Question

Let two events $A\&B$ are such that $P\left(A^c\right)=0.3,P(B)=0.4\&P\left(A{B}^c\right)=0.5$. If $P\left[B/\left(AU{B}^c\right)\right]$ is expressed as $\frac{p}{q}$ in lowest rational. Find $(p+q)$.

Answer 1

Answer: 5

$P\left(A^c\right)=0.3\Rightarrow P(A)=0.7$
$P(B)=0.4;P\left(A\cap B^c\right)=0.5$
$\text{ or }P(A)-P(A\cap B)=0.5$
$\therefore P(A\cap B)=0.7-0.5=0.2$
$\text{ Now }P\left(B/A\cup B^c\right)=\frac{P\left(B\cap \left(A\cup B^c\right)\right)}{P\left(A\cup B^c\right)}$
$=\frac{P(B\cap A)+P\left(B\cap B^c\right)}{P\left(A\cup B^c\right)}=\frac{P(A\cap B)}{P(A)+P\left(B^c\right)-P\left(A\cap B^c\right)}$
$=\frac{P(A\cap B)}{P(A)+1-P(B)-(P(A)-P(A\cap B))}=\frac{P(A\cap B)}{1-P(B)+P(A\cap B)}=\frac{0.2}{1-0.4+0.2}$
$=\frac{0.2}{0.8}=\frac{1}{4}=\frac{p}{q}\Rightarrow (p+q)=1+4=5$