If E_1 and E_2 are two events of a random exp | vedclass

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(C) Given $P(E_1) = \frac{1}{8}$,$P(E_1 \mid E_2) = \frac{1}{3}$,and $P(E_2 \mid E_1) = \frac{1}{4}$.Using the definition of conditional probability,$

Vedclass · Accepted Answer

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

Vedclass · Answer

(C) Given $P(E_1) = \frac{1}{8}$,$P(E_1 \mid E_2) = \frac{1}{3}$,and $P(E_2 \mid E_1) = \frac{1}{4}$.Using the definition of conditional probability,$P(E_1 \cap E_2) = P(E_2 \mid E_1) 	imes P(E_1) = \frac{1}{4} 	imes \frac{1}{8} = \frac{1}{32}$.Then,$P(E_2) = \frac{P(E_1 \cap E_2)}{P(E_1 \mid E_2)} = \frac{1/32}{1/3} = \frac{3}{32}$. (Matches $IV$)Now,$P(E_1 \cup E_2) = P(E_1)   P(E_2) - P(E_1 \cap E_2) = \frac{1}{8}   \frac{3}{32} - \frac{1}{32} = \frac{4 3-1}{32} = \frac{6}{32} = \frac{3}{16}$. (Matches $III$)We know $P(\bar{E}_2) = 1 - P(E_2) = 1 - \frac{3}{32} = \frac{29}{32}$.Also,$P(E_1 \cap \bar{E}_2) = P(E_1) - P(E_1 \cap E_2) = \frac{1}{8} - \frac{1}{32} = \frac{3}{32}$.Thus,$P(E_1 \mid \bar{E}_2) = \frac{P(E_1 \cap \bar{E}_2)}{P(\bar{E}_2)} = \frac{3/32}{29/32} = \frac{3}{29}$. (Matches $I$)Finally,$P(\bar{E}_1 \mid \bar{E}_2) = 1 - P(E_1 \mid \bar{E}_2) = 1 - \frac{3}{29} = \frac{26}{29}$. (Matches $II$)The correct matching is $A-III, B-IV, C-I, D-II$.

List-$I$	List-$II$
$A. P(E_1 \cup E_2)$	$I. \frac{3}{29}$
$B. P(E_2)$	$II. \frac{26}{29}$
$C. P(E_1 \mid \bar{E}_2)$	$III. \frac{3}{16}$
$D. P(\bar{E}_1 \mid \bar{E}_2)$	$IV. \frac{3}{32}$

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