Suppose I have a queue with $\lambda$ and $\mu$. I can calculate the probability that there are 2 objects in the queue trivially, but how can I compute, for example, the probability that it takes an object less than $n$ units of time to be processed?
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As you know from the comments, the processing (service) time of an M/M/1 queue is exponentially distributed, hence the probability is $P(T_{serve} < t_n) = \int_0^{t_n} \mu \mathrm e^{-\mu t} \mathrm d t = 1-\mathrm e^{-\mu t_n}$ If you want the total waiting time, you will have to add to that the queueing time. See also Little's law. |
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Since you're dealing with a $M/M/1$ queue you know that the service time is exponentially distributed with parameter $\mu$, which means that the service time $B$ has probability density function $f_{B}(t)=\mu e^{-\mu t}$. So $P(B < t) = \int_0^{t} f_{B}(\tau)d \tau$. |
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