If X is a Poisson variate and P(X=1)=2P(X=2) them P(X=3)=

Options

A. `(e^(-1))/(6)`
B. `(e^(-2))/(2)`
C. `(e^(-1))/(2)`
D. `(e^(-1))/(3)`

Answer

To solve the problem, we need to find \( P(X=3) \) given that \( P(X=1) = 2P(X=2) \) for a Poisson random variable \( X \) with parameter \( \lambda \).

### Step-by-Step Solution:

1. **Recall the Poisson Probability Formula**:
The probability mass function for a Poisson random variable is given by:
\[ P(X=k) = \frac{e^{-\lambda} \lambda^k}{k!} \]

2. **Set Up the Equations**:
From the problem statement, we have:
\[ P(X=1) = 2P(X=2) \]
Using the Poisson formula, we can express this as:
\[ \frac{e^{-\lambda} \lambda^1}{1!} = 2 \cdot \frac{e^{-\lambda} \lambda^2}{2!} \]

3. **Simplify the Equation**:
Substitute the factorials:
\[ \frac{e^{-\lambda} \lambda}{1} = 2 \cdot \frac{e^{-\lambda} \lambda^2}{2} \]
This simplifies to:
\[ e^{-\lambda} \lambda = e^{-\lambda} \lambda^2 \]

4. **Cancel Common Terms**:
Since \( e^{-\lambda} \) is non-zero, we can divide both sides by \( e^{-\lambda} \):
\[ \lambda = \lambda^2 \]

5. **Rearrange the Equation**:
Rearranging gives:
\[ \lambda^2 - \lambda = 0 \]
Factor this equation:
\[ \lambda(\lambda - 1) = 0 \]

6. **Find Possible Values for \( \lambda \)**:
This gives us two solutions:
\[ \lambda = 0 \quad \text{or} \quad \lambda = 1 \]
Since \( \lambda = 0 \) does not make sense in the context of a Poisson distribution, we take \( \lambda = 1 \).

7. **Calculate \( P(X=3) \)**:
Now, we can find \( P(X=3) \) using \( \lambda = 1 \):
\[ P(X=3) = \frac{e^{-1} \cdot 1^3}{3!} = \frac{e^{-1}}{6} \]

8. **Final Answer**:
Therefore, the value of \( P(X=3) \) is:
\[ P(X=3) = \frac{e^{-1}}{6} \]