Consider the $\mathrm{M} / \mathrm{M} / \mathrm{c} / \mathrm{K} / \mathrm{K}$ machine repair queueing system. Let $p_n$ be the probability that $n$ of the $\mathrm{K}$ machines are inoperable (either undergoing or awaiting repair) and let $q_n$ be the probability that a machine that breaks down finds $n$ inoperable machines in the repair facility.
(a) Prove that
$$
q_n=\frac{(K-n) p_n}{K-L}, \quad n=0,1, \ldots, K-1 .
$$
(b) For an $M / M / 1 / K / K$ queueing system, prove that
$$
q_n=\frac{\frac{z^{K-n-1}}{(K-n-1) !}}{\sum_{k=0}^{K-1} \frac{z^k}{k !}}, \quad n=0,1, \ldots, K-1,
$$
where $z=E[O] / W_s$, and thus, $q_n$ has the same value as $p_n$ for the $\mathrm{M} / \mathrm{M} / 1 / \mathrm{K}-1 / \mathrm{K}-1$ system; that is, the number of machines found by arriving machines is the same as that which would be seen at a randomly chosen instant in a system with one less machine.