Question

Let $E^c$ denotes the complement of an event E. If E, F, G are pairwise independent evens with P(G) > 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 $P\left(E^c\right)-P\left(F\right)$

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