Let $A,B$ and $C$ be three events, which are pairwise independence and $\overline E $ denotes the complement of an event $E$. If $P\left( {A \cap B \cap C} \right) = 0$ and $P\left( C \right) > 0$, then $P\left[ {\left( {\overline A \cap \overline B } \right)|C} \right]$ is equal to:
(A) $P\left( A \right) + P\left( {\overline B } \right)$
(B) $P\left( {\overline A } \right) - P\left( {\overline B } \right)$
(C) $P\left( {\overline A } \right) - P\left( B \right)$
(D) $P\left( {\overline A } \right) + P\left( {\overline B } \right)$

Hint: The events are called pairwise independent if any two events in the collection are independent of each other. If three events $X,Y$ and $Z$ are pairwise independent, then $P\left( {X \cap Y} \right) = P\left( X \right) \cdot P\left( Y \right)$, $P\left( {Y \cap Z} \right) = P\left( Y \right) \cdot P\left( Z \right)$ and $P\left( {X \cap Z} \right) = P\left( X \right) \cdot P\left( Z \right)$.

Complete step-by-step answer:
Given, $P\left( {A \cap B \cap C} \right) = 0$ and $A,B$ and $C$ are pair-wise independent events, therefore,
$P\left( {A \cap B} \right) = P\left( A \right) \cdot P\left( B \right)$
$P\left( {B \cap C} \right) = P\left( B \right) \cdot P\left( C \right)$
\[P\left( {A \cap C} \right) = P\left( A \right) \cdot P\left( C \right)\]
$P\left[ {\dfrac{{\left( {\overline A \cap \overline B } \right)}}{C}} \right]$ is of the form of $P\left( {\dfrac{A}{B}} \right) = \dfrac{{P\left( {A \cap B} \right)}}{{P\left( B \right)}}$.
Therefore, $P\left[ {\dfrac{{\left( {\overline A \cap \overline B } \right)}}{C}} \right]$$ = \dfrac{{P\left[ {\left( {\overline A \cap \overline B } \right) \cap C} \right]}}{{P\left( C \right)}}$
$ = \dfrac{{P\left( C \right) - P\left( {A \cap C} \right) - P\left( {B \cap C} \right) + P\left( {A \cap B \cap C} \right)}}{{P\left( C \right)}}$
Substitute \[P\left( {A \cap C} \right) = P\left( A \right) \cdot P\left( C \right)\], $P\left( {B \cap C} \right) = P\left( B \right) \cdot P\left( C \right)$and $P\left( {A \cap B \cap C} \right) = 0$,
$ = \dfrac{{P\left( C \right) - P\left( A \right) \cdot P\left( C \right) - P\left( B \right) \cdot P\left( C \right) + 0}}{{P\left( C \right)}}$
$ = \dfrac{{P\left( C \right)}}{{P\left( C \right)}} - \dfrac{{P\left( A \right) \cdot P\left( C \right)}}{{P\left( C \right)}} - \dfrac{{P\left( B \right) \cdot P\left( C \right)}}{{P\left( C \right)}}$
$ = 1 - P\left( A \right) - P\left( B \right)$
$ = P\left( {\overline A } \right) - P\left( B \right)$

Hence, option (C) is the correct answer.

Note: The term $1 - P\left( A \right) - P\left( B \right)$ may also be equal to $P\left( A \right) - P\left( {\overline B } \right)$. So, if we get $P\left( A \right) - P\left( {\overline B } \right)$ as an option, then it would be the correct. While if we got both $P\left( {\overline A } \right) - P\left( B \right)$ and $P\left( A \right) - P\left( {\overline B } \right)$ as options, the both will be correct.