3}{4}=\frac{1}{3}+P\left(B\right)-\frac{1}{4}$

$\Rightarrow P\left(B\right)=\frac{3}{4}+\frac{1}{4}-\frac{1}{3}$

$=1-\frac{1}{3}$

$=\frac{2}{3}$

$\therefore P\left(B\right)=\frac{2}{3}$