Question

Let $E,F,G$ be pairwise independent events with $P\left(G\right)>0$ and $P(E\cap F\cap G)=0$. Then $P(E^{\prime }\cap F^{\prime }|G)$ equals

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

Answer 1

Answer: (C)

$P\left(E^{\prime }\right)-P\left(F\right)$

$P(E^{\prime }\cap F^{\prime }|G)=\frac{P(E^{\prime }\cap F^{\prime }\cap G)}{P\left(G\right)}$
$=\frac{P\left(G\right)-P\left[\right(E\cap G)\cup (F\cap G\left)\right]}{P\left(G\right)}$
$=\frac{P\left(G\right)-P(E\cap G)-P(F\cap G)}{P\left(G\right)}=\frac{P\left(G\right)[1-P(E)-P(F\left)\right]}{P\left(G\right)}$
$=P\left(E^{\prime }\right)-P\left(F\right)$.