Could please indicate theory behind I was solving a probability problem, and found two ways of wolving it.
Both came to right same answer (according to test answer).
The first way gives a little more work but I understood the backgroud theory.

My doubt is about the second way which is much simpler, but I don't know if it's valid or the theory behind.
So that's what I want to know: If the simplest way is valid and if some one could explain the meaning behind or indicate the theory that explains.

Problem: Given 10 students, and making a group with 3 of then, what is the probability that a specific student will be chosen?
First way (more work but all the theory is explained):
The number of all diferent possible groups is given by the combination C(10,3) = 120; This is the total sample space.
The number of all possible groups with that specific student will be C(9,2) = 36, because one position of the three is already taken;
So the probability of the specific student be chosen is 36/120 = 0,3
The second way to solve, I was told that as it's 3 positions in 10 total possibilities I could simple do 3/10... 
 A: Suppose you are the particular student. You are all ten placed into ten positions, three of which are the chosen ones. What is your probability of being in one of those three positions? Three out of ten.
A: You can think of it by the symmetry of the problem. More formally, let $X_i$ be the indicator function of the $i$th student getting picked. Then $\mathbb{E}[X_i]=\mathbb{P}(\text{child $i$ got picked})$. We also know that $\mathbb{E}[\sum_{i=1}^{10} X_i]=3$, as in every possibility, we pick 3 students. By linearity of expectation, $\mathbb{E}[\sum_{i=1}^{10} X_i]=\sum_{i=1}^{10}\mathbb{E}[X_i]$. By symmetry, $\mathbb{E}[X_i]=\mathbb{E}[X_j]$ for any two students $i$ and $j$, so letting, for instance, $x=\mathbb{E}[X_1]$, we get $3=10x$. Divide through to get $x$.
A: You have a population of $10$ where $3$ are true and $7$ are false,
A random draw from this population therefore has $3/10$ chance of being true. 
A: In order to apply Sample Space Theory to word problems concerning odds, you have to do two things:
Describe the problem as a random experiment
Select a probability model 
There may be more than one valid way to do this, so you can't just plug into a formula or apply a simple technique - you have to understand what you are doing. 
In what follows we will give two additional solutions to the MODEL 1: C(10,3) = 120 design that the OP used. The discrete uniform distribution will play a role in the sample space design. Also, the following basic concept from probability theory,

To find the probability of two independent events that occur in sequence, find the probability of each event occurring separately, and
  then multiply the probabilities.

will be developed.

Model 2
The random experiment is to select one student out of 10, then to select another student out of the remaining 9, and to then select one final student out of the remaining 8 students. This experiment can be viewed as performing a sequence of three independent 'selection' experiments. We want to know the probability of a specific student being selected.
Let $T = \{1,2,3,4,5,6,7,8,9,10\}$. The sample space $S$ will be the ordered $\text{3-tuples}$ taken from $T \times T \times T$ that are observed when (mentally) performing the experiment. 
The sample space will contain $10 \, 9 \, 8 = 720$ outcomes. The probability of any outcome will be $ \frac{1}{10}\frac{1}{9}\frac{1}{8}  = \frac{1}{720}$.
To answer the question, we can assign our student to $1 \in T$ and look at the event $A$ in $S$ of all $\text{3-tuples}$ containing student $1$. If the student is selected right away, there are $9$ ways to select the next student and $8$ ways for the third, or a $72$ count. Continuing this counting argument, we see that $A$ has $72 + 72 + 72$ elements.
Ans: $\frac{3\; 72}{720} = 30\%$
It is not necessary to use this detail or to even count the number of outcomes in $S$ if we use this trick: 
What is probability that the student is NOT selected?
The chance of not being selected $\text{first$\;\;\;\,$is}$  $\frac{9}{10}$.
The chance of not being selected $\text{second is}$  $\frac{8}{9}$.
The chance of not being selected $\text{third$\;\;\,$ is}$  $\frac{7}{8}$.
Since each selection pick is independent from the prior pick, you can multiply, which is easy since things cancel out, and you get $\frac{7}{10}$. So the probability of our student being selected is $1 - \frac{7}{10}$, or again, $30\%$.

Model 3
The random experiment is a teacher has 10 new students coming to class and they will be seated randomly in exactly 10 chairs. He is a nasty fellow, and places tacks on three of the seats in the front row. As the students come into the class, they will choose a chair. After they sit down they will either jump up or remain seated. The first student comes into the class and sits down, and we want to know how often he jumps up. 
Our sample space $S = \{j, n\}$ where $j$ means the student jumps up and $n$ means the student is not fazed. Let $T = \{1,2,3,4,5,6,7,8,9,10\}$ represent the chairs. If we set it up so that the teacher puts the tacks on chairs $J = \{1,2,3\}$, then we can map $T$ onto $S$ and assign probabilities to the outcomes in $S$. Here, $J$ gets mapped to $j$, so $P(j) = .3$ and $P(n) = .7$ in $S$.
The first student walks in the door. We want to know the probability of event $J$ occurring (i.e. the student picks chair 1 OR chair 2 OR chair 3), so that we see the student jump up.
Ans: $30\%$
