If a random variable X has a Poisson distribution with parameter `1//2`, then P(X=2)=

Answer

To find the probability \( P(X = 2) \) for a random variable \( X \) that follows a Poisson distribution with parameter \( \lambda = \frac{1}{2} \), we can use the formula for the probability mass function of a Poisson distribution:

\[ P(X = k) = \frac{e^{-\lambda} \lambda^k}{k!} \]

where:
- \( \lambda \) is the average rate (in this case, \( \frac{1}{2} \)),
- \( k \) is the number of occurrences (in this case, \( 2 \)),
- \( e \) is the base of the natural logarithm (approximately equal to 2.71828).

### Step-by-Step Solution:

1. **Identify the parameters**:
- Here, \( \lambda = \frac{1}{2} \) and \( k = 2 \).

2. **Substitute the values into the formula**:
\[ P(X = 2) = \frac{e^{-\frac{1}{2}} \left(\frac{1}{2}\right)^2}{2!} \]

3. **Calculate \( 2! \)**:
\[ 2! = 2 \times 1 = 2 \]

4. **Calculate \( \left(\frac{1}{2}\right)^2 \)**:
\[ \left(\frac{1}{2}\right)^2 = \frac{1}{4} \]

5. **Substitute these values back into the equation**:
\[ P(X = 2) = \frac{e^{-\frac{1}{2}} \cdot \frac{1}{4}}{2} \]

6. **Simplify the expression**:
\[ P(X = 2) = \frac{e^{-\frac{1}{2}}}{8} \]

7. **Calculate \( e^{-\frac{1}{2}} \)**:
- The value of \( e^{-\frac{1}{2}} \) is approximately \( 0.6065 \).

8. **Final calculation**:
\[ P(X = 2) = \frac{0.6065}{8} \approx 0.0758125 \]

### Final Answer:
\[ P(X = 2) \approx 0.0758125 \]