The probability of appointment of A at a post is `1/2` and probability of appointment of B is `1/3`. If only one can be appointment, then find its probability.

Explanation

To solve the problem, we need to find the probability of either A or B being appointed, given that only one can be appointed.

### Step-by-Step Solution:

1. **Identify the given probabilities:**
- Probability of A being appointed, \( P(A) = \frac{1}{2} \)
- Probability of B being appointed, \( P(B) = \frac{1}{3} \)

2. **Understand the scenario:**
- Since only one person can be appointed, we need to find the probability that either A is appointed or B is appointed, but not both.

3. **Use the formula for the probability of the union of two events:**
- The formula for the probability of either A or B being appointed is:
\[ P(A \cup B) = P(A) + P(B) - P(A \cap B) \]
- Here \( P(A \cap B) \) is the probability that both A and B are appointed, which is not possible in this scenario. Therefore, \( P(A \cap B) = 0 \).

4. **Substitute the known values into the formula:**
\[ P(A \cup B) = P(A) + P(B) - P(A \cap B) \]
\[ P(A \cup B) = \frac{1}{2} + \frac{1}{3} - 0 \]

5. **Find a common denominator to add the fractions:**
- The least common multiple of 2 and 3 is 6.
- Convert the fractions:
\[ P(A) = \frac{1}{2} = \frac{3}{6} \]
\[ P(B) = \frac{1}{3} = \frac{2}{6} \]

6. **Add the fractions:**
\[ P(A \cup B) = \frac{3}{6} + \frac{2}{6} = \frac{5}{6} \]

7. **Conclusion:**
- The probability that either A or B is appointed (but not both) is:
\[ P(A \cup B) = \frac{5}{6} \]

### Final Answer:
The probability that either A or B is appointed is \( \frac{5}{6} \).