Question

If $ X $ is a Poisson variate such that $ P(X=1) = \frac{3}{10} $ and $ P(X=2) = \frac{1}{5} $, then the mean of the distribution is:

• A. \( \frac{3}{2} \)
• B. \( \frac{3}{4} \)
• C. \( \frac{2}{3} \)
• D. \( \frac{4}{3} \)

Hint:

Use ratio of probabilities in Poisson to solve for mean \( \lambda \).

Answer 1

The Correct Option is D

For a Poisson distribution with mean \( \lambda \): \[ P(X = k) = \frac{\lambda^k e^{-\lambda}}{k!} \] Given: \[ P(X=1) = \lambda e^{-\lambda} = \frac{3}{10}, \quad P(X=2) = \frac{\lambda^2}{2} e^{-\lambda} = \frac{1}{5} \] Divide the second by the first: \[ \frac{P(X=2)}{P(X=1)} = \frac{\lambda}{2} = \frac{1/5}{3/10} = \frac{2}{3} \] \[ \Rightarrow \lambda = \frac{4}{3} \]