Probability Distribution

Question

Let X have a Poisson distribution with parameter λ = 1. What is the probability that X ≥ 2 given that X ≤ 4?

Answer 1

P(X>=2) given X<=4 and λ = 1 :

Poisson Distribution falls under Discrete Random Variable so considering all the discrete points in the range {2,3,4}.

= P(X=2) + P(X=3) + P(X=4)

= (e^(-1) * 1^2)/2! + (e^(-1) * 1^3)/3! + (e^(-1) * 1^4)/4!

= (1/e) * (½ + 1/6 + 1/24)

= 17/24e

= 0.26058

Now we need to find out

P(X>=2 / X<=4)

=  ( P(X=2) + P(X=3) + P(X=4) ) / ( P(X=0) + P(X=1) + P(X=2) + P(X=3) + P(X=4) )

=  0.26058 / 2.70834

=  0.0962

Comments

In question, it is asked for the conditional probability.

$\mathbb{P}(X=k) = \frac{e^{-\lambda} \lambda^k}{k!}$ where $\lambda \in (0,\infty)$ and $k \in N_0$ (Natural numbers starts from 0)

For $\lambda=1,$ $\mathbb{P}(X=k) = \frac{1}{ek!}$

Now, $\mathbb{P}(X \geq 2 \  | \ X \leq 4 ) = \frac{\mathbb{P} (X \geq 2 \cap X \leq 4)}{\mathbb{P} (x \leq 4)}$

$= \frac{\mathbb{P}(X=2)+ \mathbb{P}(X=3) + \mathbb{P}(X=4)}{\mathbb{P}(X=0)+ \mathbb{P}(X=1)+ \mathbb{P}(X=2)+\mathbb{P}(X=3) + \mathbb{P}(X=4)}$

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@ankitgupta.1729 sir can you edit the answer ?

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I can but without @Sunnidhya Roy’s confirmation I will not edit the answer. Previously I faced some issue while editing someone’s answer.

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@ankitgupta.1729 Thanks for pointing out the mistake. I will update it.