Mathematics - Probability Question with Solution | TestHub

Using formula$P\left(\mathrm{X }|\mathrm{ Y}\right)=\frac{P\left(X\cap Y\right)}{P\left(Y\right)}$and$P\left(\mathrm{Y }|\mathrm{ X}\right)=\frac{P\left(Y\cap X\right)}{P\left(X\right)}$
$P\left(X\right)=\frac{1}{3};\frac{P\left(X\cap Y\right)}{P\left(Y\right)}=\frac{1}{2};\frac{P\left(Y\cap X\right)}{P\left(X\right)}=\frac{2}{5}$
From this information, we get
$P\left(X\cap Y\right)=\frac{2}{15};P\left(Y\right)=\frac{4}{15}$
Using formula$P(X\cup Y)=P(X)+P(Y)-P(X\cap Y)$
$\therefore P\left(X\cup Y\right)=\frac{1}{3}+\frac{4}{15}-\frac{2}{15}=\frac{7}{15}$
$P\left(\overline{X}|\mathrm{ Y}\right)=\frac{P\left(\stackrel{-}{X}\cap Y\right)}{P\left(Y\right)}=\frac{P\left(Y\right)-P\left(X\cap Y\right)}{P\left(Y\right)}$
$\Rightarrow P\left(\overline{X}|\mathrm{ Y}\right)=1-\frac{\frac{2}{15}}{\frac{4}{15}}=\frac{1}{2}$