# Poisson problem - Help!

• Oct 29th 2007, 06:29 PM
Poisson problem - Help!
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?
• Oct 30th 2007, 11:35 PM
CaptainBlack
Quote:

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?

$\displaystyle p(k,l)=\frac{\lambda^k e^{-\lambda}}{k!} ~\frac{\mu^l e^{-\mu}}{l!}$

$\displaystyle P(X=n)=\sum_{k=1}^n p(k,n-l) =\sum_{k=1}^n \frac{\lambda^k e^{-\lambda}}{k!} ~\frac{\mu^{n-k} e^{-\mu}}{(n-k)!} $$\displaystyle =\frac{e^{-\lambda-\mu}}{n!}\sum_{k=1}^n \frac{n!}{k!(n-k)! }\lambda^k \mu^{n-k}$$\displaystyle =\frac{e^{-\lambda-\mu}}{n!} \left( (\lambda + \mu)^n-\mu^n\right)$

The last step uses a rearrangement of the binomial theorem.

RonL