Question

Marty's Barber Shop has one barber. Customers have an arrival rate of 2.2 customers per hour, and haircuts are given with a service rate of 5 per hour. Use the Poisson arrivals and exponential service times model to answer the following questions:

What is the probability that one customer is receiving a haircut and one customer is waiting?

What is the probability that one customer is receiving a haircut and two customers are waiting?

What is the probability that more than two customers are waiting?

Answer 1

Answer:

Step-by-step 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

3. Probability that more than two customers are waiting =

Answer 2

The probability that one customer is receiving a haircut and one customer is waiting is 0.108416, the probability that one customer is receiving a haircut and two customers are waiting is 0.04770304 and the probability that more than two customers are waiting is 0.0375

Given :

Marty's Barber Shop has one barber. Customers have an arrival rate of 2.2 customers per hour, and haircuts are given with a service rate of 5 per hour.

The arrival rate is 2.2 customers per hour and the service rate is 5 customers per hour.

A) The probability that one customer is receiving a haircut and one customer is waiting is given by:

$\rm P(2\; customers) =\left( \dfrac{2.2}{5}\right)^2 \times \left (1-\dfrac{2.2}{5}\right)$

$\rm P(2\; customers) =0.1936 \times 0.56=0.108416$

B) The probability that one customer is receiving a haircut and two customers are waiting is given by:

$\rm P(3\; customers) =\left( \dfrac{2.2}{5}\right)^3 \times \left (1-\dfrac{2.2}{5}\right)$

$\rm P(3\; customers) =0.085184\times 0.56=0.04770304$

C) The probability that more than two customers are waiting is given by:

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

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

                                           $=1-(\dfrac{2.2}{5}\times 0.56+0.108416+0.04770304)$

                                           = 0.0375

For more information, refer to the link given below:

https://brainly.com/question/23017717