If P(A) = 2/3,P(B) = 1/2 and P(A cup B) = 5/6 | vedclass

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(C) We use the addition theorem of probability: $P(A \cup B) = P(A) + P(B) - P(A \cap B)$.Substituting the given values: $\frac{5}{6} = \frac{2}{3} +

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(C) We use the addition theorem of probability: $P(A \cup B) = P(A) + P(B) - P(A \cap B)$.Substituting the given values: $\frac{5}{6} = \frac{2}{3} + \frac{1}{2} - P(A \cap B)$.Calculating the sum: $\frac{2}{3} + \frac{1}{2} = \frac{4+3}{6} = \frac{7}{6}$.Thus,$\frac{5}{6} = \frac{7}{6} - P(A \cap B)$,which implies $P(A \cap B) = \frac{7}{6} - \frac{5}{6} = \frac{2}{6} = \frac{1}{3}$.Since $P(A \cap B) 
eq 0$,the events are not mutually exclusive.Now,check for independence: $P(A) 	imes P(B) = \frac{2}{3} 	imes \frac{1}{2} = \frac{1}{3}$.Since $P(A \cap B) = P(A) 	imes P(B) = \frac{1}{3}$,the events $A$ and $B$ are independent.

If $P(A) = 2/3$,$P(B) = 1/2$ and $P(A \cup B) = 5/6$,then events $A$ and $B$ are

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