If two events A and B are such that P(A^c) = | vedclass

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(C) 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 that $P(A) = P(A \c

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$1/4$

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(C) 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 that $P(A) = P(A \cap B) + P(A \cap B^c)$,so $0.7 = P(A \cap B) + 0.5 \implies P(A \cap B) = 0.2$.We need to find $P\left[ \frac{B}{A \cup B^c} \right] = \frac{P(B \cap (A \cup B^c))}{P(A \cup B^c)}$.Numerator: $P(B \cap (A \cup B^c)) = P((B \cap A) \cup (B \cap B^c)) = P(B \cap A) = 0.2$.Denominator: $P(A \cup B^c) = P(A) + P(B^c) - P(A \cap B^c) = 0.7 + 0.6 - 0.5 = 0.8$.Therefore,$P\left[ \frac{B}{A \cup B^c} \right] = \frac{0.2}{0.8} = \frac{1}{4}$.

If two events $A$ and $B$ are such that $P(A^c) = 0.3$,$P(B) = 0.4$,and $P(A \cap B^c) = 0.5$,then $P\left[ \frac{B}{A \cup B^c} \right]$ is equal to

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