I have the Poisson process $\{N(t)\}_{t\geq 0}$ with rate $\lambda=2$. Given that four events occur during the time interval $[0,2]$, what is the probability that the first event occurs before time $t=1$?
From what I understand, I need to calculate $\mathbb{P}(N(1)\geq1\mid N(2)-N(0)=4).$
So I assume I must use the conditional probability formula \begin{equation}\frac{\mathbb{P}(N(1)\geq 1,N(2)-N(0)=4)}{\mathbb{P}(N(2)-N(0)=4)} \end{equation}
I struggle now to see the intersection between the two parts of my numerator. I also am not too confident my workings for the denominator are correct. \begin{equation} \mathbb{P}(N(2)-N(0)=4)=e^{-2}\frac{(2)^4}{4!}=e^{-2}\frac{2}{3} \end{equation} Could someone explain to me how to identify the intersection in the numerator and if my calculation for the denominator is correct?
In the definition of Poisson Process it is assumed that $N(0)=0$. [Ref. https://en.wikipedia.org/wiki/Poisson_point_process ]
$N(2)-N(0)=N(2)$ is Poisson with parameter $4$. So the denominator is $e^{-4}\frac {4^{4}} {4!}$.
Hint for the numerator: Let $X=N(1)$ and $Y=N(2)-N(1)$. Then $X$ and $Y$ are independent with $Poiss(2)$ distribution. . Hence $P(X \geq 1, X+Y=4)= \sum\limits_{n=1}^{4} P(X=n) P(Y=4-n)=\sum\limits_{n=1}^{4}e^{-2} \frac {2^{n}} {n!} e^{-2}\frac {2^{4-n}} {(4-n)!}$. .