Find probability of a Poisson process. Given that $N=\{N(t)\mid t\geq 0\}$ is a Poisson process with parameter $\lambda>0$
I need to find $P(N(3)=2\mid N(1)=0, N(5)=4)$
So this is a conditional probability (can anyone clarify if this is correct) and I assume that the mass function for a Poisson process is:
$p(N(t)=k)=e^{-\lambda t}\frac{(\lambda t)^k}{k!}$
But if I think in the way $P(A\mid B)=\frac{P(A\cap B)}{P(B)}$, I get wrong answer (the answer should not depend on $\lambda$).
Can anyone clarify how this calculation should be done? (note my coursebook is extremely abstract with no examples whatsoever) :(
 A: The crucial property of Poisson processes that you should have mentioned in the comments since @LutzL prodded you for it (and even in the question itself) is the independence of the increments, that is, the fact that the number of events of the process in disjoint intervals are independent. 
Here, three disjoint intervals are involved, namely $(0,1]$, $(1,3]$ and $[3,5]$, hence one should try to write down everything in terms of $X=N(1)$, $Y=N(3)-N(1)$ and $Z=N(5)-N(3)$. Then, $X$, $Y$ and $Z$ are Poisson with the parameters you know and independent, and you are interested in the events
$$
B=[N(1)=0,N(5)=4]=[X=0,Y+Z=4],
$$
and
$$
A=[N(1)=0,N(3)=2,N(5)=4]=[X=0,Y=2,Z=2].
$$
Thus, 
$$
P[B]=p_\lambda(0)p_{4\lambda}(4),\qquad P[A]=p_\lambda(0)p_{2\lambda}(2)p_{2\lambda}(2),
$$
where $p_\mu(n)$ denotes the probabiolity that a Poisson random variable with parameter $\mu$ is $n$.
Thus, you are looking for
$$
\frac{P[A]}{P[B]}=\frac{p_\lambda(0)p_{2\lambda}(2)p_{2\lambda}(2)}{p_\lambda(0)p_{4\lambda}(4)}.
$$
If you compute this ratio, you should see that $\lambda$ disappears...
