May I use a grade of service problem to calculate how many doctors are needed in a hospital? I created the following problem (based on grade of service problem of Intro to Prob by Tsitsiklis book):
original problem:

My analogy problem:
A Hospital unit has 15 doctors to serve the needs of a population of 100. It is estimated that at a given time, each inhabitant will need to go to the doctor with probability .1,
independent of the others. What is the probability that there are more inhabitants
needing to be served than there are doctors?
Is it correct to use this model to answer a different problem like the one I created?
 A: Yes, basically you've just changed the objects:


*

*modem $\rightarrow$ doctor

*customer $\rightarrow$ patient


And, since $n,c,p$ are the same, your probability is $\approx 0.04$ as well.
A: As @gunes (+1) has said: Taken exactly as stated, you have the same question
either way. $X \sim \mathsf{Binom}(n=100, p=0.1)$ is
the probability a modem/doctor will be busy at a given
time. And you seek $P(X > 15) = 1 - P(X \le 15) = 0.0399.$
In R, where pbinom is a binomial CDF, the computation
is shown below:
1 - pbinom(15,100,.1)
[1] 0.03989053

However, neither model deals explicitly with the period
of time over which a modem/doctor may be engaged.
Also, in the hospital situation it seems artificial to
say the the potential population served is 100 people.
It might be better to get data to estimate the rate at
which patients arrive for doctors' services and model the number needing
service at any one time as Poisson.  For example, if records show the Poisson mean to be $\lambda = 9.5,$ so that $Y \sim
\mathsf{Pois}(\lambda =9.5),$ then $P(Y > 15) = 0.0335.$
1 - ppois(15,9.5)
[1] 0.033473

If you want to take into account the average 'service time'
of doctors, then you would need a queueing model with
appropriate rate of arrival and rate of service by a doctor.
Then the issues become the percentage of time there are more than
15 patients in the system, necessitating a queue (waiting line)
of patients, and the average waiting time in the queue.
