If `E_(1) and E_(2)` are exclusive events then `P(E_(1) capE_(2))`=_____

Answer

To solve the problem, we need to understand the concept of exclusive (or mutually exclusive) events in probability.

### Step-by-Step Solution:

1. **Definition of Exclusive Events**:
- Two events \( E_1 \) and \( E_2 \) are said to be exclusive if they cannot occur at the same time. This means that if one event occurs, the other cannot.

2. **Intersection of Events**:
- The intersection of two events \( E_1 \) and \( E_2 \), denoted as \( E_1 \cap E_2 \), represents the scenario where both events occur simultaneously.

3. **Probability of Intersection**:
- For exclusive events, since they cannot occur together, the probability of their intersection is defined as:
\[ P(E_1 \cap E_2) = 0 \]

4. **Conclusion**:
- Therefore, if \( E_1 \) and \( E_2 \) are exclusive events, we conclude that:
\[ P(E_1 \cap E_2) = 0 \]

### Final Answer:
\[ P(E_1 \cap E_2) = 0 \]