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 an arriving customer will wait for service.
A)16.67%
B)34.54%
C)12.32%
D)11.56%

Question

Quizplus · Accepted Answer

The probability that an arriving customer will wait for service is equal to the probability that both employees are busy. The arrival rate is 20 per hour, and the service time is 2 minutes. Therefore, the average number of customers in the system is 20 * 2 / 60 = 2/3. The probability that both employees are busy is equal to the probability that there are at least 2 customers in the system, which is 1 - P(0) - P(1) = 1 - (1/3)^0 * (2/3)^0 - (1/3)^1 * (2/3)^1 = 1 - 1/3 - 2/3 = 16.67%.

The Owner of an Amusement Park Has Decided to Open