Erdős theorem on 2-colorable sets 
Let $\mathcal{F}$ be a family of subsets of some finite set.  Can we
color the elements of the underlying set in red and blue so that no
member of $\mathcal{F}$ will be monochromatic? Such families are
called $2$-colorable. Recall that a family is $k$-uniform if each
member has exactly $k$ elements.
Theorem. If $k$ is sufficiently large, then there exists a $k$-uniform family $\mathcal{F}$ such that $|\mathcal{F}|\leq k^22^k$
and $\mathcal{F}$ is not $2$-colorable.
Proof: Set $r=[k^2/2]$. Let $\mathbf{A}_1,\mathbf{A}_2,\dots$ be independent random members of $\binom{[r]}{k}$, that is,
$\mathbf{A}_i$ ranges over the set of all $A\subseteq \{1,\dots,r\}$
with $|A|=k$, and $\text{Pr}[\mathbf{A}_i=A]=\binom{r}{k}^{-1}$.
Consider the family $\mathcal{F}=\{\mathbf{A}_1,\dots,\mathbf{A}_b\}$,
where $b$ is a parameter to be specified later. Let $\chi$ be a
coloring of $\{1,\dots,r\}$ in red and blue, with $a$ red points and
$r-a$ blue points. Using Jensen's inequality, for any such coloring
and any $i$, we have $$\text{Pr}[\mathbf{A}_i \ \text{is
 monochromatic}]=\text{Pr}[\mathbf{A}_i \ \text{is
 red}]+\text{Pr}[\mathbf{A}_i \ \text{is
 blue}]=$$$$=\dfrac{\binom{a}{k}+\binom{r-a}{k}}{\binom{r}{k}}\geq
 \frac{2\binom{r/2}{k}}{\binom{r}{k}}:=p,$$ where, by the asymptotic
formula $(1.9)$ for the binomial coefficients, $p$ is about
$e^{-1}2^{1-k}$. Since the members $\mathbf{A}_i$ of $\mathcal{F}$ are
independent, the probability that a given coloring $\chi$ is legal for
$\mathcal{F}$ equals $$\prod_{i=1}^b(1-\text{Pr}[\mathbf{A}_i\
 \text{is monochromatic}])\leq (1-p)^b.$$ Hence, the probability that
at least one of all $2^r$ possible colorings will be legat for
$\mathcal{F}$ does not exceed $2^r(1-p)^b<e^{r\ln 2-pb}$, which is
less than $1$ for $b=(r\ln 2)/p=(1+o(1))k^22^{k-2}e\ln 2$. But this
means that there must be at least one realization of the random family
$\mathcal{F}$, which has only $b$ sets and which cannot be colored
legally.

This is an excerpt from Jukna's "Extremal combinatorics" and I guess that there are some technical flaws in the reasoning, however, I understood the underlying idea of the proof. So let me ask my questions.

*

*Am I right that the author considers the finite probability space $(\Omega,\text{Pr})$, where $\Omega=\{A\subset [r]: |A|=k\}$, i.e. $|\Omega|=\binom{r}{k}$ with uniform distribution, i.e. $\text{Pr}[\omega]=\frac{1}{|\Omega|}$ for each $\omega\in \Omega$?


