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

$E^c = \text{complement of } E$

$E,\ F,\ G \text{ pairwise independent}, \quad P(G) > 0$

$P(E \cap F \cap G) = 0$

$\text{To find: } P(E^c \cap F^c \mid G)$

$P(A \mid G) = \dfrac{P(A \cap G)}{P(G)}$

$\text{Take } A = E^c \cap F^c$

$N = P(E^c \cap F^c \cap G)$

$E^c \cap F^c \cap G = G \setminus \big[(E \cap G) \cup (F \cap G)\big]$

$N = P(G) - P(E \cap G) - P(F \cap G) + P(E \cap F \cap G)$

$P(E \cap F \cap G) = 0 \Rightarrow N = P(G) - P(E \cap G) - P(F \cap G)$

$P(E \cap G) = P(E)P(G), \quad P(F \cap G) = P(F)P(G)$

$N = P(G) - P(E)P(G) - P(F)P(G) = P(G)\,[1 - P(E) - P(F)]$

$\text{Within } G,\ E \text{ and } F \text{ never occur together}$

$P(E^c \cap F^c \mid G) = \dfrac{N}{P(G)} = 1 - P(E) - P(F)$

$1 - P(E) - P(F) = (1 - P(E)) - P(F) = P(E^c) - P(F)$

$\Rightarrow \text{option } (3)$

$\boxed{P(E^c \cap F^c \mid G) = P(E^c) - P(F)}$