Poisson question here thats causing me problems...
A mountain hut is available for walkers to stay overnight. Members of the organisation which owns the hut have keys; others can turn up in the hope a keyholder will let them in. So, on any given night, if at least one keyholder turns up, everyone who turns up gets in; otherwise no-one gets in.
Suppose that keyholders arrive randomly at a hut, so that on a given night the number arriving follows a Poisson distribution with mean λ. Suppose also that the number of non-keyholders arriving follows a Poisson distribution with mean µ, and is independent of the keyholders. Let X be the number of people occupying the hut (on a given night). Write down an expression for the probability that k keyholders and l non-keyholders occupy the hut, where k > 0, and by doing an appropriate summation show that
P(X = n) = [(λ + µ)^n - µ^n . e^-λ-µ]/n! (sorry dont have any proper math symbol program.)
Consider the situation described above, with λ = 3 and µ = 5
1) On average, how many people are expected to be in the hut per night?
2) On a given night, 4 people are in the hut. What is the probability that they are all keyholders?
The last step uses a rearrangement of the binomial theorem.