Poisson process - calculating probabilities 
Let $\{X(t), t \geq 0 \}$ be a Poisson process with  $\lambda$ as parameter. Calculate:


*

*$\mathbb{P}(X(5)=3 | X(1)+X(2)=3)$

*$\mathbb{P}(X(5)-X(3)=1 | X(4)-X(2)=1)$

I did the first one as follows:
$$
\mathbb{P}(X(5)=3 | X(1)+X(2)=3) = \\ = \frac{\mathbb{P}(X(5)=3 , X(1)+X(2)=3)}{\mathbb{P}(X(1)+X(2)=3)} = \\ = \frac{\mathbb{P}(X(5) - (X(1)+X(2))=3 - 3 , X(1)+X(2)=3)}{\mathbb{P}(X(1)+X(2)=3)} = \\ =  \frac{\mathbb{P}(X(5 - (1+2))=0)\mathbb{P}(X(1)+X(2)=3)}{\mathbb{P} =(X(1)+X(2)=3)} = \\ = \mathbb{P}(X(2)=0)  = e^{-2\lambda}
$$
Is this correct?
What about the second one? I got kinda stuck there:
$$
\mathbb{P}(X(5)-X(3)=1 | X(4)-X(2)=1) = \frac{\mathbb{P}(X(5)-X(3)=1 , X(4)-X(2)=1)}{\mathbb{P}(X(4)-X(2)=1)} = \\
= \frac{\mathbb{P}(X(5)-X(3) - (X(4)-X(2))=1 - 1 , X(4)-X(2)=1)}{\mathbb{P}(X(4)-X(2)=1)} = \\ =
\frac{\mathbb{P}(X((5-3) - (4-2))=1 - 1 , X(4)-X(2)=1)}{\mathbb{P}(X(4)-X(2)=1)} = \frac{\mathbb{P}(X(0)=0)\mathbb{P}( X(4)-X(2)=1)}{\mathbb{P}(X(4)-X(2)=1)} = \mathbb{P}(X(0)=0) = 1
$$
This seems fishy. I appreciate any help or affirmation that it is actually correct.
 A: You should note that $X(1)$, $X(2)$, and $X(5)$ are not independent random variables, since they count Poisson point-events occurring in overlapping intervals.
However, $X(1)$, $X(2){-}X(1)$, and $X(5){-}X(2)$, are independent.   Further, they are integer-valued, being Poisson distributed with rate parameters $\lambda, \lambda,$ and $3\lambda$, respectively.
To save type-space, let us define $X(s,t)$ as $X(t){-}X(s)$.
$$~~~X(1)\sim\mathcal P(\lambda)\\X(1,2)\sim\mathcal P(\lambda)\\~~X(2,5)\sim\mathcal P(3\lambda)$$
The event of $\{X(1){+}X(2){=}3\}$ is thus the event that $\{2X(1){+}X(1,2){=}3\}$ and as such, this event is :
$$\{X(1){=}0, X(1,2){=}3\}{\,\cup\,}\{X(1){=}1, X(1,2){=}1\}$$
Likewise the intersection of this event and $\{X(5){=}3\}$ will be :
$$\{X(1){=}0, X(1,2){=}3,X(2,5){=}0\}{\,\cup\,}\{X(1){=}1,X(1,2){=}1,X(2,5){=}1\}$$
Then you can now evaluate $\mathsf P(X(5){=}3\mid X(1){+}X(2){=}3)$ using Bayes' rule.

Similar reasoning will let you evaluate $\mathsf P(X(5){−}X(3){=}1\mid X(4){−}X(2){=}1)$, using the independent random variables $X(2,3)$, $X(3,4)$, and $X(4,5)$.
