1. ## poisson 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

2. Originally Posted by mathcore
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
You're expected to understand conditional probability: $\displaystyle \displaystyle \Pr(A | B) = \frac{\Pr(A \cap B)}{\Pr(B)}$.

Note that $\displaystyle \Pr(X = 2 \cap X > 1)$ is equivalent to $\displaystyle \Pr(X = 2)$ ....

3. so this will just be p(x=2)/p(x>1) then? just divide my own calculations

i get 0.28 is thAT RIGHT

4. Originally Posted by mathcore
so this will just be p(x=2)/p(x>1) then? just divide my own calculations Mr F says: Yes.

i get 0.28 is thAT RIGHT
If you have calculated each probability correctly and then done the arithmetic correctly then your answer will be correct.

5. yesh but thats what im asking if i did it correctly or not... is my result correct or not...?

6. Originally Posted by mathcore
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