1. Given information

We are told that events $A,B,C$ are pairwise independent. Hence:

$P(A\cap B)=P(A)P(B),\quad P(B\cap C)=P(B)P(C),\quad P(C\cap A)=P(C)P(A).$

Also,

$P(A\cap B\cap C)=0,\qquad P(C)>0.$

We need to find

$P\big((\overline A\cap \overline B)\mid C\big).$

2. Use the condition $P(A\cap B\cap C)=0$

Since $A$ and $B$ are pairwise independent,

$P(A\cap B)=P(A)P(B).$

But this alone does not directly give triple independence. However, observe:

$P(A\cap B\cap C) \le P(A\cap C).$

A cleaner way is to use conditional probability with respect to $C$.

Because $A$ and $C$ are independent and $B$ and $C$ are independent,

$P(A\mid C)=P(A),\qquad P(B\mid C)=P(B).$

Now,

$P(A\cap B\mid C)=\frac{P(A\cap B\cap C)}{P(C)}=\frac{0}{P(C)}=0.$

So, within the conditional sample space $C$, events $A$ and $B$ cannot occur together.

3. Compute $P((\overline A\cap \overline B)\mid C)$

Using the complement formula inside the conditional space:

$P((\overline A\cap \overline B)\mid C)=1-P(A\cup B\mid C).$

Now,

$P(A\cup B\mid C)=P(A\mid C)+P(B\mid C)-P(A\cap B\mid C).$

Substitute the values:

$P(A\cup B\mid C)=P(A)+P(B)-0=P(A)+P(B).$

Therefore,

$P((\overline A\cap \overline B)\mid C)=1-[P(A)+P(B)].$

Rewrite in terms of complements:

$1-P(A)-P(B)=(1-P(A))-P(B)=P(\overline A)-P(B).$

Also,

$1-P(A)-P(B)=P(\overline B)-P(A).$

Among the given options, this matches:

$\boxed{P(\overline A)-P(B)}.$

4. Check options

• A: $P(\overline A)-P(B)=1-P(A)-P(B)$ ✔
• B: $P(A)+P(\overline B)=1-P(B)+P(A)$ ✘
• C: $P(\overline A)-P(\overline B)=P(B)-P(A)$ ✘
• D: $P(\overline A)+P(\overline B)=2-P(A)-P(B)$ ✘

So the correct option is A.

5. Comparison with stored answer

Stored correct answer: A

Our derived answer: A

Hence, they agree.