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(D) Given $P(\bar{A}) = \frac{1}{4}$,so $P(A) = 1 - \frac{1}{4} = \frac{3}{4}$.Given $P(\overline{A \cup B}) = \frac{1}{6}$,so $P(A \cup B) = 1 - \fra

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Independent but not equally likely

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(D) Given $P(\bar{A}) = \frac{1}{4}$,so $P(A) = 1 - \frac{1}{4} = \frac{3}{4}$.Given $P(\overline{A \cup B}) = \frac{1}{6}$,so $P(A \cup B) = 1 - \frac{1}{6} = \frac{5}{6}$.Using the formula $P(A \cup B) = P(A) + P(B) - P(A \cap B)$,we have $\frac{5}{6} = \frac{3}{4} + P(B) - \frac{1}{4}$.$\frac{5}{6} = \frac{2}{4} + P(B) = \frac{1}{2} + P(B)$.$P(B) = \frac{5}{6} - \frac{1}{2} = \frac{5-3}{6} = \frac{2}{6} = \frac{1}{3}$.Since $P(A) = \frac{3}{4}$ and $P(B) = \frac{1}{3}$,$P(A) 
eq P(B)$,so they are not equally likely.Check for independence: $P(A) 	imes P(B) = \frac{3}{4} 	imes \frac{1}{3} = \frac{1}{4}$.Since $P(A \cap B) = \frac{1}{4}$,we have $P(A \cap B) = P(A) 	imes P(B)$,which means $A$ and $B$ are independent.Thus,the events are independent but not equally likely.

If $A$ and $B$ are any two events such that $P(\overline{A \cup B}) = \frac{1}{6}$,$P(A \cap B) = \frac{1}{4}$,and $P(\bar{A}) = \frac{1}{4}$,then the events $A$ and $B$ are

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