Sample space for a simple probability problem of drawing balls without putting them back When probability problem is solved, the first thing we need to define is sample space $\Omega$, $\sigma$-algebra on $\Omega$ and probability function.
Simple problem:
6 red balls and 9 white balls in the box. For each trial, one ball is drawn blindly from the box without putting in back into the box. When the red ball is drawn we stop the experiment.
Y is a random variable that gives the number of the trial when we get red ball. Hence $Y(\Omega) = \{1,2,..,10\}$ and $P\{Y=1\} = \frac{6}{15}$ ...
The task of the problem was to calculate $P\{Y=k\}$ and that I know how to do.
What would I like to know is what is $\Omega$ for this problem?
Is it $\omega_i$ = "red ball at trial i" and $\Omega = \{\omega_{1} ...\omega_{10}\}$?
Can some other sample space be used?
 A: Sample space for your problem is how many ways you can take all the 15 balls from the bag which is given by $\binom{15}{9}$.
Event(A): first red ball drawn at 'i' turn
Find the probability of event A?
Lets say first red ball drawn at turn 1
So number of ways this can be done is $\binom{14}{9}$
Probability = $\frac{\binom{14}{9}}{\binom{15}{9}}$=$
\frac{6}{15}$
Similarly for i=2 ..i.e.  first red ball drawn at turn 2
is given by $\binom{13}{8}$
Probability = $\frac{\binom{13}{8}}{\binom{15}{9}}$=$
\frac{9*6}{15*14}$
Sample space(S): How many ways you can put 9 white balls and 6 balls on 15 blank spaces available
Event(A) Y=1: You put red ball to first blank, then how many ways you can arrange 9 white balls and 5 red balls to remaining 14 spaces
Evevt(A) Y=2: Put white ball to first place and red ball to second place, then how many ways you can arrange 8 white balls and 5 red balls to remaining 13 places
A: 
When probability problem is solved, the first thing we need to define is sample space $\Omega$, $\sigma$-algebra on $\Omega$ and probability function.

No, we often don't need to do that at all.   We usually just need to define a random variable and its probability measure (and support), which requires a model for its generation.   That is not quite the same thing as a constructing a probability space. 
Delving too deep on what is the $\{\Omega,\mathcal F,\mathsf P\}$ is a distraction from what you really want to find.
$Y$ is the count of trials until we encounter a red ball, when each trial extracts balls without replacement (or bias) from a population of six red and nine white balls.
$\mathsf P(Y=y)$ is then the probability for extracting $y-1$ from $9$ white balls, consecutively, and then one from $6$ red balls when extracting $y$ balls.   Alternatively; the probability that the last ball selected is one from the six red balls and that the $15-y$ balls not selected contain the remaining $5$ red balls.   Either is an appropriate model.
$$\begin{split}\mathsf P(Y=y) & =\dfrac{\binom{9}{y-1}\binom{6}{1}}{\binom{15}{y-1,1,15-y}}\mathbf 1_{y\in\{1,2,3,4,5,6,7,8,9,10\}} \\ &=\frac{\binom{15-y}{5}}{\binom{15}{6}}\mathbf 1_{y\in\{1,2,3,4,5,6,7,8,9,10\}}\end{split}$$
