Question

A and B are events such that ${\displaystyle P(A\cup B)=\frac{3}{4}}$, ${\displaystyle P(A\cap B)=\frac{1}{4}}$, ${\displaystyle P\left(\overline{A}\right)=\frac{2}{3}}$ then ${\displaystyle P(\overline{A}\cap B)}$ is ______.

विकल्प

• ${\displaystyle \frac{5}{12}}$

• ${\displaystyle \frac{3}{8}}$

• ${\displaystyle \frac{5}{8}}$

• ${\displaystyle \frac{1}{4}}$

Answer 1

A and B are events such that ${\displaystyle P(A\cup B)=\frac{3}{4}}$, ${\displaystyle P(A\cap B)=\frac{1}{4}}$, ${\displaystyle P\left(\overline{A}\right)=\frac{2}{3}}$ then ${\displaystyle P(\overline{A}\cap B)}$ is ${\displaystyle \underline{\frac{5}{12}}}$.

Explanation:

We know that P(A ∪ B) = P(A) + P(B) – P(A ∩ B)

⇒ ${\displaystyle \frac{3}{4}=1-P\left(\overline{A}\right)+P\left(B\right)-\frac{1}{4}}$  ...${\displaystyle [\because P\left(A\right)=1-P\left(\overline{A}\right)]}$

⇒ 1 = ${\displaystyle 1-\frac{2}{3}+P\left(B\right)}$

⇒ P(B) = ${\displaystyle \frac{2}{3}}$;

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

= ${\displaystyle \frac{2}{3}-\frac{1}{4}-\frac{5}{12}}$