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.]
$=1-P(A)-P(B)=P\left(A^C\right)-P(B)$