Calculating Poisson Probability for Car Rental Income

[thereddevils]

Homework Statement

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.

Homework Equations

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

But my answer is wrong.

 

[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?

 

[thereddevils]

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?

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.