Can I think of both arrival times and service times in a Markov chain as Poisson processes?

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According to the Wikipedia article about M/M/1 queues, the rate at which new jobs arrive is a Poisson process with parameter $\lambda$, and the rate at which the jobs are finished is an exponential distribution process with mean service time $\mu$.

To quote directly:

Arrivals occur at rate λ according to a Poisson process and move the process from state i to i + 1. Service times have an exponential distribution with parameter 1/μ in the M/M/1 queue, where μ is the mean service rate.

This seems strange to me. Why not just be consistent and say they are both Poisson processes? After all, exponential distribution is essentially the same thing as a Poisson distribution. Why confuse people and call them by different variations of the same thing?

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No. Customers arrive according to a Poisson process with rate $\lambda$. It is true that when the server is busy the service rate is $\mu$. But $\lambda < \mu$ if we have steady state. The server is idle some of the time.

Strictly speaking, Wikipedia is right. But it is more like an encyclopedia than a textbook.