Let $X$ and $Y$ be two events such that $P(X \mid Y)=\frac{1}{2}$, $P(Y / X)=\frac{1}{3}$ and $P(X \cap…
Let $X$ and $Y$ be two events such that $P(X \mid Y)=\frac{1}{2}$, $P(Y / X)=\frac{1}{3}$ and $P(X \cap Y)=\frac{1}{6}$. Which of the following is (are) correct?
$P(X \cup Y)=\frac{2}{3}$
$X$ and $Y$ are independent
$X$ and $Y$ are not independent
$P\left(X^{c} \cap Y\right)=\frac{1}{3}$
Solution
$\because P(X / Y)=\frac{P(X \cap Y)}{P(Y)} \Rightarrow \frac{1}{2}=\frac{1 / 6}{P(Y)} \Rightarrow P(Y)=\frac{1}{3}$
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.