Question

If two events $A$ and $B$ are such that $P\left(A '\right)=0.3,P\left(B\right)=0.4$ , and $P\left(A \cap B '\right)=0.5$ then $P\left(\frac{B}{A \cup B '}\right)$ equals $\frac{m}{n}$ , where $m$ and $n$ are relatively prime, then $m+n$ is equal to

Answer 1

It is given that $P\left(\bar{A}\right)=0.3$ and $P\left(B\right)=0.4$ and $P\left(A \cap B '\right)=0.5$
where, $P\left(A\right)=1-P\left(\bar{A}\right)=0.7$
$P\left(A \cap \bar{B}\right)=P\left(A\right)-P\left(A \cap B\right)=0.5$
$P\left(A \cap B\right)=0.5-0.3=0.2$
$P(B / A \cup \bar{B})=\frac{P\left(B \cap\left(A \cup \bar{B}^{\prime}\right)\right)}{P(A \cup \bar{B})}$
$=\frac{P \left(A \cap \bar{B} '\right)}{1 - P \left(B\right) + P \left(A \cap B\right)}$
$=\frac{0 . 2}{1 - 0 . 4 + 0 . 2}=\frac{1}{4}$
Here $m=1$ and $n=4$
$\therefore m+n=1+4=5$