Let A,B and C be three events, which are pair-wise independent and `bar(E )` denotes the complement of an event E. If `P (A nn B nn C) = 0 and PC) gt 0`, then `P[(bar(A) nn bar(B)) |C]` is equal to :

Options

A. `P(A) + P (bar(B))`
B. `P(bar(A)) - P (bar(B))`
C. `P(bar(A)) P (B)`
D. `P(bar(A)) + P (bar(B))`

Answer

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

### Step-by-Step Solution:

1. **Understanding the Conditional Probability**:
The conditional probability can be expressed using the formula:
\[ P(\bar{A} \cap \bar{B} | C) = \frac{P(\bar{A} \cap \bar{B} \cap C)}{P(C)} \]

2. **Using Independence**:
Since \( A, B, \) and \( C \) are pairwise independent, we can express the joint probability of the complements:
\[ P(\bar{A} \cap \bar{B} \cap C) = P(\bar{A}) \cdot P(\bar{B}) \cdot P(C) \]

3. **Finding the Complements**:
The probabilities of the complements can be expressed as:
\[ P(\bar{A}) = 1 - P(A) \quad \text{and} \quad P(\bar{B}) = 1 - P(B) \]

4. **Substituting in the Formula**:
Substitute these into the expression for \( P(\bar{A} \cap \bar{B} \cap C) \):
\[ P(\bar{A} \cap \bar{B} \cap C) = (1 - P(A))(1 - P(B)) P(C) \]

5. **Substituting Back into the Conditional Probability**:
Now substitute this back into the conditional probability formula:
\[ P(\bar{A} \cap \bar{B} | C) = \frac{(1 - P(A))(1 - P(B)) P(C)}{P(C)} \]
The \( P(C) \) cancels out:
\[ P(\bar{A} \cap \bar{B} | C) = (1 - P(A))(1 - P(B)) \]

6. **Using the Given Information**:
We know that \( P(A \cap B \cap C) = 0 \). Since \( A, B, \) and \( C \) are independent, this implies:
\[ P(A) \cdot P(B) \cdot P(C) = 0 \]
Given \( P(C) > 0 \), it follows that:
\[ P(A) \cdot P(B) = 0 \]
Therefore, at least one of \( P(A) \) or \( P(B) \) must be zero.

7. **Conclusion**:
If either \( P(A) = 0 \) or \( P(B) = 0 \), then:
\[ P(\bar{A}) = 1 \quad \text{or} \quad P(\bar{B}) = 1 \]
Thus, we find:
\[ P(\bar{A} \cap \bar{B} | C) = (1 - P(A))(1 - P(B)) = 1 \]

### Final Answer:
\[ P(\bar{A} \cap \bar{B} | C) = 1 \]