Let A, B, C be pariwise independent events with `P(C) >0 and P(AnnBnnC) = 0`. Then `P(A^c nn B^c/C)`.

Options

A. `P(A^c)-P(B)`
B. `P(A)-P(B^c)`
C. `P(A^c)+P(B^c)`
D. `P(A^c)-P(B^c)`

Answer

To solve the problem, we need to find \( P(A^c \cap B^c | C) \) given that \( A, B, C \) are pairwise independent events, \( P(C) > 0 \), and \( P(A \cap B \cap C) = 0 \).

### Step-by-step Solution:

1. **Understanding the Conditional Probability**:
We start with the definition of conditional probability:
\[ P(A^c \cap B^c | C) = \frac{P(A^c \cap B^c \cap C)}{P(C)} \]

2. **Using Independence**:
Since \( A, B, C \) are pairwise independent, we can express \( P(A^c \cap B^c \cap C) \) in terms of their individual probabilities:
\[ P(A^c \cap B^c \cap C) = P(A^c | C) \cdot P(B^c | C) \cdot P(C) \]
However, we need to find \( P(A^c \cap B^c \cap C) \) directly.

3. **Finding \( P(A^c \cap B^c) \)**:
Using the independence of \( A \) and \( B \):
\[ P(A^c \cap B^c) = P(A^c) \cdot P(B^c) \]
where \( P(A^c) = 1 - P(A) \) and \( P(B^c) = 1 - P(B) \).

4. **Using the Given Information**:
We know \( P(A \cap B \cap C) = 0 \). This means that at least one of the events \( A \), \( B \), or \( C \) must not occur simultaneously. Therefore, we can express:
\[ P(A \cap B) = P(A) \cdot P(B) = 0 \]
which implies either \( P(A) = 0 \) or \( P(B) = 0 \).

5. **Calculating \( P(A^c \cap B^c \cap C) \)**:
Given the independence and the fact that \( P(A \cap B \cap C) = 0 \), we can conclude:
\[ P(A^c \cap B^c \cap C) = P(C) - P(A \cap C) - P(B \cap C) + P(A \cap B \cap C) \]
Since \( P(A \cap B \cap C) = 0 \), we have:
\[ P(A^c \cap B^c \cap C) = P(C) - P(A \cap C) - P(B \cap C) \]

6. **Final Calculation**:
Substitute back into the conditional probability formula:
\[ P(A^c \cap B^c | C) = \frac{P(C) - P(A \cap C) - P(B \cap C)}{P(C)} \]
Simplifying gives:
\[ P(A^c \cap B^c | C) = 1 - \frac{P(A \cap C) + P(B \cap C)}{P(C)} \]

### Conclusion:
Thus, the final expression for \( P(A^c \cap B^c | C) \) is:
\[ P(A^c \cap B^c | C) = 1 - \frac{P(A \cap C) + P(B \cap C)}{P(C)} \]