Question

Let $A,B,C$ be pairwise independent events with $P(C)>0$ and $P(A\cap B\cap C)=0$. Then, $P\left(A^C\cap B^C\mid C\right)$ is equal to

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

Answer 1

Answer: (A)

$P\left(\frac{A^C\cap B^C}{C}\right)=\frac{P\left(A^C\cap B^C\cap C\right)}{P(C)}$
$=\frac{P(C)-P(A\cap C)-P(B\cap C)+P(A\cap B\cap C)}{P(C)}$ ...(i)
Given, $P(A\cap B\cap C)=0$ and $A,B,C$ are pairwise
$\therefore P(A\cap C)=P(A)\cdot P(C)$
and $P(B\cap C)-P(B)\cdot P(C)$
$\therefore P\left(\frac{A^C\cap B^C}{C}\right)$
$=\frac{P(C)-P(A)\cdot P(C)-P(B)\cdot P(C)+0}{P(C)}$[ from Eq.]
$=1-P(A)-P(B)=P\left(A^C\right)-P(B)$