Question

Let A and B be independent events with P (A ) = 0.3 and P (B) = 0.4 . Find
$P(A\cap B)$

$P(A\cup B)$

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

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

Answer 1

It is given that P A ) = 0.3 and P(B) = 0.4
If A and B are independent events, then
$P(A\cap B)=P\left(A\right)\times P\left(B\right)=0.3\times 0.4=0.12$

We know that
$P(A\cup B)=P\left(A\right)+P\left(B\right)-P(A\cap B)=P\left(A\right)+P\left(B\right)-P\left(A\right)P\left(B\right)=0.3+0.4-0.3\times 0.4=0.7-0.12=0.58$

We know that $P\left(\frac{A}{B}\right)=\frac{P(A\cap B)}{P\left(B\right)}=\frac{P\left(A\right)P\left(B\right)}{P\left(B\right)}\Rightarrow P\left(\frac{A}{B}\right)=\frac{0.12}{0.4}=0.3$

We know that $P\left(\frac{B}{A}\right)=\frac{P(B\cap A)}{P\left(A\right)}=\frac{P\left(B\right)P\left(A\right)}{P\left(A\right)}\Rightarrow P\left(\frac{B}{A}\right)=\frac{0.12}{0.3}=0.4$