Question

Let $E^c$ denote the complement of an event Let$E,F,G$ be pairwise independent events with $P\left(G\right)>0$ and $P(E\cap F\cap G)=0$ Then $P(E^c\cap F^c|G)$ equals

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

Answer 1

The correct option is C $P\left(E^c\right)-P\left(F\right)$

$P(\overline{E}\cap \overline{F}/G)=\frac{P(\overline{E}\cap \overline{F}\cap G)}{P\left(G\right)}=\frac{P\left(\overline{E}\right)P\left(\overline{F}\right)P\left(G\right)}{P\left(G\right)}=P\left(\overline{E}\right)P\left(\overline{F}\right)=P\left(\overline{E}\right)\{1-P\left(F\right)\}=P\left(\overline{E}\right)-P\left(\overline{E}\right)P\left(F\right)=P\left(\overline{E}\right)-\{1-P\left(E\right)\}P\left(F\right)=P\left(\overline{E}\right)-P\left(F\right)+P\left(E\right)P\left(F\right)$
now given that $E,F,GareindependentandP(E\cap F\cap G)=0\Rightarrow P\left(E\right)P\left(F\right)P\left(G\right)=0butP\left(G\right)\ne 0\Rightarrow P\left(E\right)P\left(F\right)=0henceP(\overline{E}\cap \overline{F}/G)=P\left(\overline{E}\right)-P\left(F\right)$