Question

If \(3\) is the variance of Poisson distribution, then \[ P(1\lt x\lt 4)= \] is:

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

Hint:

For Poisson distribution, \[ \text{Mean}=\text{Variance}=\lambda \] Always identify the value of \(\lambda\) first before calculating probabilities.

Answer 1

The Correct Option is C

Step 1: Identify $\lambda$ from the variance.
For Poisson distribution, mean $=$ variance $= \lambda$. Given variance $= 3$, so $\lambda = 3$.

Step 2: Interpret $P(1 < x < 4)$.
Since $x$ takes non-negative integer values, $1 < x < 4$ means $x \in \{2, 3\}$. So: \[ P(1<x<4) = P(x=2)+P(x=3) \]

Step 3: Use the Poisson formula $P(x=r) = \frac{e^{-\lambda}\lambda^r}{r!}$.
$P(x=2) = \frac{e^{-3} \cdot 9}{2} = \frac{9e^{-3}}{2}$. $P(x=3) = \frac{e^{-3} \cdot 27}{6} = \frac{27e^{-3}}{6} = \frac{9e^{-3}}{2}$.

Step 4: Add the two probabilities.
\[ P(1<x<4) = \frac{9e^{-3}}{2} + \frac{9e^{-3}}{2} = 9e^{-3} \]

Step 5: Note the coincidence that $P(x=2)=P(x=3)$.
This happens when $\frac{\lambda}{r} = 1$, i.e., $\lambda = r$. Here $\lambda=3$ and $r=3$: $P(x=3) = \frac{\lambda}{3}P(x=2) = P(x=2)$. So they are equal.

Step 6: State the answer.
\[ \boxed{9e^{-3}} \]