For two events A and B,if P(A cup B) = 5/6,P( | vedclass

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(B) Given $P(A^c) = 5/6$. Since $P(A) = 1 - P(A^c)$,we have $P(A) = 1 - 5/6 = 1/6$. Using the addition theorem of probability: $P(A \cup B) = P(A) + P

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Mutually exclusive

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(B) Given $P(A^c) = 5/6$. Since $P(A) = 1 - P(A^c)$,we have $P(A) = 1 - 5/6 = 1/6$. Using the addition theorem of probability: $P(A \cup B) = P(A) + P(B) - P(A \cap B)$. Substituting the given values: $5/6 = 1/6 + 2/3 - P(A \cap B)$. $5/6 = (1+4)/6 - P(A \cap B) = 5/6 - P(A \cap B)$. This implies $P(A \cap B) = 0$. Since the probability of the intersection of the two events is $0$,the events $A$ and $B$ are mutually exclusive.

For two events $A$ and $B$,if $P(A \cup B) = 5/6$,$P(A^c) = 5/6$,and $P(B) = 2/3$,then $A$ and $B$ are

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