A waiting-line system that meets the assumptions of M/M/1 has λ = 1,μ = 4.For this system,what is the probability of fewer than two units in the system?
A)0.0625
B)0.25
C)0.75
D)0.9375
E)1.00

Question

Quizplus · Accepted Answer

The probability of having fewer than two units in the system is the sum of the probabilities of having 0 and 1 unit. For an M/M/1 system, P(n) = (1 - ρ) * ρ^n, where ρ = λ/μ = 1/4. So, P(0) = (1 - 1/4) * (1/4)^0 = 0.75 and P(1) = (1 - 1/4) * (1/4)^1 = 0.1875. Thus, P(<2) = P(0) + P(1) = 0.75 + 0.1875 = 0.9375.

A Waiting-Line System That Meets the Assumptions of M/M/1 Has