Relationship between conditional probability and probability mass function 
Your University has 4 computers for external users. Urs is an external
user. If an external user logs into an computer, he gets automatically
assigned to one (without any dependence between the computers) by a
chance of $1/4$. So there can be several users logged into the same
computer.
Now assume that each computer, independently of the others, only has a
chance of 0.9 to actually work. Now three people and Urs are logged in
and further assume that Urs and two out of those three people are
working on the same
machine. How big is the chance, that all computers
work?
Hint: Let $X$ be the number of working computers. Let $S_3$ be the
number of Persons, which work on the same machine as Urs (not counting
Urs). Then calculate $P(S_3=0| X=k)$ for $k=2,3,4$. Then use the Law
of Total Probability to calculate $P(S_3=2)$. Then use the Baye's
Theorem to compute the wanted probability.

Now, what they do is:
We have$X ~ Bin(4, 0.9)$ which means $P[X=4]=0.9^4, \ P[X=3]=4\cdot0.9^3\cdot 0.1, P[X=2]=6\cdot 0.9^2\cdot 0.1^2$
Fruther it is $S_n|X=k ~ Bin(n, 1/k)$ so:
$$\begin{align}
    P(S_3=2|X=4) &= {3 \choose 2}(\frac{1}{4})^2\frac{3}{4}=\frac{9}{64}\\
    P(S_3=2|X=3) &= {3 \choose 2}(\frac{1}{3})^2\frac{2}{3}=\frac{2}{9}\\
    P(S_3=2|X=2) &= {3 \choose 2}(\frac{1}{2})^2\frac{1}{2}=\frac{3}{8}\\
\end{align}$$
Using bayes Theorem, we get:
$P(x=4 | S_3=2) = \frac{P(S_3=2|X=4)P(X=4)}{P(S_3=2|X=4)P(X=4) + \dots + P(S_3=2|X=2)P(X=2)} = 0.526$
Now I'm mostly confused on what exactly $P(S_3=2 | X=4)$ represents. We usually denoted conditional probability like that but here it looks more like it's used for the binomial probability mass function. So is it just bad notation and it doesn't have anything to do with conditional probability?
Further: When we use Baye's Theorem, we actually do have a conditional probability. So looks like $P(S_3=2 | X=4)$ is actually describing a conditional probability.
So the question arises: What is the relationship between "Conditional Probability" and "(Binomial) Probability Mass Function"?
 A: You have four computers, $c_1, c_2, c_3$ and $c_4$.  Let´s say Urs is automatically assigned to one of the 4 working computers. Then the question is what is the probability that two users are assigned to Urs computer, while the remaining user is assigned to another computer. The remaining user are denoted by $u_1, u_2$ and $u_3$. Next we can evaluate how the users can be distributed if they fulfill the condition.
$u_1, u_2$ are assigned to the computer of Urs and $u_3$ is not. The probability for that is $\frac14\cdot \frac14\cdot \frac34$
$X=4$ means that all $4$ computers are available. That is the condition. Thus the probability that one of the three users is assigned  to the computer of Urs is  $\frac14$.
$u_1, u_3$ are assigned to the computer of Urs and $u_2$ is not. The probability for that is $\frac14\cdot \frac34\cdot \frac14$
$u_2, u_3$ are assigned to the computer of Urs and $u_2$ is not. The probability for that is $\frac34\cdot \frac14\cdot \frac14$
Basically we are looking for the number of ways to order $u_1, u_2$ and $u_3$ at the first two places (out of 3). This can be done with the binomial coefficient.
So given that all 4 computers do work the probability that two users are assigned to the computer of Urs is $P(S_3=2|X=4)=\binom{3}{2}\cdot \frac14\cdot \frac14\cdot \frac34=\binom{3}{2}\cdot \left(\frac14\right)^2\cdot \frac34=\frac9{64}$
