Let `E^(@)` denotes the complement of an event E. If E, F, G are pairwise independent events with `P(G) gt 0` and `P (E nn F nn G) =0`. Then, `P(E^(@) nn F^(@)|G)` equals :

Options

A. `P(E^(@)) + P(F^(@))`
B. `P(E^(@)) - P(F^(@))`
C. `P(E^(@)) - P(F)`
D. `P€ - P(F^(@))`

Answer

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

### Step-by-Step Solution:

1. **Understanding the Conditional Probability**:
The conditional probability \( P(E^c \cap F^c | G) \) can be expressed using the formula:
\[ P(E^c \cap F^c | G) = \frac{P(E^c \cap F^c \cap G)}{P(G)} \]

2. **Using Independence of Events**:
Since \( E, F, G \) are pairwise independent, we can express \( P(E^c \cap F^c \cap G) \) as:
\[ P(E^c \cap F^c \cap G) = P(E^c) \cdot P(F^c) \cdot P(G) \]

3. **Calculating the Complements**:
The probabilities of the complements can be calculated as:
\[ P(E^c) = 1 - P(E) \quad \text{and} \quad P(F^c) = 1 - P(F) \]

4. **Substituting into the Equation**:
Now substituting these into our expression for \( P(E^c \cap F^c \cap G) \):
\[ P(E^c \cap F^c \cap G) = (1 - P(E))(1 - P(F)) P(G) \]

5. **Substituting Back into the Conditional Probability**:
Now substitute this back into the conditional probability formula:
\[ P(E^c \cap F^c | G) = \frac{(1 - P(E))(1 - P(F)) P(G)}{P(G)} \]

6. **Simplifying the Expression**:
The \( P(G) \) cancels out:
\[ P(E^c \cap F^c | G) = (1 - P(E))(1 - P(F)) \]

7. **Using the Given Condition**:
We are given that \( P(E \cap F \cap G) = 0 \). Since the events are independent, this implies:
\[ P(E) \cdot P(F) \cdot P(G) = 0 \]
Given \( P(G) > 0 \), it follows that \( P(E) \cdot P(F) = 0 \). Therefore, at least one of \( P(E) \) or \( P(F) \) must be zero.

8. **Final Calculation**:
If either \( P(E) = 0 \) or \( P(F) = 0 \), then:
- If \( P(E) = 0 \), then \( P(E^c) = 1 \) and \( P(F^c) = 1 - P(F) \).
- If \( P(F) = 0 \), then \( P(F^c) = 1 \) and \( P(E^c) = 1 - P(E) \).
In either case, we find that:
\[ P(E^c \cap F^c | G) = 1 \]

### Conclusion:
Thus, the final answer is:
\[ P(E^c \cap F^c | G) = 1 \]