Probability 3 Question 22
22. If $X$ and $Y$ are two events such that $P(X / Y)=\frac{1}{2}, P(Y / X)=\frac{1}{3}$ and $P(X \cap Y) \frac{1}{6}$. Then, which of the following is/are correct?
(2012)
(a) $P(X \cup Y)=2 / 3$
(b) $X$ and $Y$ are independent
(c) $X$ and $Y$ are not independent
(d) $P\left(X^{c} \cap Y\right)=1 / 3$
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Solution:
- PLAN
(i) Conditional probability, i.e. $P(A / B)=\frac{P(A \cap B)}{P(B)}$
(ii) $P(A \cup B)=P(A)+P(B)-P(A \cap B)$
(iii) Independent event, then $P(A \cap B)=P(A) \cdot P(B)$
Here, $P(X / Y)=\frac{1}{2}, P \frac{Y}{X}=\frac{1}{3}$
and $P(X \cap Y)=6$
$\therefore \quad P \frac{X}{Y}=\frac{P(X \cap Y)}{P(Y)}$
$\Rightarrow \quad \frac{1}{2}=\frac{1 / 6}{P(Y)} \quad \Rightarrow \quad P(Y)=\frac{1}{3}$
$P \frac{Y}{X}=\frac{1}{3} \Rightarrow \frac{P(X \cap Y)}{P(X)}=\frac{1}{3}$
$\Rightarrow \quad \frac{1}{6}=\frac{1}{3} P(X)$
$\therefore \quad P(X)=\frac{1}{2}$
$P(X \cup Y)=P(X)+P(Y)-P(X \cap Y)$
$=\frac{1}{2}+\frac{1}{3}-\frac{1}{6}=\frac{2}{3}$
$P(X \cap Y)=\frac{1}{6}$ and $P(X) \cdot P(Y)=\frac{1}{2} \cdot \frac{1}{3}=\frac{1}{6}$
$\Rightarrow \quad P(X \cap Y)=P(X) \cdot P(Y)$
i.e. independent events
$\therefore \quad P\left(X^{c} \cap Y\right)=P(Y)-P(X \cap Y)$
$$ =\frac{1}{3}-\frac{1}{6}=\frac{1}{6} $$
