If A and B are any two events having `P(AnnB)=1/2` an `P(A)=2/3`, then the probability of `AnnbarB` is :

Options

A. `1/2`
B. `2/3`
C. `1/6`
D. `1/3`

Answer

To solve the problem, we need to find the probability of the event \( A \cap B' \) (the intersection of event A and the complement of event B). We are given the following:

- \( P(A \cap B) = \frac{1}{2} \)
- \( P(A) = \frac{2}{3} \)

### Step-by-Step Solution:

1. **Understanding the Events**:
- We have two events A and B.
- \( A \cap B \) represents the probability that both events A and B occur.
- \( A \cap B' \) represents the probability that event A occurs while event B does not occur.

2. **Using the Formula**:
- We can use the formula:
\[ P(A \cap B') = P(A) - P(A \cap B) \]
- This formula states that the probability of A occurring without B is equal to the probability of A minus the probability of A occurring with B.

3. **Substituting the Values**:
- We know \( P(A) = \frac{2}{3} \) and \( P(A \cap B) = \frac{1}{2} \).
- Substitute these values into the formula:
\[ P(A \cap B') = \frac{2}{3} - \frac{1}{2} \]

4. **Finding a Common Denominator**:
- The least common multiple of 3 and 2 is 6. We will convert both fractions to have a denominator of 6:
\[ P(A) = \frac{2}{3} = \frac{4}{6} \]
\[ P(A \cap B) = \frac{1}{2} = \frac{3}{6} \]

5. **Performing the Subtraction**:
- Now we can perform the subtraction:
\[ P(A \cap B') = \frac{4}{6} - \frac{3}{6} = \frac{1}{6} \]

6. **Final Result**:
- Therefore, the probability of \( A \cap B' \) is:
\[ P(A \cap B') = \frac{1}{6} \]

### Conclusion:
The probability of \( A \cap B' \) is \( \frac{1}{6} \).