Question

If P $\left(A\right)=0.8,P\left(B\right)=0.5andP\left(\frac{B}{A}\right)=0.4,$ find
$P(A\cap B)$

$P\left(\frac{A}{B}\right)$

$P(A\cup B)$

Answer 1

Given, $P\left(A\right)=0.8,P\left(B\right)=0.5,P\left(\frac{B}{A}\right)=0.4\Rightarrow \frac{P(A\cap B)}{P\left(A\right)}=0.4[P\left(\frac{B}{A}\right)=\frac{P(A\cap B)}{P\left(A\right)}]\Rightarrow \frac{P(A\cap B)}{0.8}=0.4\Rightarrow P(A\cap B)=0.8\times 0.4=0.32$

$P\left(\frac{A}{B}\right)=\frac{P(A\cap B)}{P\left(B\right)}=\frac{0.32}{0.5}=\frac{32}{50}=0.64$

$FromtherelationP(A\cup B)=P\left(A\right)+P\left(B\right)-P(A\cap B)=0.8+0.5-0.32=1.3-0.32=0.98$