A bus with infinite capacity runs on an infinite route with stops indexed $n = 1, 2, 3,\dots$ The bus arrives empty at stop $1$. When the bus arrives at stop $n$, each passenger on it gets off with probability $p$, independent of each other. Then all the waiting passengers get on. Let $W(n)$ be the number of passengers waiting at stop $n$, and suppose $\{W(n), n\ge0\}$ are iid Possion$(λ)$ random variables. Let $R(n)$ be the number of passengers on the bus when it leaves stop $n$. What is the distribution of $R(n)$?
The answer is given as: $$Poisson(λ/p*(1-(1-p)^n))$$
My attempt: $$R(n)=Binomial(R(n-1),1-p)+Poisson(λ) = Poisson(R(n-1)*(1-p)+λ)$$ I don't know what to do past this, or if I'm thinking about this the right way.
For $i=1,\ldots,n,\;$ let $R_i(n)$ be the number of passengers on the bus when it leaves stop $n$ that got on at stop $i$.
So $R(n) = \sum\limits_{i=1}^n R_i(n).$
Any passenger getting on a stop $i$ must "survive" $n-i$ stops to still be on the bus when leaving stop $n$. This has probability $(1-p)^{n-i}$.
Thus, $R_i(n)$ has a Poisson distribution with rate $\lambda(1-p)^{n-i}$.
Now, the sum of independent Poisson variables also is Poisson with rate equal to the sum of the individual Poisson rates. Here, this sum of rates is
$$\sum\limits_{i=1}^n \lambda(1-p)^{n-i} = \lambda\sum\limits_{i=0}^{n-1} (1-p)^{n-i} = \lambda\dfrac{1-(1-p)^n}{1-(1-p)} = \dfrac{\lambda}{p}(1-(1-p))^n.$$
So, $R(n)\sim \text{Poisson}\left(\dfrac{\lambda}{p}(1-(1-p))^n\right).$