# Probability Poisson Distribution

• May 3rd 2010, 10:40 PM
Qwertyuiop23
Probability Poisson Distribution
I just got a question in my exercise book that I tried to attempt but continually got the wrong answer. I am asking here to see if it was my problem or the books. The question is really quite simple:

Y is distributed as a poisson distribution. When m = 2.7 (mean) find $\displaystyle P(X<=4|X>=2)$

I thought it would just be the probability of $\displaystyle P(X=2) + P(X=3) + P(X=4)$ as all events are independent. However, this is not the correct answer.

I tried to do it using the formula for conditional probability but of course as they are independent this did not work.

A progression of the solution would be appreciated.

Cheers
Qwertyuiop23
• May 4th 2010, 01:56 AM
downthesun01
What's the correct answer?

Have you tried $\displaystyle \frac{pr(2\leq X \leq 4)}{pr(X\geq 2)}$$\displaystyle =\frac{pr(X=2)+pr(X=3)+pr(X=4)}{1-[pr(X=1)+pr(X=0)]}$?

This would make sense to me since $\displaystyle pr(C|A)=\frac{pr(A\cap C)}{pr(A)}$