Question

Let $A$ and $B$ be two events with $P\left(A\right)=\frac{1}{3},P\left(B\right)=\frac{1}{6}$ and $P(A\cap B)=\frac{1}{12}$. What is $P\left(B|\overline{A}\right)$ equal to?

• A. $\frac{1}{4}$
• B. $\frac{1}{7}$
• C. $\frac{1}{8}$
• D. $\frac{1}{10}$

Answer 1

The correct option is A $\frac{1}{4}$

$P(A\cap B)=P\left(B\right)-P(\overline{A}\cap B)$ and $P\left(\overline{A}\right)=1-P\left(A\right)=\frac{2}{3}$

Also, $P(A\cup B)=P\left(A\right)+P\left(B\right)-P(A\cap B)⟹P\left(B\right)=\frac{1}{3}$

$\therefore P(\overline{A}\cap B)=\frac{1}{6}$

$⟹P\left(B|\overline{A}\right)=\frac{P(\overline{A}\cap B)}{P\left(\overline{A}\right)}=\frac{\frac{1}{6}}{\frac{2}{3}}=\frac{1}{4}$