We have an infinite-server queue. Customers arrive as per a Poisson process with rate $\lambda$ per hour, but in groups of various random sizes that are i.i.d and also independent of the arrival times. The distribution of group sizes is geometric with parameter $a$. If it is given that the service time distribution is Exponential$(c)$, we want to calculate the expected number of customers whose service finished by time $t > 0$ hours.
My approach :
I started by finding out the probability of getting a group of size $k$ is $(1-a)^{k-1}a$.
If we take the Poisson process to be $N(t)$, given that $N(t)=N$, then the number of $k$-sized groups arrived by time $t$ is a binomial random variable with parameters $(N,(1-a)^{k-1}a)$.
The total number of individuals to have arrived by time $t$ has a distribution equivalent to $\sum_{k=1}^{N(t)} G_k$ where $G_k$ is the number of groups of size $k$. But I can't proceed after this.
I want to show that the arrival time of a randomly chosen individual who has arrived by time t is $U(0,t)$.