Let `E^c` denote the complement of an event `E.` Let `E,F,G` be pairwise independent events with `P(G) gt 0` and `P(E nn F nn G)=0` Then `P(E^c nn F^c nn G)` equals (A) `P(E^c)+P(F^c)` (B) `P(E^c)-P(F^c)` (C) `P(E^c)-P(F)` (D) `P(E)-P(F^c)`

Options

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

Answer

To solve the problem, we need to find \( P(E^c \cap F^c \cap G) \) given the conditions about the events \( E, F, \) and \( G \).

### Step-by-Step Solution:

1. **Understanding the Problem**: We are given that \( E, F, G \) are pairwise independent events, \( P(G) > 0 \), and \( P(E \cap F \cap G) = 0 \). We need to find \( P(E^c \cap F^c \cap G) \).

2. **Using the Complement**: We can express \( P(E^c \cap F^c \cap G) \) in terms of the probabilities of the complements:
\[ P(E^c \cap F^c \cap G) = P(G) - P(E \cap G) - P(F \cap G) + P(E \cap F \cap G) \]
Since \( P(E \cap F \cap G) = 0 \), we can simplify this to:
\[ P(E^c \cap F^c \cap G) = P(G) - P(E \cap G) - P(F \cap G) \]

3. **Using Independence**: Since \( E, F, G \) are pairwise independent, we can express the intersections:
\[ P(E \cap G) = P(E) \cdot P(G) \quad \text{and} \quad P(F \cap G) = P(F) \cdot P(G) \]
Substituting these into our equation gives:
\[ P(E^c \cap F^c \cap G) = P(G) - P(E) \cdot P(G) - P(F) \cdot P(G) \]

4. **Factoring Out \( P(G) \)**: We can factor \( P(G) \) out of the equation:
\[ P(E^c \cap F^c \cap G) = P(G) \left(1 - P(E) - P(F)\right) \]

5. **Expressing in Terms of Complements**: We know that \( P(E^c) = 1 - P(E) \) and \( P(F^c) = 1 - P(F) \). Thus:
\[ P(E^c \cap F^c \cap G) = P(G) \left(P(E^c) + P(F^c) - 1\right) \]

6. **Final Expression**: Since we are looking for \( P(E^c \cap F^c \cap G) \) without \( P(G) \), we can express it as:
\[ P(E^c \cap F^c) = P(E^c) + P(F^c) - P(E^c \cap F^c) \]
However, from the independence and the given conditions, we can directly conclude:
\[ P(E^c \cap F^c \cap G) = P(E^c) + P(F^c) - 1 \]

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