If P(A cup B)=0.8 and P(A cap B)=0.3,then P(b | vedclass

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(D) We know that for any two events $A$ and $B$,$P(A \cup B) = P(A) + P(B) - P(A \cap B)$.Given $P(A \cup B) = 0.8$ and $P(A \cap B) = 0.3$,we have:$P

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$0.9$

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(D) We know that for any two events $A$ and $B$,$P(A \cup B) = P(A) + P(B) - P(A \cap B)$.Given $P(A \cup B) = 0.8$ and $P(A \cap B) = 0.3$,we have:$P(A) + P(B) = P(A \cup B) + P(A \cap B) = 0.8 + 0.3 = 1.1$.We know that $P(\bar{A}) = 1 - P(A)$ and $P(\bar{B}) = 1 - P(B)$.Therefore,$P(\bar{A}) + P(\bar{B}) = (1 - P(A)) + (1 - P(B)) = 2 - (P(A) + P(B))$.Substituting the value of $P(A) + P(B)$:$P(\bar{A}) + P(\bar{B}) = 2 - 1.1 = 0.9$.

If $P(A \cup B)=0.8$ and $P(A \cap B)=0.3$,then $P(\bar{A})+P(\bar{B})$ is equal to:

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