If A,B,C are three pairwise independent events such that `P(AnnBnnC)=0` then `P((B^cnnC^c)/A)` is equal to-

Options

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

Answer

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

### Step-by-Step Solution:

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

2. **Finding the Numerator**:
We need to compute \( P((B^c \cap C^c) \cap A) \). This can be rewritten using the properties of intersections:
\[ P((B^c \cap C^c) \cap A) = P(A) - P(A \cap B) - P(A \cap C) \]
Since \( P(A \cap B \cap C) = 0 \), we know that:
\[ P(A \cap B) = P(A) \cdot P(B) \quad \text{(due to independence)} \]
\[ P(A \cap C) = P(A) \cdot P(C) \quad \text{(due to independence)} \]

3. **Substituting the Values**:
Substitute the values into the numerator:
\[ P((B^c \cap C^c) \cap A) = P(A) - P(A) \cdot P(B) - P(A) \cdot P(C) \]
Factor out \( P(A) \):
\[ P((B^c \cap C^c) \cap A) = P(A) \left(1 - P(B) - P(C)\right) \]

4. **Putting it Back into the Conditional Probability**:
Now substitute this back into the conditional probability formula:
\[ P((B^c \cap C^c) | A) = \frac{P(A) \left(1 - P(B) - P(C)\right)}{P(A)} \]
The \( P(A) \) in the numerator and denominator cancels out:
\[ P((B^c \cap C^c) | A) = 1 - P(B) - P(C) \]

5. **Final Expression**:
We can express this result in terms of complements:
\[ P((B^c \cap C^c) | A) = P(B^c) + P(C^c) - 1 \]
This can be rewritten as:
\[ P((B^c \cap C^c) | A) = P(C^c) - P(B) \]

### Conclusion:
Thus, the final answer is:
\[ P((B^c \cap C^c) | A) = P(C^c) - P(B) \]