I am trying to solve the following problem.
Customers arrive at a service facility in groups. The groups arrive according to a homogeneous Poisson process at the rate of 5 per minute. The number N of individuals in each group is described by the distribution $P_N(1) =0.3$, $P_N(2) =0.6$ and $P_N(3) =0.1$
- What are the probabilities of zero, one and two arrivals in one minute?
- Suppose a customer in the system is chosen at random. What is the probability that he/she arrived in a group of size n?
Thoughts towards a solution $P(S) = 0.3s+0.6s^{2}+0.1s^{3}$ So the generating function would be $G(s) = e^{5t(0.3s+0.6s^{2}+0.1s^{3})}$
Firstly i am unsure how to use the generating function to evaluate the first part. I think the probability is given by the coefficients of the z terms so the answers to the three questions should be 0, $e^{0.15t}$ and $e^{3t}$ respectively.
As per the second part i suspect the answer would be obtained by a conditional probability, as in the product of he/she arriving in a group of size n(1/3), times the probability of such a group coming in to the store, as stated above (assuming that is correct) but i am unsure how to put this into a formula
Any help would be much appreciated.