Let an insurance company be such that it receives two types of claims:
Type 1 has Poisson$(\lambda_1)$ arrival and Exp$(\mu_1)$ size distributions (iid)
Type 2 has Poisson$(\lambda_2)$ arrival and Exp$(\mu_2)$ size distributions (iid)
Given a claim has just arrived:
a) Find the probability the claim is type 1.
b) find probability the claim is larger than 1.
For (a) from experience with similar problems it seems the probability should be proportional to the sum of both arrival rates so something like $\displaystyle \frac{\lambda_1}{\lambda_1+\lambda_2}$ but I am not sure if this is correct because I find it difficult to formalize why this would be the case. Namely, how to set up the probability.
For (b) I believe this would be
$P(C > 1) = P(C > 1 \mid C \text{ is type 1})P(\text{arrival of type 1} )+P(C > 1 \mid C \text{ is type 2})P(\text{arrival of type 2} )$
$\\= 1 - P(C \le 1 \mid C \text{ is type 1})P(\text{arrival of type 1} )+1- P(C \le 1 \mid C \text{ is type 2})P(\text{arrival of type 2} )$
Let $\{N_1(t):t\geqslant0\}$ and $\{N_2(t):t\geqslant0\}$ be independent Poisson processes with rates $\lambda_1$ and $\lambda_2$, respectively. Let $T^{(i)}_n$ be the $n^{\mathrm{th}}$ arrival time of the $i^{\mathrm{th}}$ process. Problem (a) is to compute the probability of the event $$ \left\{T^{(1)}_1<T^{(2)}_1\right\} $$ First we need the result that if $X\sim\mathrm{Expo}(\lambda)$ and $Y\sim\mathrm{Expo}(\mu)$ are independent, then $$ \mathbb P(X<Y) = \frac\lambda{\lambda+\mu}. $$ We compute this probability by integrating over the joint density: \begin{align} \mathbb P(X<Y) &= \int_{(0,\infty)^2}f_{X,Y}(x,y)\ \mathsf d(x\times y)\\ &= \int_0^\infty \int_0^y \lambda e^{-\lambda x}\mu e^{-\mu y}\\ &= \frac\lambda{\lambda+\mu}. \end{align} It follows then, that $$ \mathbb P\left(T^{(1)}_1<T^{(2)}_1\right) = \frac{\lambda_1}{\lambda_1+\lambda_2}. $$
For problem (b), let $C_n^{(i)}$ be the size of the $n^{\mathrm{th}}$ claim of the $i^{\mathrm{th}}$ process, and $E=\mathsf 1_{C_1^{(1)}}+2\cdot\mathsf 1_{C_1^{(2)}}$. Then by the law of total probability, \begin{align} \mathbb P(C_1^E > 1) &= \mathbb P(C_1^E > 1\mid E=1)\mathbb P(E=1) + \mathbb P(C_1^E > 1\mid E=2)\mathbb P(E=2)\\ &= \int_1^\infty \lambda_1 e^{-\lambda_1 t}\ \mathsf dt \left(\frac{\lambda_1}{\lambda_1+\lambda_2}\right)+\int_1^\infty \lambda_1 e^{-\lambda_2 t}\ \mathsf dt \left(\frac{\lambda_2}{\lambda_1+\lambda_2}\right)\\ &=\frac{\lambda _1e^{-\lambda _1} +\lambda _2e^{-\lambda _2} }{\lambda _1+\lambda _2}. \end{align}