The owner of an amusement park has decided to open a second entry booth and hire another employee to service customers entering the park.Customers arrive at the rate of 20 per hour and will wait in a single line until one of the two employees is available to provide service.The average service time of both employees is 2 minutes to provide service.The arrival rate follows Poisson distribution,and the service time follows a negative exponential distribution.Determine the probability that no customers are there in the system.
A)50%
B)60%
C)70%
D)45%

Question

Quizplus · Accepted Answer

To solve this problem, we need to use the M/M/2 queuing system, where:

- M stands for the Poisson arrival process
- M stands for the negative exponential service time distribution
- 2 stands for the number of service channels (i.e., employees)

The following formula can be used to calculate the probability of zero customers in the system:

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

Where:
p = arrival rate / (number of service channels * service rate)
service rate = 1 / average service time

Using the given values, we can calculate:

p = 20 / (2 * (1 / 2)) = 0.5
service rate = 1 / 2 = 0.5

Plugging these values into the formula, we get:

P(0) = [(1 - 0.5)(0.5)^2] / [1 - 3(0.5) + 3(0.5)^2 - (0.5)^3] = 0.5

Therefore, the probability of no customers in the system is 50%, which corresponds to choice A.

The Owner of an Amusement Park Has Decided to Open