Question

A Poisson variate \( X \) satisfies \( P(X=1) = P(X=2) \). \( P(X=6) \) is equal to

• A. \( \frac{4}{45}e^{-2} \)
• B. \( \frac{4}{45}e^{-1} \)
• C. \( \frac{1}{9}e^{-2} \)
• D. \( \frac{1}{4}e^{-2} \)
• E. \( \frac{1}{45}e^{-2} \)

Hint:

Equal probabilities in Poisson often help determine \( \lambda \).

Answer 1

The Correct Option is A

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

Step 1: Use given condition.
\[ P(1) = P(2) \] \[ \frac{e^{-\lambda}\lambda}{1!} = \frac{e^{-\lambda}\lambda^2}{2!} \]

Step 2: Cancel common terms.
\[ \lambda = \frac{\lambda^2}{2} \]

Step 3: Solve for \( \lambda \).
\[ 2\lambda = \lambda^2 \Rightarrow \lambda = 2 \]

Step 4: Compute \( P(6) \).
\[ P(6) = \frac{e^{-2}2^6}{6!} \]

Step 5: Simplify.
\[ = \frac{64e^{-2}}{720} = \frac{4}{45}e^{-2} \]