Construct a sample space $\Omega$ when interested in the probability of having conducting $i$ number of trails for $r$ times of success I know that for a random variable $X$ follows Pascal distribution $PA(r,\theta)$, the random variable refers to the number of trails needed for $r$ success. I have no problem in writing the probability distribution, as it is easily thinkable that if we want $r$ success, suppose we conduct $i$ times of the experiment, the $i$-th trail must be the $r$-th success, and for the $r-1$ previous trails, it is nothing but a binomial probability,  for the $i-1$ number of trails, select $r-1$ number of trails for being a success, the total number of combinations $\left(\begin{array}{l}
i-1 \\
r-1
\end{array}\right)$ times the probability of having $r-1$ success in a total of $i-1$ trails, which is $\theta^{r-1}(1-\theta)^{i-r}$, therefore $$P(X=i)=\left(\begin{array}{c}
i-1 \\
r-1
\end{array}\right) \theta^{r}(1-\theta)^{i-r}$$
However, I always have difficulty to link these distributions to what I have told in the measure-theoretic probability, for example, in this case, if I go back to the definition of a random variable, $X:(\Omega,\mathcal{A},\mathbb{P})\rightarrow(R,\mathcal{B})$, what $\Omega$, $\mathcal{A}, \mathbb{P}$ would look like? Would it be problematic if I write $P(X=i)=\mathbb{P}(\{w:X(w)=i\})$? In that case, what would $\Omega$ look like?
My first instinct is to suppose $r=1$, denotes failure as $0$, success as $1$, $\Omega=\{\{1\},\{0,1\},\{0,0,1\},...\}$, however the problem is that if I change $r$, the whole $\Omega$ would look complete different! Wouldn't that suggest the distribution of a random variable depends on the specification of the measure space?
 A: Here is one possibility.
Consider $\Omega=\{0,1\}^{\mathbb{N}}$ (infinite sequences of $0$s and $1$s) endowed with the product $\sigma$-algebra $\mathscr{F}:=\bigotimes_n\mathcal{P}(\{0,1\})$ and the product measure $P(\{(\omega_1,\ldots,\omega_n)\})=\prod^n_{j=1}\theta^{\omega_j}(1-\theta)^{1-\omega_j}$
This is basically the abstract space where Bernoulli trials are defined so that $P(\{\omega:\omega_1=1\})=\theta$ and $P(\{\omega:\omega_1=0\})=1-\theta$.
For each $n\in\mathbb{N}$, define the function $X_j=\boldsymbol{\omega}=(\omega_1,\omega_2,\ldots)\mapsto \omega_j$, that is, $X_j(\boldsymbol{\omega})=\omega_j$ is the projection onto the $j$-th component.
Define $S_0=0$ and $S_n=\mathbb{1}_{\{1\}}(X_1)+\ldots+ \mathbb{1}_{\{1\}}(X_n)$ for $n\in\mathbb{N}$. For each $n\in\mathbb{Z}_+$, $S_n$ gives the number of ones that are in the first $n$ components of $\boldsymbol{\omega}$ (number of successes in $n$ consecutive experiments if you will).
Define
$$X_r(\boldsymbol{\omega}):=\min\{n\geq1: S_n(\boldsymbol{\omega})=r\}$$
It is not difficult to check that $X_r$ is an $\mathscr{F}$-$\mathbb{B}(\mathbb{R})$ measurable function (random variable) taking values on $\mathbb{N}$. The law (distribution) of $X$ is what you described in the OP as the Pascal distribution.
