Question

Let $E’$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 $P\left(E’\cap F’|G\right)$equals

• A. $P\left(E’\right)+P\left(F’\right)$
• B. $P\left(E’\right)-P\left(F’\right)$
• C. $P\left(E’\right)-P\left(F\right)$
• D. $P\left(E\right)–P\left(F’\right)$

Answer 1

The correct option is C

$P\left(E’\right)-P\left(F\right)$

Explanation for correct option:

Finding the value of $P\left(E’\cap F’|G\right)$:

Consider the given equation as,

$$\begin{gathered} P\left(E’\cap F’|G\right)=\frac{P\left(E’\cap F’\cap G\right)}{P\left(G\right)} \\ P\left(E’\cap F’|G\right)=\frac{P\left(G\right)-P\left(E\cap G\right)-P\left(G\cap F\right)}{P\left(G\right)} \end{gathered}$$

Since, $P\left(G\right)$ is not Equal to $0$

Then,

$$\begin{gathered} P\left(E\text{'}\cap F\text{'}|G\right)=\frac{P\left(G\right)\left[1-P\left(E\right)-P\left(F\right)\right]}{P\left(G\right)} \\ P\left(E\text{'}\cap F\text{'}|G\right)=1-P\left(E\right)-P\left(F\right) \\ P\left(E\text{'}\cap F\text{'}|G\right)=P\left(E\text{'}\right)-P\left(F\right)\left[1-P\left(E\right)=P\left(E\text{'}\right)\right] \end{gathered}$$

Hence, the correct answer is Option (C).