A city has 3 doctors and 4 sick people. What is the probability that every doctor is called?
I have no idea how to start thinking about this problem, I was trying to build a probability model for this problem, but I am not sure if this is right and how this will help me formalize (and understand) this apparently simple exercise.
$D = \{d_1,...,d_D\}$ : set of doctors
$M = \{m_1,...,m_M\}$ : set of sick people
$$\Omega = D^M=\{\omega :M\rightarrow D\}$$ $$\mathcal{A}=\mathcal{P}(\Omega)$$ $$P = \text{uniform over } \mathcal{A} $$
With $\omega$ the array describing the doctor of each person.
If these are the only information you are given, then I would assume that a sick patient immediately calls randomly any of the four doctors, and that a doctor can take multiple calls at the same time (receptionists with more than one phone, something fancy like that).
Suppose we have $n$ patients and $k$ doctors. Then with the assumptions it is a Combination-with-repetitions-problem, i.e. "how many ways are there to distribute $n$ objects (patients) into $k$ categories (doctors) without respecting order". The formula for how many possibilities there are is
$$\binom{n+k-1}{n}$$
In your example, there are $n=4$ patients, $k = 3$ doctors, so a total of $\binom{6}{4} = 15$ possibilities. How many possibilities are there to get all 3 doctors called without respecting the order? There are 3 possibilities: Each doctor might be called twice. Then the final probability is $3/15 = 1/5$.