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Let $E^c$ denotes the complement of an event $E$. Let $E, F, G$ be pairwise independent events with $P(G)>0$ and $P(E \cap F \cap G)=0$. Then, $P\left(E^c \cap F^c \mid G\right)$ equals
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Verified Answer
The correct answer is:
$P\left(E^c\right)-P(F)$
$P\left(E^c\right)-P(F)$
$$
\text { } \begin{aligned}
P\left(\frac{E^c \cap F^c}{G}\right) & =\frac{P\left(E^c \cap F^c \cap G\right)}{P(G)} \\
& =\frac{P(G)-P(E \cap G)-P(G \cap F)}{P(G)} \\
& =\frac{P(G)[1-P(E)-P(F)]}{P(G)} \\
& =1-P(E)-P(F) \\
& =P\left(E^c\right)-P(F) .
\end{aligned}
$$
\text { } \begin{aligned}
P\left(\frac{E^c \cap F^c}{G}\right) & =\frac{P\left(E^c \cap F^c \cap G\right)}{P(G)} \\
& =\frac{P(G)-P(E \cap G)-P(G \cap F)}{P(G)} \\
& =\frac{P(G)[1-P(E)-P(F)]}{P(G)} \\
& =1-P(E)-P(F) \\
& =P\left(E^c\right)-P(F) .
\end{aligned}
$$
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