# Homework Help: Poisson probabiltiy

1. Jul 14, 2010

### thereddevils

1. The problem statement, all variables and given/known data

A car rental shop has four cars to be rented out on a daily basis at $50 per car. The average daily demand for cars is four. (1) Calculate the expected daily income received from the rentals (2) If the shop wishes to have one more car, the additional cost incurred is$ 20 per day DEtermine whether the shop should buy another car for rental.

2. Relevant equations

3. The attempt at a solution

(1) X-p(4)

P(X=0)=0.0183
P(X=1)=0.07326
P(X=2)=0.1465
P(X=3)=0.195
P(X=4)=0.195

The expected number of cars rented out is about 1.73 so the expected income is

1.73 x $50=$ 86.6

2. Jul 14, 2010

### CompuChip

Why do you stop calculating probabilities at 4... they hardly add up to the total of 1, so you are using an incomplete distribution here.

I'm not really into Poisson distributions, but if the average is 4, isn't the answer simply 4 x \$50 as it would with a binomial distribution?

3. Jul 14, 2010

### thereddevils

thanks Compuchip, i figured that out. I calculated P(X=4) instead of P(X>=4), that's why the probabilities do not add up to 1.