I need help with a, b, and c. I don't know how to do them. Please, help.

1. A fast-food franchise is considering operating a drive-up window food-service operation. Assume that customer arrivals follow a Poisson probability distribution, with an arrival rate of 24 cars per hour, and that service times follow an exponential probability distribution. Arriving customers place orders at an intercom station at the back of the parking lot and then drive to the service window to pay for and receive their orders. The following three service alternatives are being considered:

A single-server operation in which one employee fills the order and takes the money from the customer. The average service time for this alternative is 2 minutes.

A single-server operation in which one employee fills the order while a second employee takes the money from the customer. The average service time for this alternative is 1.25 minutes.

A two-server operation with two service windows and two employees. The employee stationed at each window fills the order and takes the money for customers arriving at the window. The average service time for this alternative is 2 minutes for each server.

For each of the three design alternatives, answer the following questions. Then, recommend one of the design options.

a. What is the probability that no cars are in the system?

b. What is the average number of cars waiting for service?

c. What is the average number of cars in the system?

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Step 1In System 1 where there is only one server to provide service, arrival ate is 24 cars per hour and service time is 2 minutes.So we have:   K=1\lambda=24 cars per hour\mu=\frac{60}{2}=30 cars per hourStep 2In System 2, Single Server case where there are two servers and two service providers:k=1λ=24 cars per hourμ=601.25=48 cars per hourStep 3In System 3, Two server but 2 employees so the arrival rate is the same as 24 but the service rate is changed:   k=2\lambda=24 cars per hour\mu=\frac{60}{2}=30 cars per hourStep 4a) The probability that no cars are in the system is calculated as:For single server: P0=1-λkμFor two server: P0=11+λμ+λμ2μ2μ-λThen,System 1System 2System 3P0=1-2430=1-0.8=0.2P0=1-2448=1-0.5=0.5P0=11+2430+2430230230-24=12.3333=0.4286Step 5b) The average number of cars waiting for service is calculated asFor single server: Lq=λkμλμ-λFor two server:Lq=λμλμ2P02μ-λ2 Then,System 1System 2System 3Lq=24302430-24=0.84=3.2Lq=24482448-24=0.51=0.5Lq=2430243020.4286230-242=197.49891296=0.1524Step 6c) The average number of cars in the system is calculated as:For single server: L=\frac{\lambda}{\mu-\lambda}For two server:L=Lq+λμ Then,System 1System 2System 3L=2430-24=4L=2448-24=1L=0.1524+2430=0.1524+0.8=0.9524System 3 is the best option. ...