Question

Consider an $\mathrm{M} / \mathrm{M} / 1 / \mathrm{K}$ queueing system. Let $q_n$ be the probability there are $n$ customers in the system just before a customer arrival that actually enters the system; that is, $q_n$ is the probability that there an $n$ customers in the system when an arrival is about to occur. Thus, $q_n=P\left[A_n \mid A\right]$ for $n=0,1,2, \ldots, K-1$, where $A_n$ is the event that there are $n$ customers in the system and $A$ is the event that an arrival is about to occur. Use Bayes' theorem (Theorem 2.4.3) to prove that $$ q_n=\frac{p_n}{1-p_K}, \quad n=0,1, \ldots, K-1 . $$

   Consider an $\mathrm{M} / \mathrm{M} / 1 / \mathrm{K}$ queueing system. Let $q_n$ be the probability there are $n$ customers in the system just before a customer arrival that actually enters the system; that is, $q_n$ is the probability that there an $n$ customers in the system when an arrival is about to occur. Thus, $q_n=P\left[A_n \mid A\right]$ for $n=0,1,2, \ldots, K-1$, where $A_n$ is the event that there are $n$ customers in the system and $A$ is the event that an arrival is about to occur. Use Bayes' theorem (Theorem 2.4.3) to prove that
$$
q_n=\frac{p_n}{1-p_K}, \quad n=0,1, \ldots, K-1 .
$$
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Probability, Statistics, and Queuing Theory with Computer Science Applications, Second Edition (Computer Science and Scientific Computing)
Probability, Statistics, and Queuing Theory with Computer Science Applications, Second Edition (Computer Science and Scientific Computing)
Arnold O. Allen 2nd Edition
Chapter 5, Problem 12 ↓

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In an $\mathrm{M} / \mathrm{M} / 1 / \mathrm{K}$ queueing system, there is a single server, arrivals follow a Poisson process, service times are exponentially distributed, and there is a maximum of $K$ customers allowed in the system (including those being  Show more…

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Consider an $\mathrm{M} / \mathrm{M} / 1 / \mathrm{K}$ queueing system. Let $q_n$ be the probability there are $n$ customers in the system just before a customer arrival that actually enters the system; that is, $q_n$ is the probability that there an $n$ customers in the system when an arrival is about to occur. Thus, $q_n=P\left[A_n \mid A\right]$ for $n=0,1,2, \ldots, K-1$, where $A_n$ is the event that there are $n$ customers in the system and $A$ is the event that an arrival is about to occur. Use Bayes' theorem (Theorem 2.4.3) to prove that $$ q_n=\frac{p_n}{1-p_K}, \quad n=0,1, \ldots, K-1 . $$
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