## Poisson Distribution Question-is this ok?

Customers arrive randomly at a toll at an average rate of 4 per minute.calculate probability that:
(i) exactly three customers arrive in any one minute period
(ii) two or less customers arrive in any one minute period
(iii) more that five customers arrive in any one minute period.

My workings:

(i)Formula:

$\frac{(e^-\lambda)(\lambda^x)}{x!}$

(i) exactly 3

$\frac{(e^-4)(4^3)}{3!}=0.19537$ to be continued

(ii) two or less

$\frac{(e^-4)(4^0)}{0!}+\frac{(e^-4)(4^1)}{1!}+\frac{(e^-4)(4^2)}{2!}=0.0183156388+0.07326256+0.146525=0.23 81$

(iii)more than five = 1-P(less than 5)

$0.2381+\frac{(e^-4)(4^3)}{3!}+\frac{(e^-4)(4^4)}{4!}+\frac{(e^-4)(4^5)}{5!}=0.195367+0.195367+0.156293=0.785127$

Therefore P(more than 5)

$1-0.785127=0.214873$