Given P(A)=0.5, P(B)=0.4, P(A cap B)=0.3,then | vedclass

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(C) We are given $P(A)=0.5, P(B)=0.4$,and $P(A \cap B)=0.3$. We need to find $P(A^{\prime} \mid B^{\prime})$. Using the definition of conditional prob

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$2/3$

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(C) We are given $P(A)=0.5, P(B)=0.4$,and $P(A \cap B)=0.3$. We need to find $P(A^{\prime} \mid B^{\prime})$. Using the definition of conditional probability,$P(A^{\prime} \mid B^{\prime}) = \frac{P(A^{\prime} \cap B^{\prime})}{P(B^{\prime})}$. By De Morgan's Law,$A^{\prime} \cap B^{\prime} = (A \cup B)^{\prime}$,so $P(A^{\prime} \cap B^{\prime}) = P((A \cup B)^{\prime}) = 1 - P(A \cup B)$. First,calculate $P(A \cup B) = P(A) + P(B) - P(A \cap B) = 0.5 + 0.4 - 0.3 = 0.6$. Then,$P(A^{\prime} \cap B^{\prime}) = 1 - 0.6 = 0.4$. Also,$P(B^{\prime}) = 1 - P(B) = 1 - 0.4 = 0.6$. Therefore,$P(A^{\prime} \mid B^{\prime}) = \frac{0.4}{0.6} = \frac{2}{3}$.

Given $P(A)=0.5, P(B)=0.4, P(A \cap B)=0.3$,then $P(A^{\prime} / B^{\prime})$ is equal to

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