If X is a poisson distribution such that P(X=1)=P(X=2)then,P(X=4)=

Options

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

Answer

To solve the problem, we need to find \( P(X = 4) \) given that \( P(X = 1) = P(X = 2) \) for a Poisson distribution.

### Step-by-Step Solution:

1. **Understand the Poisson Distribution Formula**:
The probability mass function of a Poisson distribution is given by:
\[ P(X = k) = \frac{e^{-\mu} \mu^k}{k!} \]
where \( \mu \) is the average rate (mean) of occurrences.

2. **Set Up the Equation**:
From the problem, we know:
\[ P(X = 1) = P(X = 2) \]
Using the Poisson formula, we can write:
\[ P(X = 1) = \frac{e^{-\mu} \mu^1}{1!} = e^{-\mu} \mu \]
\[ P(X = 2) = \frac{e^{-\mu} \mu^2}{2!} = \frac{e^{-\mu} \mu^2}{2} \]

3. **Equate the Two Probabilities**:
Setting the two probabilities equal gives:
\[ e^{-\mu} \mu = \frac{e^{-\mu} \mu^2}{2} \]

4. **Cancel \( e^{-\mu} \)**:
Since \( e^{-\mu} \) is common on both sides and is not zero, we can cancel it:
\[ \mu = \frac{\mu^2}{2} \]

5. **Solve for \( \mu \)**:
Rearranging the equation gives:
\[ 2\mu = \mu^2 \]
\[ \mu^2 - 2\mu = 0 \]
Factoring out \( \mu \):
\[ \mu(\mu - 2) = 0 \]
This gives us two solutions: \( \mu = 0 \) or \( \mu = 2 \). Since \( \mu = 0 \) does not make sense in this context, we take \( \mu = 2 \).

6. **Calculate \( P(X = 4) \)**:
Now that we have \( \mu = 2 \), we can find \( P(X = 4) \):
\[ P(X = 4) = \frac{e^{-2} 2^4}{4!} \]
Calculate \( 2^4 = 16 \) and \( 4! = 24 \):
\[ P(X = 4) = \frac{e^{-2} \cdot 16}{24} \]
Simplifying gives:
\[ P(X = 4) = \frac{2}{3} e^{-2} \]

### Final Answer:
Thus, the final result is:
\[ P(X = 4) = \frac{2}{3} e^{-2} \]