Q. Let $E^c$ denote the complement of an event $E$. Let $E, F, G$ be pairwise independent events with $P(G) > 0$ and $P(E\cap F\cap G) = 0$. Then $P(E^c \cap F^c | G)$ equals
Probability - Part 2
Solution:
We have
$\because E\cap F\cap G=\phi$
$P\left(E^{c}\cap F^{c}/G\right)=\frac{P\left(E^{c}\cap F^{c}\cap G\right)}{P\left(G\right)}$
$=\frac{P\left(G\right)-P\left(E\cap G\right)-P\left(G\cap F\right)}{P\left(G\right)}$
[From venn diagram $E^c \cap F^c \cap G = G - E \cap G - F \cap G$]
$=\frac{P\left(G\right)-P\left(E\right)P\left(G\right)-P\left(G\right)P\left(F\right)}{P\left(G\right)}$ [$\because$ E, F, G are pairwise independent]
$= 1- P (E) - P (F) = P (E^c) - P (F)$
