Question

Customer arrivals at a bank are random and independent; the probability of an arrival in any one-minute period is the same as the probability of an arrival in any other one-minute period. Answer the following questions, assuming a mean arrival rate of three customers per minute.
Required:
a. What is the probability of exactly three arrivals in one-minute period?
b. What is the probability of at least three arrivals in a one-minute period?

Answer 1

Answer:

a)0.2240

b)0.5768

Step-by-step explanation:

Given:

µ=3

Poison probability is given by :

$f_k=\frac{\mu^ke^-^\mu}{k!}$

a) Evaluating at k=3

$f(3)=\frac{3^3e^-^3}{3!} \approx 0.2240$

b)Evaluating at k=0,1,2:

$f(0)=\frac{3^0e^-^3}{0!} \approx 0.0498$

$f(1)=\frac{3^1e^-^3}{1!} \approx 0.1494$

$f(2)=\frac{3^2e^-^3}{2!} \approx 0.2240$

Use complement rule:

P(x≥3)= 1 - f(0) - f(1) - f(2)= 1- 0.0498 - 0.1494 - 0.2240 =0.5768