Suppose that E_1 and E_2 are two events of a | vedclass

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(C) Given,$P(E_1) = \frac{1}{4}$,$P(E_2 / E_1) = \frac{1}{2}$,and $P(E_1 / E_2) = \frac{1}{4}$.$(A)$ Since $P(E_2 / E_1) = \frac{P(E_1 \cap E_2)}{P(E_

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$(A)$-(iv),$(B)$-(ii),$(C)$-(vi),$(D)$-$(i)$

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(C) Given,$P(E_1) = \frac{1}{4}$,$P(E_2 / E_1) = \frac{1}{2}$,and $P(E_1 / E_2) = \frac{1}{4}$.$(A)$ Since $P(E_2 / E_1) = \frac{P(E_1 \cap E_2)}{P(E_1)}$,we have $\frac{1}{2} = \frac{P(E_1 \cap E_2)}{1/4}$,so $P(E_1 \cap E_2) = \frac{1}{8}$.Also,$P(E_1 / E_2) = \frac{P(E_1 \cap E_2)}{P(E_2)} = \frac{1}{4}$.Substituting $P(E_1 \cap E_2) = \frac{1}{8}$,we get $\frac{1/8}{P(E_2)} = \frac{1}{4}$,which implies $P(E_2) = \frac{1}{2}$. Thus,$(A)$ matches with $(iv)$.$(B)$ $P(E_1 \cup E_2) = P(E_1) + P(E_2) - P(E_1 \cap E_2) = \frac{1}{4} + \frac{1}{2} - \frac{1}{8} = \frac{2+4-1}{8} = \frac{5}{8}$. Thus,$(B)$ matches with $(ii)$.$(C)$ $P(\bar{E}_1 / \bar{E}_2) = \frac{P(\bar{E}_1 \cap \bar{E}_2)}{P(\bar{E}_2)} = \frac{1 - P(E_1 \cup E_2)}{1 - P(E_2)} = \frac{1 - 5/8}{1 - 1/2} = \frac{3/8}{1/2} = \frac{3}{4}$. Thus,$(C)$ matches with $(vi)$.$(D)$ $P(E_1 / \bar{E}_2) = \frac{P(E_1 \cap \bar{E}_2)}{P(\bar{E}_2)} = \frac{P(E_1) - P(E_1 \cap E_2)}{1 - P(E_2)} = \frac{1/4 - 1/8}{1 - 1/2} = \frac{1/8}{1/2} = \frac{1}{4}$. Thus,$(D)$ matches with $(i)$.

List-$I$	List-$II$
$(A)$ $P(E_2)$	$(i)$ $1/4$
$(B)$ $P(E_1 \cup E_2)$	$(ii)$ $5/8$
$(C)$ $P(\bar{E}_1 / \bar{E}_2)$	$(iii)$ $1/8$
$(D)$ $P(E_1 / \bar{E}_2)$	$(iv)$ $1/2$
	$(v)$ $3/8$
	$(vi)$ $3/4$

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