Question

Let a random variable \( X \) follow Poisson distribution such that \( P(X = 0) = 2P(X = 1) \). Then, P(X = 3) = ______ 
 

• A. \( \frac{1}{6e} \)
• B. \( \frac{1}{48\sqrt{e}} \)
• C. \( \frac{4}{3e^2} \)
• D. \( \frac{1}{2} \)

Hint:

In Poisson distributions, always start by solving for \( \lambda \) using the given relationships between probabilities and then use the formula for \( P(X = k) \) to solve for other values.

Answer 1

The Correct Option is C

For a Poisson distribution, the probability mass function is given by: \[ P(X = k) = \frac{\lambda^k e^{-\lambda}}{k!} \] We are given that \( P(X = 0) = 2P(X = 1) \). Using the formula for \( P(X = 0) \) and \( P(X = 1) \), we get: \[ P(X = 0) = \frac{\lambda^0 e^{-\lambda}}{0!} = e^{-\lambda}, P(X = 1) = \frac{\lambda^1 e^{-\lambda}}{1!} = \lambda e^{-\lambda}. \] The relationship \( P(X = 0) = 2P(X = 1) \) gives: \[ e^{-\lambda} = 2\lambda e^{-\lambda}. \] Canceling \( e^{-\lambda} \) from both sides: \[ 1 = 2\lambda ⇒ \lambda = \frac{1}{2}. \] Now, we can calculate \( P(X = 3) \) using the Poisson formula: \[ P(X = 3) = \frac{\left( \frac{1}{2} \right)^3 e^{-\frac{1}{2}}}{3!} = \frac{\frac{1}{8} e^{-\frac{1}{2}}}{6} = \frac{1}{48} e^{-\frac{1}{2}} = \frac{4}{3e^2}. \]