= 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 rates
$$
\phi_f(f)=\nu\mathbf1_{\{f>0\}},\qquad
\phi_h(h)=\lambda\mathbf1_{\{h>0\}},\qquad
\phi_a(a)=\mu a.
$$
Rejected 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 $h+a\leq N$. Its transitions are
$$
(h,a)\to(h+1,a)\text{ at rate }\nu\quad(h+a<N),
$$
$$
(h,a)\to(h-1,a+1)\text{ at rate }\lambda\quad(h>0),\qquad
(h,a)\to(h,a-1)\text{ at rate }\mu a.
$$
The directed cycle has equal traffic weights; choosing each weight to be $\nu$ gives the unnormalized <product-form stationary distribution of a closed migration process>
$$
w(h,a)=\left(\frac\nu\lambda\right)^h
\frac{(\nu/\mu)^a}{a!}.
$$
We 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,
$$
w(h,a)\lambda\mathbf1_{\{h>0\}},\qquad
w(h,a)\mu a,\qquad
w(h,a)\nu\mathbf1_{\{h+a<N\}}.
$$
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 gives
$$
\sum_{a=0}^n w(n-a,a)
=\left(\frac\nu\lambda\right)^n
\sum_{a=0}^n\left(\frac\lambda\mu\right)^a\frac1{a!}=H(n).
$$
The <normalizing constant> is $G_N=\sum_{n=0}^NH(n)$. 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. Consequently
$$
\boxed{P_{\mathrm{loss}}=\frac{H(N)}{\sum_{n=0}^NH(n)}.}
$$
The 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. Define
$$
D_2(h)=\prod_{j=1}^h\min(j,2)
=\begin{cases}1,&h=0,\\2^{h-1},&h\geq1.\end{cases}
$$
The 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. Thus
$$
H_2(n)=\sum_{a=0}^n
\frac{(\nu/\lambda)^{n-a}(\nu/\mu)^a}{D_2(n-a)a!},
\qquad
\boxed{P_{\mathrm{loss}}^{(2)}=\frac{H_2(N)}{\sum_{n=0}^NH_2(n)}.}
$$
This 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.
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