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Let $A$ and $B$ be two events with $P\left(A^{C}\right)=0.3$ $P(B)=0.4 \quad$ and $\quad P\left(A \cap B^{C}\right)=0.5 . \quad$ Then
$P\left(B \mid A \cup B^{c}\right)$ is equal to
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$P\left(B \mid A \cup B^{c}\right)$ is equal to
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Verified Answer
The correct answer is:
$\frac{1}{4}$
$P\left(\frac{B}{A \cup B^{c}}\right)=\frac{P\left(B \cap\left(A \cup B^{e}\right)\right)}{P\left(A \cup B^{c}\right)}$
$=\frac{P(A \cap B)}{P(A)+P\left(B^{c}\right)-P\left(A \cap B^{c}\right)}$
$=\frac{P(A)-P\left(A \cap B^{c}\right)}{P(A)+P\left(B^{c}\right)-P\left(A \cap B^{c}\right)}$
$=\frac{0.7-0.5}{0.7+0.6-0.5}=\frac{0.2}{0.8}=\frac{1}{4}$
$=\frac{P(A \cap B)}{P(A)+P\left(B^{c}\right)-P\left(A \cap B^{c}\right)}$
$=\frac{P(A)-P\left(A \cap B^{c}\right)}{P(A)+P\left(B^{c}\right)-P\left(A \cap B^{c}\right)}$
$=\frac{0.7-0.5}{0.7+0.6-0.5}=\frac{0.2}{0.8}=\frac{1}{4}$
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