If A, B, C are pair-wise mutually exclusive, and exhaustive, events of a sample space S, then ` P(A) + P(B) +P(C) =`

Options

A. 0
B. `1/3`
C. 1
D. `2/3`

Answer

To solve the problem, we need to understand the concepts of mutually exclusive and exhaustive events in probability.

### Step-by-Step Solution:

1. **Understanding Mutually Exclusive Events**:
- Events A, B, and C are said to be mutually exclusive if the occurrence of one event excludes the occurrence of the others. This means that if A occurs, B and C cannot occur, and vice versa.

2. **Understanding Exhaustive Events**:
- Events A, B, and C are exhaustive if they cover the entire sample space S. This means that at least one of the events A, B, or C must occur when an experiment is conducted.

3. **Using the Properties of Probability**:
- Since A, B, and C are mutually exclusive, the probability of their union is given by:
\[ P(A \cup B \cup C) = P(A) + P(B) + P(C) \]

4. **Considering the Exhaustiveness**:
- Since A, B, and C are also exhaustive, the probability of their union equals 1 (the total probability of the sample space):
\[ P(A \cup B \cup C) = 1 \]

5. **Combining the Two Properties**:
- From the above two points, we can set up the equation:
\[ P(A) + P(B) + P(C) = P(A \cup B \cup C) = 1 \]

6. **Conclusion**:
- Therefore, we conclude that:
\[ P(A) + P(B) + P(C) = 1 \]

### Final Answer:
Thus, the answer is:
\[ P(A) + P(B) + P(C) = 1 \]