A randam variable X has Poisson's distribution with mean 2. Then , `P(X)gt 1.5)` is equal to

Options

A. `1-(3)/(e^2)`
B. `(3)/(e^2)`
C. `(2)/(e^2)`
D. 0

Answer

To solve the problem, we need to find the probability \( P(X > 1.5) \) for a random variable \( X \) that follows a Poisson distribution with a mean \( \lambda = 2 \).

### Step-by-Step Solution:

1. **Understanding the Poisson Distribution**:
The probability mass function (PMF) of a Poisson random variable is given by:
\[ P(X = k) = \frac{e^{-\lambda} \lambda^k}{k!} \]
where \( \lambda \) is the mean of the distribution and \( k \) is a non-negative integer.

2. **Identify the Mean**:
In this case, the mean \( \lambda = 2 \).

3. **Convert the Probability**:
We need to find \( P(X > 1.5) \). Since \( X \) can only take integer values, we can rewrite this as:
\[ P(X > 1.5) = P(X \geq 2) \]

4. **Use the Complement Rule**:
To find \( P(X \geq 2) \), we can use the complement rule:
\[ P(X \geq 2) = 1 - P(X < 2) = 1 - (P(X = 0) + P(X = 1)) \]

5. **Calculate \( P(X = 0) \)**:
Using the PMF:
\[ P(X = 0) = \frac{e^{-2} \cdot 2^0}{0!} = e^{-2} \]

6. **Calculate \( P(X = 1) \)**:
Using the PMF:
\[ P(X = 1) = \frac{e^{-2} \cdot 2^1}{1!} = 2e^{-2} \]

7. **Combine the Probabilities**:
Now we can combine these to find \( P(X < 2) \):
\[ P(X < 2) = P(X = 0) + P(X = 1) = e^{-2} + 2e^{-2} = 3e^{-2} \]

8. **Final Calculation**:
Now substitute this back into the equation for \( P(X \geq 2) \):
\[ P(X \geq 2) = 1 - P(X < 2) = 1 - 3e^{-2} \]

### Conclusion:
Thus, the final result is:
\[ P(X > 1.5) = 1 - 3e^{-2} \]