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

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(B) 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)P(E_2) = \

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$A-iii, B-i, C-v, D-ii$

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(B) 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,$P(E_1 \cap E_2) = P(E_1)P(E_2) = \frac{1}{2}P(E_2)$.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} + P(E_2) - \frac{1}{2}P(E_2)$$\frac{2}{3} - \frac{1}{2} = \frac{1}{2}P(E_2)$$\frac{1}{6} = \frac{1}{2}P(E_2) \implies P(E_2) = \frac{1}{3}$. (Matches $A-iii$)Now,$P(E_1 \cap E_2) = \frac{1}{2} 	imes \frac{1}{3} = \frac{1}{6}$.$P(\frac{E_1}{E_2}) = \frac{P(E_1 \cap E_2)}{P(E_2)} = \frac{1/6}{1/3} = \frac{1}{2}$. (Matches $B-i$)$P(\frac{\bar{E}_2}{E_1}) = \frac{P(\bar{E}_2 \cap E_1)}{P(E_1)} = \frac{P(E_1) - P(E_1 \cap E_2)}{P(E_1)} = \frac{1/2 - 1/6}{1/2} = \frac{1/3}{1/2} = \frac{2}{3}$. (Matches $C-v$)$P(\bar{E}_1 \cup \bar{E}_2) = P(\overline{E_1 \cap E_2}) = 1 - P(E_1 \cap E_2) = 1 - \frac{1}{6} = \frac{5}{6}$. (Matches $D-ii$)Thus,the correct match is $A-iii, B-i, C-v, D-ii$.

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

$Face$	$1$	$2$	$3$	$4$	$5$	$6$
$P(F)$	$0.2$	$0.22$	$0.11$	$0.25$	$0.05$	$0.17$

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