*I was able to show that for each coloring $\chi$ we have $$\text{Pr}[\{A\in \Omega: A \ \text{is monochromatic under coloring}\  \chi\}]=\binom{r}{k}^{-1}\left[\binom{a}{k}+\binom{r-a}{k} \right].$$ But I am really confused what does the author mean by $\text{Pr}[\mathbf{A}_i \ \text{is monochromatic}]$. Can anyone explain what does this event mean?
 A: I'll try to give really detailed explanation of the details.
Consider the following probability space: $(\Omega, \text{Pr})$ where $\Omega=\{A\subset [r]: |A|=k\}$ with uniform distribution, i.e. $\text{Pr}(A)=\frac{1}{|\Omega|}$ for each $A\in \Omega$. Easy to see that $|\Omega|=\binom{r}{k}$ where $r=[k^2/2]$. The set $\{1,2,\dots,r\}$ can be colored in $2^r$ ways since each number can be colored red or blue and we have $r$ numbers. Fix some coloring $\chi:\{1,2,\dots,r\}\to \{\text{Blue},\text{Red}\}$ such that $a$ numbers in $\{1,\dots,r\}$ are colored in red and $r-a$ numbers are in blue.
For each such coloring $\chi$ we can define the event (subset of $\Omega$) defined as follows: $\{A\in \Omega: A \ \text{is monochromatic}\}$ and it is easy to see that $$\{A\in \Omega: A \ \text{is monochromatic}\}=\{A\in \Omega: A \ \text{is red}\}\sqcup \{A\in \Omega: A \ \text{is blue}\}.$$ Let's compute its probability $$\text{Pr}[\{A\in \Omega: A \ \text{is monochromatic}\}]=\text{Pr}[\{A\in \Omega: A \ \text{is red}\}]+\text{Pr}[\{A\in \Omega: A \ \text{is blue}\}]=$$ $$=\dfrac{|\{A\in \Omega: A \ \text{is red}\}|+\{A\in \Omega: A \ \text{is blue}\}}{|\Omega|}=$$$$=\frac{\binom{a}{k}+\binom{r-a}{k}}{\binom{r}{k}}\geq 2\binom{r/2}{k}\binom{r}{k}:=p.$$ The last inequality obtained from Jensen's inequality and note that $\binom{r/2}{k}$ makes sense because $r$ which is defined as $[k^2/2]$ is always even number.
It implies that $$\text{Pr}[\{A\in \Omega: A \ \text{is NOT monochromatic}\}]=1-\text{Pr}[\{A\in \Omega: A \ \text{is monochromatic}\}]\leq 1-p.$$
Consider the "new" probability space which can be constructed from the old one and this space is called product space. For $b>1$ which will be specified later one can consider the pair $(\Omega^b, F)$ where $\Omega^b:=\underbrace{\Omega\times\dots\times \Omega}_{b \ \text{times}}$ and $F:\Omega^b\to [0,1]$ defined as $F[(A_1,\dots,A_b)]:=\text{Pr}[A_1]\times \cdots \times \text{Pr}[A_b]$. One can that this pair is actually probability space.
Fix one of the $2^r$ possible colorings $\chi:\{1,2,\dots,r\}\to \{\text{Red}, \text{Blue}\}$ and consider the event associated with $\chi$, i.e. subset of $\Omega^b$ defined as follows $$E_{\chi}:=\{(A_1,\dots,A_b)\in \Omega^b: A_1,\dots,A_b \ \text{is not monochromatic under}\ \chi\}=$$$$=\cap _{i=1}^{b}\{(A_1,\dots,A_b)\in \Omega^b: A_i \ \text{is not monochromatic under}\ \chi\}.$$ Using the definition of  the probability measure $F$ on product pace $\Omega^b$ one can check that the events $\{(A_1,\dots,A_b)\in \Omega^b: A_i \ \text{is not monochromatic under}\ \chi\}$ are independent for each $i$ because the events $\{(A_1,\dots,A_b)\in \Omega^b: A_i \ \text{is monochromatic under}\ \chi\}$ are independent. It shows that $$F(E_{\chi})=$$$$=F[\{(A_1,\dots,A_b)\in \Omega^b: A_1,\dots,A_b \ \text{is not monochromatic under}\ \chi\}]=\prod_{i=1}^{b}F[\{(A_1,\dots,A_b)\in \Omega^b: A_i \ \text{is not monochromatic under}\ \chi\}]=$$$$=\prod_{i=1}^{b}\text{Pr}[\{A_i\in \Omega: A_i \ \text{is not monochromatic under}\ \chi\}]\leq$$$$\leq\prod_{i=1}^{b}(1-p)=(1-p)^b.$$ Hence $F(E_{\chi})\leq (1-p)^b$ for each $\chi$. Hence $$F(\cup_{\chi}E_{\chi})\leq \sum \limits_{\chi} F(E_{\chi})\leq \sum \limits_{\chi}(1-p)^b=2^r(1-p)^b.$$ The remaining optimization process is clear.
