Let X and Y be two events such that P(X)=frac | vedclass

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(A) Given $P(X) = \frac{1}{3}$,$P(X \mid Y) = \frac{1}{2}$,and $P(Y \mid X) = \frac{2}{5}$.Using the definition of conditional probability,$P(Y \mid X

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$A, D$

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(A) Given $P(X) = \frac{1}{3}$,$P(X \mid Y) = \frac{1}{2}$,and $P(Y \mid X) = \frac{2}{5}$.Using the definition of conditional probability,$P(Y \mid X) = \frac{P(X \cap Y)}{P(X)}$.$\frac{2}{5} = \frac{P(X \cap Y)}{1/3} \Rightarrow P(X \cap Y) = \frac{2}{5} 	imes \frac{1}{3} = \frac{2}{15}$.Now,$P(X \mid Y) = \frac{P(X \cap Y)}{P(Y)}$.$\frac{1}{2} = \frac{2/15}{P(Y)} \Rightarrow P(Y) = \frac{2}{15} 	imes 2 = \frac{4}{15}$. (Option $D$ is correct)For $P(X^{\prime} \mid Y)$,we have $P(X^{\prime} \mid Y) = 1 - P(X \mid Y) = 1 - \frac{1}{2} = \frac{1}{2}$. (Option $A$ is correct)For $P(X \cup Y)$,$P(X \cup Y) = P(X) + P(Y) - P(X \cap Y) = \frac{1}{3} + \frac{4}{15} - \frac{2}{15} = \frac{5+4-2}{15} = \frac{7}{15}$.Thus,options $A$ and $D$ are correct.

List-$I$	List-$II$
$A$. $P(E_2)$	$(i)$ $\frac{1}{2}$
$B$. $P(\frac{E_1}{E_2})$	$(ii)$ $\frac{5}{6}$
$C$. $P(\frac{\bar{E}_2}{E_1})$	$(iii)$ $\frac{1}{3}$
$D$. $P(\bar{E}_1 \cup \bar{E}_2)$	$(iv)$ $\frac{1}{6}$
	$(v)$ $\frac{2}{3}$

$Face$	$1$	$2$	$3$	$4$	$5$	$6$
$P(F)$	$0.1$	$0.24$	$0.19$	$0.18$	$0.15$	$0.14$

$X$	$0$	$1$	$2$	$3$	$4$
$P(X)$	$k$	$2k$	$4k$	$2k$	$k$

Let $X$ and $Y$ be two events such that $P(X)=\frac{1}{3}$,$P(X \mid Y)=\frac{1}{2}$ and $P(Y \mid X)=\frac{2}{5}$. Then:
$A) P(X^{\prime} \mid Y)=\frac{1}{2}$
$B) P(X \cap Y)=\frac{1}{5}$
$C) P(X \cup Y)=\frac{2}{5}$
$D) P(Y)=\frac{4}{15}$

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Let $X$ and $Y$ be two events such that $P(X)=\frac{1}{3}$,$P(X \mid Y)=\frac{1}{2}$ and $P(Y \mid X)=\frac{2}{5}$. Then:$A) P(X^{\prime} \mid Y)=\frac{1}{2}$$B) P(X \cap Y)=\frac{1}{5}$$C) P(X \cup Y)=\frac{2}{5}$$D) P(Y)=\frac{4}{15}$