For any two events $A$ and $B$, if $P(\bar{A}) = \frac{1}{2}, P(\bar B) = \frac{2}{3}$ and $P(A \cap B) = \frac{1}{4}$, then $P\left(\frac{\bar{A}}{B}\right)$ equals: |
$\frac{3}{8}$ $\frac{8}{9}$ $\frac{5}{8}$ $\frac{1}{4}$ |
$\frac{1}{4}$ |
$P(\bar{A}) = \frac{1}{2} \Rightarrow P(A) = \frac{1}{2}$ $P(\bar{B}) = \frac{2}{3} \Rightarrow P(B) = \frac{1}{3}$ $P(A \cap B) = \frac{1}{4}$ $P(\bar{A} \cap B) = P(B) - P(A \cap B)$ $= \frac{1}{3} - \frac{1}{4} = \frac{1}{12}$ $P\left(\frac{\bar{A}}{B}\right) = P(\bar{A} \mid B)$ $= \frac{P(\bar{A} \cap B)}{P(B)}$ $= \frac{\frac{1}{12}}{\frac{1}{3}} = \frac{1}{4}$ The value is $\frac{1}{4}$. |