Customers arrive to a bank branch according to a Poisson pro

Customers arrive to a bank branch according to a Poisson process with a 12 customers per hour. The bank opens at 9:00 a.m. and has three tellers What is the probability that more than one customer arrive to the bank within half an hour? What is the probability that all of them will be idle during the first 15 minutes of operation? What is the probability that at least two tellers are busy during the first 30 minutes of operation? The bank has 10 branches in the greater Philadelphia area, what is the probability that more than 3 branches have at least two tellers busy during the first 30 minutes of operation?

Solution

Probability of time of arrival has exponential distribution

then, probabiltiy of number of arrivals has a poisson distribution

= 1/5 per minute

= 0.2 per minute

t = time in minutes

P(x) = e^(-t)*(t)^x/x!

a)P(X>1,t=30) = 1 - P(X=0,t=30) = 1 - e^(-0.2*30)*(0.2*30)^0/0!

P = 1 - 0.00247875218

P = 0.997521248

b) P(x=0,t=15) = e^(-0.2*15)*(0.2*15)^0/0! = 0.0497870684

c) P(X>=2,t=30) = 1 - P(X=1,t=30) - P(x=0,t=30) = 1 -e^(-0.2*30)*(0.2*30)^1/1! - 0.00247875218

= 1 - 0.0148725131 - 0.00247875218

= .9826487252124646

d) this will be binomial distribution with n = 10 and p = .9826487252124646

P(X>3) = 1 - P(X<=3) = 1 - P(X=0) - P(x=1) - P(x=2) - P(x=3)

= 1 - 10C0*(1-p)^10 - 10C1*p*(1-p)^9 - 10C2*p^2*(1-p)^8 - 10C3*p^3*(1-p)^7

= 1-(1-p)^10-10*p*(1-p)^9-45*p^2*(1-p)^8-120*p^3*(1-p)^7

= 1 - 5.427168792892338*10^-11

= 0.99999999995