Key Concepts and Formulas

1. Conditional Probability: The probability of event X occurring given event Y has occurred is defined as: $P(X|Y) = \frac{P(X \cap Y)}{P(Y)}, \quad \text{provided } P(Y) > 0$

2. De Morgan's Laws for Events: For any two events A and B: $\overline{A} \cap \overline{B} = \overline{A \cup B}$

3. Pairwise Independence: Three events A, B, and C are pairwise independent if the probability of their intersection in pairs equals the product of their individual probabilities: $P(A \cap B) = P(A)P(B)$ $P(A \cap C) = P(A)P(C)$ $P(B \cap C) = P(B)P(C)$

Step-by-Step Solution

Step 1: Apply the Definition of Conditional Probability We need to calculate $P\left[ {\left( {\overline A \cap \overline B } \right)\left| C \right.} \right]$. Using the definition $P(X|Y) = \frac{P(X \cap Y)}{P(Y)}$, with $X = (\overline A \cap \overline B)$ and $Y = C$: $P\left[ {\left( {\overline A \cap \overline B } \right)\left| C \right.} \right] = \frac{P\left( {(\overline A \cap \overline B ) \cap C} \right)}{P(C)}$ We are given $P(C) > 0$, so the denominator is valid.

Step 2: Simplify the Numerator using De Morgan's Law The term $(\overline A \cap \overline B)$ in the numerator can be simplified using De Morgan's Law, which states $\overline A \cap \overline B = \overline{A \cup B}$. Substituting this into the numerator: $P\left( {(\overline A \cap \overline B ) \cap C} \right) = P\left( {\overline{A \cup B} \cap C} \right)$ This expression represents the probability that event C occurs AND the event $(A \cup B)$ does NOT occur.

Step 3: Express the Numerator as a Probability of Set Difference The expression $P\left( {\overline{A \cup B} \cap C} \right)$ is of the form $P(\overline{Y} \cap X)$, which is equivalent to $P(X \cap \overline{Y})$. We know that $P(X \cap \overline{Y}) = P(X) - P(X \cap Y)$. Applying this with $X=C$ and $Y=(A \cup B)$: $P\left( {\overline{A \cup B} \cap C} \right) = P\left( {C \cap \overline{A \cup B}} \right) = P(C) - P(C \cap (A \cup B))$

Step 4: Expand $P(C \cap (A \cup B))$ using the Principle of Inclusion-Exclusion We can distribute $C$ over the union: $C \cap (A \cup B) = (C \cap A) \cup (C \cap B)$. Now, we apply the Principle of Inclusion-Exclusion for two events, $(C \cap A)$ and $(C \cap B)$: $P((C \cap A) \cup (C \cap B)) = P(C \cap A) + P(C \cap B) - P((C \cap A) \cap (C \cap B))$ The intersection $(C \cap A) \cap (C \cap B)$ simplifies to $C \cap A \cap B$. So, $P(C \cap (A \cup B)) = P(A \cap C) + P(B \cap C) - P(A \cap B \cap C)$

Step 5: Substitute Back and Use the Given Condition $P(A \cap B \cap C) = 0$ Substitute the expanded form of $P(C \cap (A \cup B))$ back into the numerator expression from Step 3: $P\left( {\overline{A \cup B} \cap C} \right) = P(C) - [P(A \cap C) + P(B \cap C) - P(A \cap B \cap C)]$ $P\left( {\overline{A \cup B} \cap C} \right) = P(C) - P(A \cap C) - P(B \cap C) + P(A \cap B \cap C)$ We are given that $P(A \cap B \cap C) = 0$. Substituting this value: $P\left( {\overline{A \cup B} \cap C} \right) = P(C) - P(A \cap C) - P(B \cap C)$ Now, our original conditional probability expression from Step 1 becomes: $P\left[ {\left( {\overline A \cap \overline B } \right)\left| C \right.} \right] = \frac{P(C) - P(A \cap C) - P(B \cap C)}{P(C)}$

Step 6: Utilize the Pairwise Independence Condition The problem states that A, B, and C are pairwise independent. This means:

• $P(A \cap C) = P(A)P(C)$
• $P(B \cap C) = P(B)P(C)$ Substitute these into the numerator: $P\left[ {\left( {\overline A \cap \overline B } \right)\left| C \right.} \right] = \frac{P(C) - P(A)P(C) - P(B)P(C)}{P(C)}$

Step 7: Simplify the Expression to Reach the Final Answer Since $P(C) > 0$, we can divide each term in the numerator by $P(C)$: $P\left[ {\left( {\overline A \cap \overline B } \right)\left| C \right.} \right] = \frac{P(C)}{P(C)} - \frac{P(A)P(C)}{P(C)} - \frac{P(B)P(C)}{P(C)}$ $P\left[ {\left( {\overline A \cap \overline B } \right)\left| C \right.} \right] = 1 - P(A) - P(B)$ To match the given options, we use the property $P(\overline A) = 1 - P(A)$: $P\left[ {\left( {\overline A \cap \overline B } \right)\left| C \right.} \right] = P(\overline A) - P(B)$

Common Mistakes & Tips

• Distinguish Pairwise vs. Mutual Independence: Pairwise independence ($P(E \cap F) = P(E)P(F)$ for all pairs) is a weaker condition than mutual independence ($P(A \cap B \cap C) = P(A)P(B)P(C)$ in addition to pairwise independence). Do not assume mutual independence unless explicitly stated.
• Interpreting $P(A \cap B \cap C) = 0$: This condition is simply a given fact. Under pairwise independence, it does not imply that any individual event probability is zero. If it were mutual independence, then $P(A)P(B)P(C)=0$ would imply at least one of $P(A), P(B), P(C)$ is zero, which is a much stronger conclusion.
• Correct Application of Set Identities: Be meticulous with De Morgan's laws and the distribution property of intersection over union. A small error can lead to a completely different result.

Summary

This problem effectively tests your foundational knowledge of probability, particularly conditional probability, De Morgan's Laws, the Principle of Inclusion-Exclusion, and the precise application of pairwise independence. The solution involved systematically breaking down the conditional probability expression, simplifying it using set identities and given conditions ($P(A \cap B \cap C) = 0$ and pairwise independence), and then performing algebraic simplification to arrive at the final form.

The final answer is $\boxed{P\left( {\overline A } \right) - P\left( B \right)}$, which corresponds to option (A).