Existence of universal $(n,k)$ sets 
A set of $0-1$ strings of length $n$ is $(n,k)$-universal if, for
any subset of $k$ coordinates $S=\{i_1,\dots,i_k\}$, the projection
$$A\upharpoonright_S:=\{(a_{i_1},\dots,a_{i_k}):(a_1,\dots,a_n)\in
 A\}$$ of $A$ onto the coordinates in $S$ contains all possible $2^k$
configurations.
Theorem 3.2 (Kleitman-Spencer 1973). If $\binom{n}{k}2^{k}(1-2^{-k})^r<1$ , then there is an $(n,k)$-universal
set of size $r$.
Proof. Let $A$ be a set of $r$ random $0-1$ strings of length $n$, each entry of which takes values $0$ or $1$ independently and with
equal probability $1/2$. For every fixed set $S$ of $k$ coordinates
and for every fixed vector $v\in \{0,1\}^k$,
$$\text{Pr}[v\notin A\upharpoonright_S]=\prod \limits_{a\in
 A}\text{Pr}[v\neq a\upharpoonright_S]=\prod \limits_{a\in
 A}(1-2^{-|S|})=(1-2^{-k})^r.$$ Since there are only $\binom{n}{k}2^k$
possibilities to choose a pair $(S,v)$, the set $A$ is not
$(n,k)$-universal with probability at most
$\binom{n}{k}2^k(1-2^{-k})^r$, which is strictly smaller than $1$.
Thus, at least one set $A$ of $r$ vectors must be $(n,k)$-universal,
as claimed.

This is an excerpt from Jukna's book "Extremal combinatorics" which I am reading to understand the essence of probabilistic method in combinatorial problems. I am not so familiar with probability theory but I know some basics so I'd like to clarify some moments from the proof of this theorem.
Questions.

*

*Usually authors do not write explicitly what the probability space is in certain examples but I am a beginner so I'd like to understand the details. We have $\{0,1\}^n$ - binary string of length $n$ and there $2^n$ such strings. From the context I see that the probability space is the following: $(\Omega,\text{Pr}),$ where $\Omega:=\{A\subset \{0,1\}^n: |A|=r\}$, i.e. $|\Omega|=\binom{2^n}{r}$ and probability distribution is uniform, i.e. $\text{Pr}[A]=\frac{1}{|\Omega|}$. Is my guess correct?


*What is $[v\notin A\upharpoonright_S]$? I guess it is some event, i.e. subset of $\Omega$, but I cannot write it explicitly. Can explain what is this please? Initially I thought that $[v\notin A\upharpoonright_S]$ is the same as $\{A\in \Omega: v\notin A\upharpoonright_S\}$ but it does make sense  since in the next equality he is taking product over all element of $A$.
I have two questions so far because I need to understand those moments in order to move forward.
I'd be very thankful for help!
EDIT: 3) I am slightly confused with the definition of $(n,k)$ universal set. Let's take $n=3, k=2$ and consider the following set of binary strings of length $3$, say $\mathcal{A}=\{(1,0,0),(0,1,0),(1,1,0),(0,0,1)\}$. Then we have the following set of coordinates $S_{12}=\{1,2\},\ S_{13}=\{1,3\},\ S_{23}=\{2,3\}.$ One can check that $$A\upharpoonright_{S_{12}}=\{(1,0),(0,1),(1,1),(0,0)\},$$ $$A\upharpoonright_{S_{13}}=\{(1,0),(0,0),(0,1)\},$$ $$A\upharpoonright_{S_{23}}=\{(0,0),(1,0),(0,1)\}.$$
Can we say $\mathcal{A}$ is $(3,2)$-universal set?
 A: Edit:
After discussion with @PlayGame, it seems that confusion arises as in the proof, $A$ is defined to be a set, which would imply that the elements are distinct (and thus we lose the independence of strings). To make the proof sensical, we need to define $A$ as a vector of strings. Note that the proof remains correct, because a universal $(n,k)$-vector of size $r$ implies a universal $(n,k)$-set even smaller.

*

*This is correct
Edit :
In fact no, in your probability space, nothing tells you that $a_i\neq a_j~ \forall a_i, a_j \in A, i\neq j$, so $|\Omega| \neq \binom{2^n}{r}$. You have $rn$ binary values to choose, so you have $2^{rn}$ possible choices, so $|\Omega| = 2^{rn}$.

*Your guess was correct, you are probably confused by the fact that the product is on $A$, since $A$ is a random variable, but in fact, it does not really rely on $A$. If we write $A = \{a_1, a_2,\dots, a_r\}$, we could rewrite the product as $\prod \limits_{i\in
 1,\dots,r}\text{Pr}[v\neq a_i\upharpoonright_S]$, and it no longer depends on $A$.

*No, because $A\upharpoonright_{S_{13}}$ does not contain every possible string of length $2$. Add $(1, 1, 1)$ and you get a universal set.

