Given that $E$ and $F$ are events such that $P\left(E\right)=0.6,$
$P\left(F\right)=0.3$ and $P(E\cap F)=0.2,$ find $P\left(E|F\right)$ and $P\left(F|E\right)$

Given:$P\left(E\right)=0.6,P\left(F\right)=0.3,$
and $P(E\cap F)=P(F\cap E)=0.2$
Now, we know that
$P\left(E|F\right)=\frac{P(E\cap F)}{P\left(F\right)}$ .....$\left(1\right)$
Substituting the value of $P(E\cap F)\&P\left(F\right)$ in $\left(1\right).$
$P\left(E|F\right)=\frac{0.2}{0.3}$
$P\left(E|F\right)=\frac{2}{3}$
Thus,$P\left(E|F\right)=\frac{2}{3}$
$P\left(F|E\right)=\frac{P(E\cap F)}{P\left(E\right)}$ .....$\left(2\right)$
Substituting the values of $P(E\cap F)\&P\left(E\right)$ in $\left(2\right).$
$P\left(F|E\right)=\frac{0.2}{0.6}$
$P\left(F|E\right)=\frac{2}{6}=\frac{1}{3}$
Thus, $P\left(F|E\right)=\frac{1}{3}$