Let E_{1} and E_{2} be two events such that t | vedclass

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(NONE) Given $P(E_{1} \mid E_{2}) = \frac{P(E_{1} \cap E_{2})}{P(E_{2})} = \frac{1}{2} \implies P(E_{2}) = 2 \cdot P(E_{1} \cap E_{2}) = 2 \cdot \frac

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$P(E_{1} \cap E_{2}^{\prime}) = P(E_{1}) \cdot P(E_{2}^{\prime})$

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(NONE) Given $P(E_{1} \mid E_{2}) = \frac{P(E_{1} \cap E_{2})}{P(E_{2})} = \frac{1}{2} \implies P(E_{2}) = 2 \cdot P(E_{1} \cap E_{2}) = 2 \cdot \frac{1}{8} = \frac{1}{4}$.Given $P(E_{2} \mid E_{1}) = \frac{P(E_{1} \cap E_{2})}{P(E_{1})} = \frac{3}{4} \implies P(E_{1}) = \frac{4}{3} \cdot P(E_{1} \cap E_{2}) = \frac{4}{3} \cdot \frac{1}{8} = \frac{1}{6}$.Now,$P(E_{1}) \cdot P(E_{2}) = \frac{1}{6} \cdot \frac{1}{4} = \frac{1}{24} 
eq P(E_{1} \cap E_{2})$. Thus,$E_{1}$ and $E_{2}$ are not independent.Check option $(B)$: $P(E_{1}^{\prime} \cap E_{2}^{\prime}) = 1 - P(E_{1} \cup E_{2}) = 1 - (P(E_{1}) + P(E_{2}) - P(E_{1} \cap E_{2})) = 1 - (\frac{1}{6} + \frac{1}{4} - \frac{1}{8}) = 1 - (\frac{4+6-3}{24}) = 1 - \frac{7}{24} = \frac{17}{24}$.$P(E_{1}^{\prime}) \cdot P(E_{2}^{\prime}) = (1 - \frac{1}{6}) \cdot (1 - \frac{1}{4}) = \frac{5}{6} \cdot \frac{3}{4} = \frac{15}{24} = \frac{5}{8} 
eq \frac{17}{24}$.Check option $(C)$: $P(E_{1} \cap E_{2}^{\prime}) = P(E_{1}) - P(E_{1} \cap E_{2}) = \frac{1}{6} - \frac{1}{8} = \frac{4-3}{24} = \frac{1}{24}$.$P(E_{1}) \cdot P(E_{2}^{\prime}) = \frac{1}{6} \cdot (1 - \frac{1}{4}) = \frac{1}{6} \cdot \frac{3}{4} = \frac{3}{24} = \frac{1}{8} 
eq \frac{1}{24}$.Check option $(D)$: $P(E_{1}^{\prime} \cap E_{2}) = P(E_{2}) - P(E_{1} \cap E_{2}) = \frac{1}{4} - \frac{1}{8} = \frac{1}{8}$.Since $P(E_{1}^{\prime} \cap E_{2}) = P(E_{2}) - P(E_{1} \cap E_{2}) = \frac{1}{8}$,and $P(E_{1}) \cdot P(E_{2}) = \frac{1}{24}$,this is not equal. Note: The question implies checking for independence properties. If $E_{1}, E_{2}$ are independent,then $E_{1}^{\prime}, E_{2}$ are independent. Here they are not.

Let $E_{1}$ and $E_{2}$ be two events such that the conditional probabilities $P(E_{1} \mid E_{2}) = \frac{1}{2}$,$P(E_{2} \mid E_{1}) = \frac{3}{4}$ and $P(E_{1} \cap E_{2}) = \frac{1}{8}$. Then:

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