E_1 and E_2 are two independent events of a r | vedclass

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(A) Given: $P(E_1) = \frac{1}{2}$ and $P(E_1 \cup E_2) = \frac{2}{3}$.Since $E_1$ and $E_2$ are independent events, $P(E_1 \cap E_2) = P(E_1) 	imes P

Vedclass · Accepted Answer

$A-III, B-IV, C-I, D-II$

Vedclass · Answer

(A) Given: $P(E_1) = \frac{1}{2}$ and $P(E_1 \cup E_2) = \frac{2}{3}$.Since $E_1$ and $E_2$ are independent events, $P(E_1 \cap E_2) = P(E_1) 	imes P(E_2)$.Let $P(E_2) = x$. Then $P(E_1 \cap E_2) = \frac{1}{2}x$.Using the formula $P(E_1 \cup E_2) = P(E_1) + P(E_2) - P(E_1 \cap E_2)$:$\frac{2}{3} = \frac{1}{2} + x - \frac{x}{2}$$\frac{2}{3} = \frac{1}{2} + \frac{x}{2}$$\frac{x}{2} = \frac{2}{3} - \frac{1}{2} = \frac{4-3}{6} = \frac{1}{6}$$x = \frac{1}{3}$. Thus, $P(E_2) = \frac{1}{3}$. $(A ightarrow III)$Now, $P(E_1 \cap E_2) = \frac{1}{2} 	imes \frac{1}{3} = \frac{1}{6}$.$P(E_1 | E_2) = \frac{P(E_1 \cap E_2)}{P(E_2)} = \frac{1/6}{1/3} = \frac{1}{2}$. $(B ightarrow IV)$$P(\bar{E}_2 | E_1) = 1 - P(E_2 | E_1) = 1 - P(E_2) = 1 - \frac{1}{3} = \frac{2}{3}$. $(C ightarrow I)$$P(\bar{E}_1 \cup \bar{E}_2) = P(\overline{E_1 \cap E_2}) = 1 - P(E_1 \cap E_2) = 1 - \frac{1}{6} = \frac{5}{6}$. $(D ightarrow II)$Therefore, the correct match is $A-III, B-IV, C-I, D-II$.

List-$I$	List-$II$
$A. P(E_2) =$	$I. 2/3$
$B. P(E_1 \| E_2) =$	$II. 5/6$
$C. P(\bar{E}_2 \| E_1) =$	$III. 1/3$
$D. P(\bar{E}_1 \cup \bar{E}_2) =$	$IV. 1/2$

List-$I$	List-$II$
$(I)$ Probability of $(X_2 \geq Y_2)$ is	$(P)$ $\frac{3}{8}$
$(II)$ Probability of $(X_2 > Y_2)$ is	$(Q)$ $\frac{11}{16}$
$(III)$ Probability of $(X_3 = Y_3)$ is	$(R)$ $\frac{5}{16}$
$(IV)$ Probability of $(X_3 > Y_3)$ is	$(S)$ $\frac{355}{864}$
	$(T)$ $\frac{77}{432}$

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