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A waiting-line system that meets the assumptions of M/M/1 has. For this system, the probability of more than zero units in the system is approximately:

a) 0
b) 0.25
c) 0.5
d) 0.75

1 Answer

3 votes

Final answer:

The question is related to probability theory, focusing on waiting-line systems, and it requires knowledge of the M/M/1 queueing model and uniform distributions to find the probability of the system having more than zero units in it.

Step-by-step explanation:

The question falls within the realm of probability theory, specifically dealing with waiting-line systems and the uniform probability distribution. An M/M/1 system is a basic queuing model which assumes a single server, Poisson arrival process, exponential service times, and first-come-first-serve queue discipline. The question asks for the probability of more than zero units in the system, which, in an M/M/1 model, is the complement of the probability of the system being empty. Without the specific parameters of the arrival rate (λ) and service rate (μ), it is not possible to give the exact answer, but typically this probability is calculated using the formula P(n > 0) = 1 - P(0), where P(0) is the probability of zero units in the system.

As for the other provided scenarios, they involve different situations where the uniform distribution is used to calculate various probabilities related to waiting times. For instance, if wait times for a bus are uniformly distributed between 0 and 15 minutes, the probability that a person waits fewer than 12.5 minutes can be found by dividing the desired interval width (12.5 minutes) by the total interval width (15 minutes).

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