13. If A and B are two events such that` P(A)=1/2 and P(B)=2/3`, then

Options

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

Explanation

To solve the problem regarding the events A and B with given probabilities \( P(A) = \frac{1}{2} \) and \( P(B) = \frac{2}{3} \), we will derive the possible values for the intersection of events A and B, denoted as \( P(A \cap B) \).

### Step-by-Step Solution:

1. **Understanding the Given Probabilities**:
We have:
- \( P(A) = \frac{1}{2} \)
- \( P(B) = \frac{2}{3} \)

2. **Using the Formula for the Intersection of Two Events**:
The probability of the intersection of two events can be bounded by the individual probabilities of the events. Specifically:
\[ P(A \cap B) \geq P(A) + P(B) - 1 \]
and
\[ P(A \cap B) \leq \min(P(A), P(B)) \]

3. **Calculating the Lower Bound**:
Let's calculate the lower bound:
\[ P(A \cap B) \geq P(A) + P(B) - 1 = \frac{1}{2} + \frac{2}{3} - 1 \]
To perform this calculation, we need a common denominator:
\[ \frac{1}{2} = \frac{3}{6}, \quad \frac{2}{3} = \frac{4}{6} \]
Therefore,
\[ P(A \cap B) \geq \frac{3}{6} + \frac{4}{6} - \frac{6}{6} = \frac{1}{6} \]

4. **Calculating the Upper Bound**:
Now, let's calculate the upper bound:
\[ P(A \cap B) \leq \min(P(A), P(B)) = \min\left(\frac{1}{2}, \frac{2}{3}\right) \]
Since \( \frac{1}{2} = \frac{3}{6} \) and \( \frac{2}{3} = \frac{4}{6} \), we have:
\[ P(A \cap B) \leq \frac{1}{2} \]

5. **Combining the Results**:
From the calculations, we find:
\[ \frac{1}{6} \leq P(A \cap B) \leq \frac{1}{2} \]

### Final Result:
Thus, the possible values for \( P(A \cap B) \) lie within the range:
\[ \frac{1}{6} \leq P(A \cap B) \leq \frac{1}{2} \]