Question

Let $E^c$ denotes the complement of an event $E$. If $E, F, G$ are pairwise independent events with $P (G) > 0$ and $P(E \cap F \cap G)=0$ then , $P(E^c \cap F^c|G) equals$

• A. $P(E^c)+P(F^c)$
• B. $P(E^c)-P(F^c)$
• C. $P(E^c)-P(F)$
• D. $P(E)-P(F^c)$

Answer 1

Answer: (C)

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

Answer 2

$P\bigg(\frac{E^c \cap F^c}{G}\bigg)=\frac{P(E^c \cap F^c \cap G)}{P(G)}$
$=\frac{P(G)-P(E \cap G) -P(G \cap F)}{P(G)}$
$=\frac{P(G)[P(E)-P(F)]}{P(G)} \, \, \, \, \, \, [\because \, P(G) \ne 0]$
$=1-P(E)-P(F)=P(E^c)-P(F)$