 Solution

Use one particle for each telephone line, including unused lines. Let $f$ be the number of free lines, $h$ the number of callers waiting for or being checked by an operator, and $a$ the number using automated service. Then $f+h+a=N$. These are the three colonies of a closed migration process, with directed cycle $f\to h\to a\to f$ and departure ratesRejected attempted calls leave the state unchanged. They must not be represented as extra customers in the waiting stage.

Equivalently the state is $(h,a)$ with $h+a\le N$. Its transitions areThe directed cycle has equal traffic weights; choosing each weight to be $\nu$ gives the unnormalized product-form stationary distribution of a closed migration processWe can verify global balance for a continuous-time Markov chain directly. Weighted incoming transitions from $(h-1,a)$, $(h+1,a-1)$ and $(h,a+1)$ contribute, respectively,Their sum equals the weight times the outgoing rate. This proves stationarity after normalization, without falsely asserting reversibility of the directed migration cycle.

For $n=h+a$ occupied lines, summing over the automated population givesThe normalizing constant is $G_N={\sum }_{n=0}^{N}H(n)$. The attempted calls form an independent Poisson process, so Poisson arrivals see time averages: the rate of attempted arrivals seeing a full state is $\nu$ times its stationary probability, and division by the total attempted-arrival rate gives the loss proportion. ConsequentlyThe numerator is essential; the PDF contains it even though the converted TeX loses it.

With two independent operators, the total check-completion rate is $\lambda \min (h,2)$. In particular it is only $\lambda$ when one caller is present. DefineThe human-stage weight becomes $(\nu /\lambda {)}^h/D_2(h)$, while the free and automated weights are unchanged. The same global-balance calculation applies using these departure rates. ThusThis is the blocking probability in a finite-line sequential-service network with two operators. It remains valid for $N=0$, when every attempted call is lost; for $N=1$ it coincides with the single-operator answer.