Question

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

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

Answer 1

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

Given: $P\left(E^e\right)=1-P\left(E\right)$ and E, F, G are pairwise independent events, P(G)>0, $P(E\cap F\cap G)=0$

To find: $P(E^e\cap F^e|G)=?$

Sol:

$P(E^e\cap F^e|G)=\frac{P(E^e\cap F^e\cap G)}{P\left(G\right)}=\frac{P\left(E^e\right)P\left(F^e\right)P\left(G\right)}{P\left(G\right)}=P\left(E^e\right)P\left(F^e\right)⟹=P\left(E^e\right)[1-P(F\left)\right]=P\left(E^e\right)-P\left(E^e\right)P\left(F\right)⟹=P\left(E^e\right)-[1-P(E\left)\right]P\left(F\right)⟹=P\left(E^e\right)-P\left(F\right)+P\left(E\right)P\left(F\right)$

E, F, G are independent and $P(E\cap F\cap G)=0$

$⟹P\left(E\right)P\left(F\right)P\left(G\right)=0$ but $P\left(G\right)\ne 0$

$⟹P\left(E\right)P\left(F\right)=0$

Hence $P(E^e\cap F^e|G)=P\left(E^e\right)-P\left(F\right)$