Problem: A company has 4 computers for an external user group. One of those users is called Urs. Several users can work on the same machine. If a user logs into a computer (think remote login), he gets automatically and randomly assigned to a computer. The chance is 1/4th and the assignment of one user doesn't depend on the others. So there might be several people working on the same machine.
How big is the probability, that at most one other person is working on the same machine as Urs, if (exluding him) 10 people are logged in.
Hint: Let $S_n$ be the random variable, which models the number of users logged into Urs's computer when there are n user logged in.
Question: The solution is: $\mathbb P(S_{10} \leq 1) = \begin{pmatrix} \frac{3}{4}\end{pmatrix}^{10} + 10\begin{pmatrix}\frac{3}{4}\end{pmatrix}^9\frac{1}{4}=\frac{3^9\cdot 13}{4^{10}}=0.244$
Now I don't fully understand that. I know that we have a binomial distribution (since I was told so) but I don't full see why we have an addition here and what was the thought process when modelling this.
I know it's a very simple problem, but if you never touched probability in your life and your solution don't have any motivation/explanation, it can be hard to get into the "proper thinking".
When dealing with elementary probability theory, it always helps to express the problem in simple words.
We're looking for the probability that at most 1 user is logged in the same computer as Urs $\mathbb P(S_{10} \leq 1)$. This means that either 1 other user is logged in the same computer or no other person is.
If Urs is using the computer alone, then the other 10 people must be assigned to another computer. The probability that a single person logs in a different computer is $\frac{3}{4}$. The probability that all 10 people will connect to a different computer is $(\frac{3}{4})^{10}$
Someone logs in Urs' computer with probability $\frac{1}{4}$. The other 9 people will connect to the other 3 machines with probability $(\frac{3}{4})^{9}$.
The probability that only the first person that connects to the network is assigned to Urs' computer is $\frac{1}{4}(\frac{3}{4})^{9}$. Similarly, the probability that the second person is the one that connects to Urs' computer is also $\frac{1}{4}(\frac{3}{4})^{9}$. Since we have 10 users $\mathbb P(S_{10} =1) = 10\frac{1}{4}(\frac{3}{4})^{9}$
So putting everything together we end up with: $\mathbb P(S_{10} \leq 1) = \mathbb P(S_{10} = 0) + \mathbb P(S_{10} = 1) =(\frac{3}{4})^{10}+10\frac{1}{4}(\frac{3}{4})^{9}$