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

Question

Quizplus · Accepted Answer

In M/M/1 system, the utilization factor (ρ) can be calculated as λ/μ. In this case, ρ = 1/4 = 0.25.

Using the formula for the probability of 0 or 1 units in the system, P(0) + P(1) = 1 - ρ = 0.75.

So, the probability of fewer than two units in the system (i.e. 0 or 1) is approximately 0.75 - P(1)

Using the formula for P(n) = (1-ρ)ρ^n, we can calculate P(1) as (1-0.25)*(0.25)^1 = 0.1875.

Therefore, the probability of fewer than two units in the system is approximately 0.75 - 0.1875 = 0.5625, which is closest to choice D (0.9375 when subtracted from 1).

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