Let E^c denote the complement of an event E. | vedclass

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(C) We are given that $E, F, G$ are pairwise independent events,which implies $P(E \cap G) = P(E)P(G)$ and $P(F \cap G) = P(F)P(G)$.We need to find $P

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$P(E^c) - P(F)$

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(C) We are given that $E, F, G$ are pairwise independent events,which implies $P(E \cap G) = P(E)P(G)$ and $P(F \cap G) = P(F)P(G)$.We need to find $P(E^c \cap F^c \mid G) = \frac{P(E^c \cap F^c \cap G)}{P(G)}$.Using the property $P(A \cap B \cap C) = P(C) - P(A^c \cap C) - P(B^c \cap C) + P(A^c \cap B^c \cap C)$ is not direct,so we use the inclusion-exclusion principle on the complement:$P(E^c \cap F^c \cap G) = P(G \setminus (E \cup F)) = P(G) - P((E \cup F) \cap G) = P(G) - P((E \cap G) \cup (F \cap G))$.By the inclusion-exclusion principle for probabilities:$P((E \cap G) \cup (F \cap G)) = P(E \cap G) + P(F \cap G) - P(E \cap F \cap G)$.Given $P(E \cap F \cap G) = 0$,we have:$P((E \cap G) \cup (F \cap G)) = P(E)P(G) + P(F)P(G) - 0 = P(G)(P(E) + P(F))$.Substituting this back:$P(E^c \cap F^c \cap G) = P(G) - P(G)(P(E) + P(F)) = P(G)(1 - P(E) - P(F))$.Therefore,$P(E^c \cap F^c \mid G) = \frac{P(G)(1 - P(E) - P(F))}{P(G)} = 1 - P(E) - P(F)$.Since $P(E^c) = 1 - P(E)$,we have $1 - P(E) - P(F) = P(E^c) - P(F)$.

$X$	$0$	$1$	$2$	$3$	$4$
$P(X)$	$k$	$2k$	$4k$	$2k$	$k$

Let $E^c$ denote 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 $P(E^c \cap F^c \mid G)$ equals

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