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

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(D) Given, $P(A \cup B)=0.8$ and $P(A \cap B)=0.3$. We know that $P(A \cup B) = P(A) + P(B) - P(A \cap B)$. Therefore, $P(A) + P(B) = P(A \cup B) + P(

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$0.9$

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(D) Given, $P(A \cup B)=0.8$ and $P(A \cap B)=0.3$. We know that $P(A \cup B) = P(A) + P(B) - P(A \cap B)$. Therefore, $P(A) + P(B) = P(A \cup B) + P(A \cap B) = 0.8 + 0.3 = 1.1$. We need to find $P(A^C) + P(B^C)$. Using the property $P(E^C) = 1 - P(E)$, we have: $P(A^C) + P(B^C) = (1 - P(A)) + (1 - P(B)) = 2 - (P(A) + P(B))$. Substituting the value of $P(A) + P(B)$: $P(A^C) + P(B^C) = 2 - 1.1 = 0.9$.

If $P(A \cup B)=0.8$ and $P(A \cap B)=0.3$, then $P(A^C)+P(B^C)$ is equal to

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