Conditional Probability of Complements of Pairwise Independent Events

By De Morgan's Law, $E^c \cap F^c = (E \cup F)^c$. Therefore:
$P(E^c \cap F^c \mid G) = 1 - P(E \cup F \mid G)$

By the addition theorem of probability conditioned on $G$:
$P(E \cup F \mid G) = P(E \mid G) + P(F \mid G) - P(E \cap F \mid G)$

1. Since $E, G$ and $F, G$ are pairwise independent:
   $P(E \mid G) = P(E), \quad P(F \mid G) = P(F)$

2. The conditional joint probability is:
   $P(E \cap F \mid G) = \frac{P(E \cap F \cap G)}{P(G)} = \frac{0}{P(G)} = 0$

Substituting these back:
$P(E \cup F \mid G) = P(E) + P(F) - 0 = P(E) + P(F)$

Therefore:
$P(E^c \cap F^c \mid G) = 1 - [P(E) + P(F)] = [1 - P(E)] - P(F) = P(E^c) - P(F)$

Correct Answer: C