Question

If X is random Poisson variate such that P(X=1) = P(X=2). What is the P(X=4) ?

• A. $(5e^{-2})/6$
• B. $(2e^{-2})/2$
• C. $(e^{-2})/3$
• D. $(2e^{-2})/3$

Hint:

For any Poisson distribution, if \( P(X = m) = P(X = m+1) \), then the parameter \( \lambda \) is simply equal to \( m + 1 \). Here, \( P(X=1) = P(X=2) \implies \lambda = 2 \).

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
The Poisson distribution is a discrete probability distribution that models the number of events occurring within a fixed interval of time or space.
Key Formula or Approach:
The probability mass function of a Poisson random variable \( X \) with parameter \( \lambda \) (\( \lambda > 0 \)) is given by:
\[ P(X = k) = \frac{e^{-\lambda} \lambda^k}{k!} \]

Step 2: Detailed Explanation:
We are given that:
\[ P(X = 1) = P(X = 2) \]
Substituting \( k = 1 \) and \( k = 2 \) into the formula:
\[ \frac{e^{-\lambda} \lambda^1}{1!} = \frac{e^{-\lambda} \lambda^2}{2!} \]
Since \( e^{-\lambda} \neq 0 \) and \( \lambda > 0 \), we can divide both sides by \( e^{-\lambda} \):
\[ \lambda = \frac{\lambda^2}{2} \]
Dividing both sides by \( \lambda \):
\[ 1 = \frac{\lambda}{2} \implies \lambda = 2 \]
Thus, the parameter of the Poisson distribution is \( \lambda = 2 \).
Now, we calculate \( P(X = 4) \) using the value of \( \lambda = 2 \):
\[ P(X = 4) = \frac{e^{-2} \cdot 2^4}{4!} \]
We expand the power and the factorial:
\[ 2^4 = 16 \]
\[ 4! = 4 \times 3 \times 2 \times 1 = 24 \]
Thus, we have:
\[ P(X = 4) = e^{-2} \cdot \frac{16}{24} = e^{-2} \cdot \frac{2}{3} = \frac{2e^{-2}}{3} \]
This matches option (D).

Step 3: Final Answer:
The value of P(X=4) is $(2e^{-2})/3$.