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