Question

If variance of a Poisson distribution is 3, find: \[ P(1<X<4) \]

• A. \( \frac{123}{8} e^{-3} \)
• B. \( 3e^{-\sqrt{3}} \)
• C. \( 9e^{-3} \)
• D. \( \left( \frac{3 + \sqrt{3}}{2} \right) e^{-3} \)

Hint:

Use the Poisson formula and compute each term precisely. For consecutive values, sum corresponding probabilities.

Answer 1

The Correct Option is C

Poisson distribution with \( \lambda = 3 \) (since variance = λ). We want: \[ P(1<X<4) = P(X = 2) + P(X = 3) \] Use Poisson probability formula: \[ \begin{align} P(X = r) = \frac{e^{-\lambda} \lambda^r}{r!} \Rightarrow P(X = 2) = \frac{e^{-3} \cdot 3^2}{2!} = \frac{9e^{-3}}{2} \Rightarrow P(X = 3) = \frac{e^{-3} \cdot 3^3}{3!} = \frac{27e^{-3}}{6} = \frac{9e^{-3}}{2} \] Add: \[ P = \frac{9e^{-3}}{2} + \frac{9e^{-3}}{2} = 9e^{-3} \]