Question

Marty's barber shop has one barber. Customers arrive at a rate of 2.2 per hour and haircuts are given at a rate of 5 customers per hour. Assume a Poisson arrival rate and an Exponential service time distribution.

Required:
a. What is the probability that one customer is receiving a haircut and one customer is waiting?
b. What is the probability that one customer is receiving a haircut and two customers are waiting?
c. What is the probability that more than two customers are waiting?

Answer 1

Explanation:

Arrival rate = ∧ = 2.2 customers per hour

Service rate = u = 5 customers per hour

1. Probability that one customer is receiving a haircut and one customer is waiting

P(2 customers)=(∧/u)^2 * (1-∧/u)=(2.2/5)^2 * (1-2.2/5)=0.1936*0.56= 0.108416

2. Probability that one customer is receiving a haircut and two customers are waiting

P(3 customers)= (∧/u)^3 * (1-∧/u)=(2.2/5)^3 * (1-2.2/5)= 0.085184

* 0.56= 0.04770304

3. Probability that more than two customers are waiting

P(more than 3 customers)=1- P(less than 3 customers) =

1- [P(0)+P(1)+P(2)+P(3)]=

= 1- [(1-2.2/5) +2.2/5*(1- 2.2/5) + 0.108416+0.04770304]=1-0.9625=0.0375