Question

If $A$ and $B$ are two events such that $P ( A )=\frac{1}{3}, P ( B )=\frac{1}{5}$ and $P ( A \cup B )=\frac{1}{2}$, then $P \left( A \mid B ^{\prime}\right)+ P \left( B \mid A ^{\prime}\right)$ is equal to

• A. $\frac{3}{4}$
• B. $\frac{5}{8}$
• C. $\frac{5}{4}$
• D. $\frac{7}{8}$

Answer 1

The Correct Option is B

$P(A\cup B)=P(A)+P(B)-P(A\cap B)$
$\frac{1}{2}=\frac{1}{3}+\frac{1}{5}-P(A\cap B)$
$P(A\cap B)=\frac{1}{30}$
$P\left(\frac{A}{\overline{B}}\right)+P\left(\frac{B}{\overline{A}}\right)=\frac{P(A\cap \overline{B})}{P(\overline{B})}+\frac{P(B\cap \overline{A})}{P(\overline{A})}$
$=\frac{P(A)-P\left(A\cap B^{\prime }\right)}{1-P(B)}+\frac{P(B)-P(A\cap B)}{1-P(A)}$
$=\frac{\frac{1}{3}-\frac{1}{30}}{\frac{4}{5}}+\frac{\frac{1}{5}-\frac{1}{30}}{\frac{2}{3}}=\frac{5}{8}$