At a 24-hour movie theater, customers arrive at a rate of 10 customers per hour and buy a horror movie ticket with probability 0.2, independent of what other customers buy.
Given that no one bought a horror movie ticket during the first 2 hours of the day, how many hours will it take to until 10 customers buy a horror movie ticket.
Find the expected value.
I believe this is a gamma distribution, so I divided 10 (lambda) by 10 by 0.2 = 2 (alpha) to get 5 hours. Can someone please confirm I did this correctly?
You are correct that the random time until the $10^{\rm th}$ horror movie purchase occurs is gamma distributed, assuming that the customer arrivals obey a homogeneous Poisson process. The parameters of such a gamma distribution are $n = 10$, which represents the number of events of interest we need to observe--$10$ horror movie purchases, and $\lambda = 2$, because the Poisson intensity for the rate of horror movie purchases per hour is $(0.2)(10) = 2$. Therefore, the expected value is $n/\lambda = 5$ hours.
This makes intuitive sense: if on average $2$ customers buy a horror ticket per hour, it will take on average $5$ hours for $10$ such tickets to be bought.
It is important to note that the above calculation does not include the $2$ hours that elapsed without a purchase; i.e., it calculates the average amount of additional time it will take.
As an exercise for the reader: