If A and B are mutually exclusive events such that `P(A)=0.25, P(B)=0.4,` then `P(A^(C )nnB^(C ))` is equal to

Options

A. 0.45
B. 0.55
C. 0.9
D. 0.35

Answer

To solve the problem, we need to find \( P(A^C \cap B^C) \), where \( A \) and \( B \) are mutually exclusive events with given probabilities \( P(A) = 0.25 \) and \( P(B) = 0.4 \).

### Step-by-Step Solution:

1. **Understanding Mutually Exclusive Events**:
- Since \( A \) and \( B \) are mutually exclusive, it means that they cannot occur at the same time. Therefore, \( P(A \cap B) = 0 \).

2. **Finding \( P(A \cup B) \)**:
- We can use the formula for the probability of the union of two events:
\[ P(A \cup B) = P(A) + P(B) - P(A \cap B) \]
- Substituting the known values:
\[ P(A \cup B) = P(A) + P(B) - 0 = 0.25 + 0.4 = 0.65 \]

3. **Finding \( P(A^C \cap B^C) \)**:
- The complement of the union of events \( A \) and \( B \) is given by:
\[ P(A^C \cap B^C) = 1 - P(A \cup B) \]
- Substituting the value we found:
\[ P(A^C \cap B^C) = 1 - 0.65 = 0.35 \]

4. **Final Answer**:
- Therefore, \( P(A^C \cap B^C) = 0.35 \).

### Summary:
The probability \( P(A^C \cap B^C) \) is equal to 0.35.