Mathematics - Probability Question with Solution | TestHub

$\because P(X/Y)=\frac{P(X\cap Y)}{P(Y)}\Rightarrow \frac{1}{2}=\frac{1/6}{P(Y)}\Rightarrow P(Y)=\frac{1}{3}$ Similarly, $\mathrm{P}(Y/X)=\frac{P(X\cap Y)}{P(X)}$ $\Rightarrow \frac{1}{3}=\frac{1/6}{P(X)}\Rightarrow P(X)=\frac{1}{2}$ (a) $P(X\cup Y)=P(X)+P(Y)-P(X\cap Y)=\frac{1}{2}+\frac{1}{3}-\frac{1}{6}=\frac{2}{3}$ $\therefore$ (a) is true. (b) $\because P(X\cap Y)=P(X)P(Y)$ $\Rightarrow X$ and $Y$ are independent events. $\therefore$ (b) is true. But (c) is not true. (d) $\mathrm{P}\left(X^C\cap Y\right)=P\left(X^C\right)\times P(Y)=\frac{1}{2}\times \frac{1}{3}=\frac{1}{6}$ $\therefore$ (d) is not true.