If E_1 and E_2 are two events of the sample s | vedclass

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(C) We have, $P(E_1 \cap E_2) = P(E_1) \cdot P(E_2 | E_1)$.$	herefore P(E_1 \cap E_2) = \frac{1}{4} \cdot \frac{1}{3} = \frac{1}{12}$.Now, $P(E_1 | E

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$\frac{1}{5}$

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(C) We have, $P(E_1 \cap E_2) = P(E_1) \cdot P(E_2 | E_1)$.$	herefore P(E_1 \cap E_2) = \frac{1}{4} \cdot \frac{1}{3} = \frac{1}{12}$.Now, $P(E_1 | E_2) = \frac{P(E_1 \cap E_2)}{P(E_2)}$.$\Rightarrow \frac{1}{2} = \frac{1/12}{P(E_2)}$.$\Rightarrow P(E_2) = \frac{1}{12} 	imes 2 = \frac{1}{6}$.$	herefore P(\bar{E}_2) = 1 - P(E_2) = 1 - \frac{1}{6} = \frac{5}{6}$.Now, $P(E_1 | \bar{E}_2) = \frac{P(E_1 \cap \bar{E}_2)}{P(\bar{E}_2)} = \frac{P(E_1) - P(E_1 \cap E_2)}{P(\bar{E}_2)}$.$= \frac{\frac{1}{4} - \frac{1}{12}}{\frac{5}{6}} = \frac{\frac{3-1}{12}}{\frac{5}{6}} = \frac{\frac{2}{12}}{\frac{5}{6}} = \frac{1}{6} 	imes \frac{6}{5} = \frac{1}{5}$.

If $E_1$ and $E_2$ are two events of the sample space such that $P(E_1) = \frac{1}{4}$, $P(E_1 | E_2) = \frac{1}{2}$ and $P(E_2 | E_1) = \frac{1}{3}$, then $P(E_1 | \bar{E}_2) = $

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