Question

If $P\left(A\right)=0.8,P\left(B\right)=0.5$ and $P\left(B|A\right)=0.4$, find
(i) $P(A\cap B)$
(ii) $P\left(A|B\right)$
(iii) $P(A\cup B)$

Answer 1

It is given that $P\left(A\right)=0.8,P\left(B\right)=0.5$ and $P\left(B|A\right)=0.4$
(i) $P\left(B|A\right)=0.4$
$\therefore \frac{P(A\cap B)}{P\left(A\right)}=0.4$
$\Rightarrow \frac{P(A\cap B)}{0.8}=0.4$
$\Rightarrow P(A\cap B)=0.32$
(ii) $P\left(A|B\right)=\frac{P(A\cap B)}{P\left(B\right)}$
$\Rightarrow P\left(A|B\right)=\frac{0.32}{0.5}=0.64$
(iii) $P(A\cup B)=P\left(A\right)+P\left(B\right)-P(A\cap B)$
$\Rightarrow P(A\cup B)=0.8+0.5-0.32=0.98$