Question

If $X$ follows Poisson distribution with variance 2, then $P(X \geq 3) = $

• A. \( \dfrac{5}{e^2} \)
• B. \( \dfrac{5 + 2}{e^2} \)
• C. \( \dfrac{e^2 - 5}{e^2} \)
• D. \( \dfrac{5 - e^2}{4} \)

Hint:

For Poisson distributions, use cumulative probabilities up to one less than the desired threshold and subtract from 1 to find tail probabilities.

Answer 1

The Correct Option is C

Step 1: Understand the given information
Given: Poisson distribution with variance \( \lambda = 2 \)
So, \( X \sim \text{Poisson}(\lambda = 2) \)
We are required to find:
\[ P(X \geq 3) = 1 - P(X < 3) = 1 - P(X = 0) - P(X = 1) - P(X = 2) \] Step 2: Use the Poisson formula
\[ P(X = k) = \frac{e^{-\lambda} \lambda^k}{k!}, \quad \lambda = 2 \] Compute individual probabilities:
\[ P(X = 0) = \frac{e^{-2} \cdot 2^0}{0!} = e^{-2} \] \[ P(X = 1) = \frac{e^{-2} \cdot 2^1}{1!} = 2e^{-2} \] \[ P(X = 2) = \frac{e^{-2} \cdot 2^2}{2!} = \frac{4e^{-2}}{2} = 2e^{-2} \] Step 3: Sum and subtract from 1
\[ P(X < 3) = e^{-2} + 2e^{-2} + 2e^{-2} = 5e^{-2} \] \[ P(X \geq 3) = 1 - 5e^{-2} \] \[ P(X \geq 3) = \frac{e^2 - 5}{e^2} \] % Final Answer
\[ \boxed{P(X \geq 3) = \frac{e^2 - 5}{e^2}} \]