Say customers arrive at an ice cream shop at an average rate of 100 customers per day. Independently, the ice cream shop receives a restock of their ice cream at an average rate of one shipment every 5 days. Say each customer buys only one serving of ice cream. If the shop begins the week with 500 servings of ice cream, what is the probability the next shipment will arrive before they run out of ice cream? Both the arrivals of the shipments and the arrivals of the customers are indeed poisson processes.
Purely intuitively, I believe the answer should be $1/5$, but I cannot figure out how to actually rigorously calculate this. I think the question is essentially asking "what is the probability one restock arrives before 5 groups of 100 customers arrive". So I believe this is a question of superposition of poisson processes, in that, the probability that, each day, a shipment comes in before 100 customers come in is $\frac{\frac{1}{5}}{\frac{1}{5} + 100}$. And then the probability that 100 customers come in on a given day before a shipment comes in is $\frac{100}{\frac{1}{5} + 100}$. At this point I get lost. I believe there is a combinatorial argument to be made here but I'm not sure what that is. I would appreciate if someone could help me walk through this problem
This answer assumes the shipment arrivals are also a Poisson process (with rate $1/5$ arrivals per day).
If you superimpose the processes, you have a Poisson process with rate $100+1/5$, where each arrival is independently either a customer or a shipment with probabilities $p := \frac{100}{100+1/5}$ and $1-p = \frac{1/5}{100+1/5}$ respectively.
You want the first arrival that is a shipment to be within the first 500 arrivals. It may be easier to consider the complement event: the first 500 arrivals are all customers. This is $p^{500} \approx 0.368$, so the probability you want is $\approx 0.632$.