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(

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

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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,$P(E_1 \cap E_2) = P(E_1) 	imes P(E_2)$.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-$A$	List-$B$
$(I)$ $A, B$ are mutually exclusive events	$(i)$ $P(A \cap B) = P(B) - P(\bar{A})$
$(II)$ $A, B$ are independent events	$(ii)$ $P(A) \leq P(B)$
$(III)$ $A \cap B = A$	$(iii)$ $P(\frac{\bar{A}}{B}) = 1 - P(A)$
$(IV)$ $A \cup B = S$	$(iv)$ $P(A \cup B) = P(A) + P(B)$
	$(v)$ $P(A) + P(B) = 2$

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