If A and B are two events with P(A cup B) = 0 | vedclass

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(C) We know that $P(A \cup B) = P(A) + P(B) - P(A \cap B)$.Rearranging the terms,we get $P(A) + P(B) = P(A \cup B) + P(A \cap B)$.Substituting the giv

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$1.2$

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(C) We know that $P(A \cup B) = P(A) + P(B) - P(A \cap B)$.Rearranging the terms,we get $P(A) + P(B) = P(A \cup B) + P(A \cap B)$.Substituting the given values,$P(A) + P(B) = 0.65 + 0.15 = 0.8$.We need to find $P(A^C) + P(B^C)$.Using the complement rule,$P(A^C) = 1 - P(A)$ and $P(B^C) = 1 - P(B)$.Therefore,$P(A^C) + P(B^C) = (1 - P(A)) + (1 - P(B)) = 2 - (P(A) + P(B))$.Substituting the sum $P(A) + P(B) = 0.8$,we get $2 - 0.8 = 1.2$.

If $A$ and $B$ are two events with $P(A \cup B) = 0.65$ and $P(A \cap B) = 0.15$,then find the value of $P(A^C) + P(B^C)$.

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