If two events A and B are such that P(A cup B | vedclass

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(A) Given $P(A \cup B) = \frac{5}{6}$,$P(A \cap B) = \frac{1}{3}$,and $P(\bar{A}) = \frac{1}{2}$.Since $P(A) = 1 - P(\bar{A}) = 1 - \frac{1}{2} = \fra

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(A) Given $P(A \cup B) = \frac{5}{6}$,$P(A \cap B) = \frac{1}{3}$,and $P(\bar{A}) = \frac{1}{2}$.Since $P(A) = 1 - P(\bar{A}) = 1 - \frac{1}{2} = \frac{1}{2}$.Using the addition theorem: $P(A \cup B) = P(A) + P(B) - P(A \cap B)$.$\frac{5}{6} = \frac{1}{2} + P(B) - \frac{1}{3}$.$\frac{5}{6} = \frac{1}{6} + P(B) \Rightarrow P(B) = \frac{4}{6} = \frac{2}{3}$.Now,check for independence: $P(A) 	imes P(B) = \frac{1}{2} 	imes \frac{2}{3} = \frac{1}{3}$.Since $P(A \cap B) = P(A) 	imes P(B) = \frac{1}{3}$,the events $A$ and $B$ are independent.

If two events $A$ and $B$ are such that $P(A \cup B) = \frac{5}{6}$,$P(A \cap B) = \frac{1}{3}$,and $P(\bar{A}) = \frac{1}{2}$,then the events $A$ and $B$ are:

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