Question

If $X$ and $Y$ are two events such that $P (X | Y )=\frac{1}{2}, P(X | Y)=\frac{1}{3.} \, and \, P(X \cap Y)\frac{1}{6}$ Then, which of the following is/are correct?

• A. $P(X \cup Y) = \frac {2}{3}$
• B. $X$ and $Y$ are independent
• C. $X$ and $Y$ are not independent
• D. $P(X^c \cap Y) = \frac{1}{3}$

Answer 1

Answer: (A), (B)

PLAN
(i) Conditional probability, i.e. $P(A/B) =\frac{P(A \cap B)}{P(B)}$
(ii) $P (A \cup B) = P(A) + P(B) -P (A \cup B)$
(iii) Independent event, then $P(A \cap B) = P(A)-P(B) $
Here, $P(X/Y) =\frac{1}{2}, P\bigg(\frac{Y}{X}\bigg)=\frac{1}{3} $ and $ P(X \cap \, Y)=6$
$\therefore \, \, \, \, \, \, P\bigg(\frac{x}{y}\bigg) =\frac{P(X \cap Y)}{P(Y)}$
$\Rightarrow \, \, \, \, \, \, \, \frac{1}{2}=\frac{1/6}{P(Y)} \, \, \Rightarrow \, \, P(Y)=\frac{1}{3} \, \, \, \, \, \, \, $ ....( i)
$ \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, P\bigg(\frac{Y}{X})=\frac{1}{3} \, \, \Rightarrow \, \, \frac{P(X \cap Y)}{P(X)}=\frac{1}{3}$
$\Rightarrow \, \, \, \, \, \, \frac{1}{6}=\frac{1}{3}P(X)$
$\therefore \, \, \, \, \, P(X)=\frac{1}{2} \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, $ ...(ii)
$P(X \cup Y ) = P(X) + P(Y) - P ( X \cap Y ) $
$ \, \, \, \, \, \, \, \, \, = \frac{1}{2}+\frac{1}{3}-\frac{1}{6}=\frac{2}{3} \, \, \, \, \, \, \, \, \, \, \, \, \, $ ...(iii)
$P(X \cap Y)=\frac{1}{6} $ and $ \, P(X).P(Y)=\frac{1}{2}.\frac{1}{3}=\frac{1}{6}$
$\Rightarrow \, \, \, \, P(X \cap Y) = P(X).P(Y)$
i.e. independent events
$\therefore \, \, \, \, \, \, P(X^c \cap Y)=P(Y)-P(X \cap Y)$
$\frac{1}{3}-\frac{1}{6}=\frac{1}{6}$