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Let $\mathrm{X}$ and $\mathrm{Y}$ be two events such that $\mathrm{P}(\mathrm{X})=\frac{1}{3}, \mathrm{P}(\mathrm{X} \mid \mathrm{Y})=\frac{1}{2}$ and $\mathrm{P}(\mathrm{Y} \mid \mathrm{X})=\frac{2}{5}$. Then
$[A]$ $\mathrm{P}\left(\mathrm{X}^{\prime} \mid \mathrm{Y}\right)=\frac{1}{2}$ $[B]$ $\mathrm{P}(\mathrm{X} \cap \mathrm{Y})=\frac{1}{5}$ $[C]$ $\mathrm{P}(\mathrm{X} \cup \mathrm{Y})=\frac{2}{5}$ $[D]$ $\mathrm{P}(\mathrm{Y})=\frac{4}{15}$
$A,D$
$A,C$
$A,B$
$A,C,D$
Solution
$\mathrm{P}\left(\frac{\mathrm{X}}{\mathrm{Y}}\right)=\frac{\mathrm{P}(\mathrm{X} \cap \mathrm{Y})}{\mathrm{P}(\mathrm{Y})}=\frac{1}{2}$
$\mathrm{P}\left(\frac{\mathrm{Y}}{\mathrm{X}}\right)=\frac{\mathrm{P}(\mathrm{X} \cap \mathrm{Y})}{\mathrm{P}(\mathrm{X})}=\frac{2}{5}$
$\Rightarrow \mathrm{P}(\mathrm{X} \cap \mathrm{Y})=\frac{2}{15}$
$\mathrm{P}(\mathrm{Y})=\frac{4}{15} \text { also } \mathrm{P}(\mathrm{X} \cup \mathrm{Y})=\frac{7}{15}$
$\mathrm{P}\left(\frac{\mathrm{X}^{\prime}}{\mathrm{Y}}\right)=1-\mathrm{P}\left(\frac{\mathrm{X}}{\mathrm{Y}}\right)=1-\frac{1}{2}=\frac{1}{2}$
