Question
Let $${E^c}$$ denote the complement of an event $$E.$$ Let $$E, F, G$$ be pairwise independent events with $$P\left( G \right) > 0$$ and $$P\left( {E \cap F \cap G} \right) = 0.$$ Then $$P\left( {{E^c} \cap \frac{{{F^c}}}{G}} \right)$$ 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 :
$$P\left( {{E^c}} \right) - P\left( {{F}} \right)$$
Solution :
$$\eqalign{
& P\left( {{E^c} \cap \frac{{{F^c}}}{G}} \right) = \frac{{P{{\left( {E \cup F} \right)}^c}}}{G} \cr
& 1 - P\left( {\frac{{E \cup F}}{G}} \right) \cr
& = 1 - P\left( {\frac{E}{G}} \right) - P\left( {\frac{F}{G}} \right) + P\left( {E \cap \frac{F}{G}} \right) \cr
& = 1 - P\left( E \right) - P\left( F \right) + O \cr} $$
(∵ $$E, F, G$$ are pairwise independent and
$$P\left( {E \cap F \cap G} \right) = 0$$
$$ \Rightarrow \,\,P\left( E \right).P\left( F \right) = 0\,\,{\text{as }}P\left( G \right) > 0)$$
$$ = P\left( {{E^C}} \right) - P\left( F \right)$$