Question

If A and B are events such that $P(\overline{A}\cup \overline{B})=\frac{3}{4},P(\overline{A}\cap \overline{B})=\frac{1}{4}$ and $P\left(A\right)=\frac{1}{3}$, then $P(\overline{A}\cap B)$ is

• A. $\frac{5}{12}$
• B. $\frac{3}{8}$
• C. $\frac{5}{8}$
• D. $\frac{1}{4}$

Answer 1

The correct option is A $\frac{5}{12}$

$P(\overline{A}\cup \overline{B})=\frac{3}{4}\Rightarrow P\left(\overline{A\cap B}\right)=\frac{3}{4}\Rightarrow P(A\cap B)=\frac{1}{4}$

$P(\overline{A}\cap \overline{B})=\frac{1}{4}\Rightarrow P\left(\overline{A\cup B}\right)=\frac{1}{4}\Rightarrow P(A\cup B)=\frac{3}{4}$

$P\left(A\right)=\frac{1}{3},P\left(\overline{A}\right)=\frac{2}{3}$

$P(\overline{A}\cap B)=P\left(B\right)-P(A\cap B)$

$P(A\cup B)=P\left(A\right)+P\left(B\right)-P(A\cap B)$

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

$P\left(B\right)=\frac{3}{4}-\frac{1}{12}=\frac{9-1}{12}=\frac{8}{12}=\frac{2}{3}$

$\therefore P(\overline{A}\cap B)=\frac{2}{3}-\frac{1}{4}=\frac{8-3}{12}=\frac{5}{12}$