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
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