If two events E_1 and E_2 are such that P(E_1 | vedclass

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(A) Given that $P(E_1 \cup E_2) = \frac{5}{8}$,$P(\bar{E}_1) = \frac{3}{4}$,and $P(E_2) = \frac{1}{2}$.First,find $P(E_1)$ using the complement rule:

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(A) Given that $P(E_1 \cup E_2) = \frac{5}{8}$,$P(\bar{E}_1) = \frac{3}{4}$,and $P(E_2) = \frac{1}{2}$.First,find $P(E_1)$ using the complement rule: $P(E_1) = 1 - P(\bar{E}_1) = 1 - \frac{3}{4} = \frac{1}{4}$.Using the addition theorem of probability: $P(E_1 \cup E_2) = P(E_1) + P(E_2) - P(E_1 \cap E_2)$.Substituting the values: $\frac{5}{8} = \frac{1}{4} + \frac{1}{2} - P(E_1 \cap E_2)$.$\frac{5}{8} = \frac{3}{4} - P(E_1 \cap E_2)$.$P(E_1 \cap E_2) = \frac{3}{4} - \frac{5}{8} = \frac{6-5}{8} = \frac{1}{8}$.Now,check for independence: $P(E_1) 	imes P(E_2) = \frac{1}{4} 	imes \frac{1}{2} = \frac{1}{8}$.Since $P(E_1 \cap E_2) = P(E_1) 	imes P(E_2)$,the events $E_1$ and $E_2$ are independent events.

If two events $E_1$ and $E_2$ are such that $P(E_1 \cup E_2) = \frac{5}{8}$,$P(\bar{E}_1) = \frac{3}{4}$,and $P(E_2) = \frac{1}{2}$,then $E_1$ and $E_2$ are:

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