If `P(A)=1//3, P(B)=1//2` and A and B are mutually exclusive, find `P(Acup B)` and `P(A cap B)`.

Answer

To solve the problem, we need to find the probabilities \( P(A \cup B) \) and \( P(A \cap B) \) given that \( P(A) = \frac{1}{3} \), \( P(B) = \frac{1}{2} \), and that events A and B are mutually exclusive.

### 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, the probability of both A and B happening together, \( P(A \cap B) \), is 0.
\[ P(A \cap B) = 0 \]

2. **Using the Formula for Union of Two Events**:
- The formula for the probability of the union of two events is given by:
\[ P(A \cup B) = P(A) + P(B) - P(A \cap B) \]
- Since we know \( P(A \cap B) = 0 \), we can simplify this formula to:
\[ P(A \cup B) = P(A) + P(B) \]

3. **Substituting the Values**:
- Now we substitute the known values into the formula:
\[ P(A) = \frac{1}{3}, \quad P(B) = \frac{1}{2} \]
- Therefore:
\[ P(A \cup B) = \frac{1}{3} + \frac{1}{2} \]

4. **Finding a Common Denominator**:
- To add \( \frac{1}{3} \) and \( \frac{1}{2} \), we need a common denominator. The least common multiple of 3 and 2 is 6.
- Convert \( \frac{1}{3} \) and \( \frac{1}{2} \) to have a denominator of 6:
\[ \frac{1}{3} = \frac{2}{6}, \quad \frac{1}{2} = \frac{3}{6} \]

5. **Adding the Probabilities**:
- Now, we can add the two fractions:
\[ P(A \cup B) = \frac{2}{6} + \frac{3}{6} = \frac{5}{6} \]

6. **Final Answers**:
- Thus, the final answers are:
\[ P(A \cup B) = \frac{5}{6}, \quad P(A \cap B) = 0 \]

### Summary of Results:
- \( P(A \cup B) = \frac{5}{6} \)
- \( P(A \cap B) = 0 \)