Let A and B be two events with P(A^{C}) = 0.3 | vedclass

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(A) Given: $P(A^{C}) = 0.3 \implies P(A) = 1 - 0.3 = 0.7$. $P(B) = 0.4 \implies P(B^{C}) = 1 - 0.4 = 0.6$. $P(A \cap B^{C}) = 0.5$. We know $P(A) = P(

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$\frac{1}{4}$

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(A) Given: $P(A^{C}) = 0.3 \implies P(A) = 1 - 0.3 = 0.7$. $P(B) = 0.4 \implies P(B^{C}) = 1 - 0.4 = 0.6$. $P(A \cap B^{C}) = 0.5$. We know $P(A) = P(A \cap B) + P(A \cap B^{C})$,so $P(A \cap B) = P(A) - P(A \cap B^{C}) = 0.7 - 0.5 = 0.2$. We need to find $P(B \mid A \cup B^{C}) = \frac{P(B \cap (A \cup B^{C}))}{P(A \cup B^{C})}$. Numerator: $P(B \cap (A \cup B^{C})) = P((B \cap A) \cup (B \cap B^{C})) = P(A \cap B) = 0.2$. Denominator: $P(A \cup B^{C}) = P(A) + P(B^{C}) - P(A \cap B^{C}) = 0.7 + 0.6 - 0.5 = 0.8$. Thus,$P(B \mid A \cup B^{C}) = \frac{0.2}{0.8} = \frac{1}{4}$.

Let $A$ and $B$ be two events with $P(A^{C}) = 0.3$,$P(B) = 0.4$,and $P(A \cap B^{C}) = 0.5$. Then $P(B \mid A \cup B^{C})$ is equal to

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