i know the poisson formula but how does a given condition affect it?

i.e. E(X)=3, find P(X=2|X>1).

i have X>1, as 1 - 4e^-3, and i can work out the normal P(X=2) but how does the given X>1 affect it? i dont know what to do

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- Dec 14th 2010, 09:32 AMmathcorepoisson with given
i know the poisson formula but how does a given condition affect it?

i.e. E(X)=3, find P(X=2|X>1).

i have X>1, as 1 - 4e^-3, and i can work out the normal P(X=2) but how does the given X>1 affect it? i dont know what to do - Dec 14th 2010, 10:38 AMmr fantastic
- Dec 14th 2010, 11:26 AMmathcore
so this will just be p(x=2)/p(x>1) then? just divide my own calculations

i get 0.28 is thAT RIGHT - Dec 14th 2010, 06:30 PMmr fantastic
- Dec 15th 2010, 01:46 AMmathcore
yesh but thats what im asking if i did it correctly or not... is my result correct or not...?

- Dec 15th 2010, 02:51 AMCaptainBlack
Bayes' theorem can be used here:

$\displaystyle P(x=2|X>1)=\dfrac{P(X>1|X=2)P(X=2)}{P(X>1)}=\dfrac {P(X=2)}{1-(P(X=0)+P(X+1))}$

where:

$\displaystyle P(X=k)=\dfrac{3^ke^{-3}}{k!}$

To two significant firures this gives $\displaystyle 0.28$ which is your answer, but you should give more digits (even if they are zeros)

CB