If E_1 and E_2 are two events of a sample spa | vedclass

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(D) Given that $P(E_1) = \frac{1}{4}$ and $P(E_2 \mid E_1) = \frac{1}{2}$.Using the definition of conditional probability,$P(E_1 \cap E_2) = P(E_1)

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$\frac{3}{4}$

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(D) Given that $P(E_1) = \frac{1}{4}$ and $P(E_2 \mid E_1) = \frac{1}{2}$.Using the definition of conditional probability,$P(E_1 \cap E_2) = P(E_1) 	imes P(E_2 \mid E_1) = \frac{1}{4} 	imes \frac{1}{2} = \frac{1}{8}$.We are also given $P(E_1 \mid E_2) = \frac{1}{4}$.By definition,$P(E_1 \mid E_2) = \frac{P(E_1 \cap E_2)}{P(E_2)}$,so $\frac{1}{4} = \frac{1/8}{P(E_2)}$.This implies $P(E_2) = \frac{1/8}{1/4} = \frac{1}{2}$.We need to find $P(\bar{E}_1 \mid E_2)$.Using the property of conditional probability,$P(\bar{E}_1 \mid E_2) = 1 - P(E_1 \mid E_2)$.Substituting the given value,$P(\bar{E}_1 \mid E_2) = 1 - \frac{1}{4} = \frac{3}{4}$.

If $E_1$ and $E_2$ are two events of a sample space such that $P(E_1) = \frac{1}{4}$,$P(E_2 \mid E_1) = \frac{1}{2}$,and $P(E_1 \mid E_2) = \frac{1}{4}$,then $P(\bar{E}_1 \mid E_2) = $

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