Find the random event set and c.d.f. for a random variable. So, I need to find random event set for as well as find the c.d.f. for a random variable that describes the chance that there are no black balls left to pull from the bag of total 5 balls, 2 of them black and 3 of them white. So, I think there must be 9 possible events that can happen. Then I would write a chance that all the black balls are already pulled out. I would start with $X=2$ and the chance that two black balls will be pulled out at second pulling would be $\frac{1}{9}$. Then the chance that all the black balls will be pulled out after third time would be $\frac{2}{9}$ and the chance that there won't be left any black balls after fourth time of pulling a random ball is $\frac{2}{9}$ as well. Finally the chance that at fifth time the second black ball will be pulled out is $\frac{4}{9}$. I want to check whether my reasoning is correct and if I can start with $X=2$? Then about the c.f.d.: is it noted as $F(x)$? And secondly, when writing the function $F(x)$, do I need to say that the chance that all the black balls will be pulled out after fifth time as $x=5$ or can I write $x \ge 5$? The same goes with $x < 2$, do I need to indicate that $x > 0$? 
 A: It seems that you select the balls without replacement. Then the probability to select 2 balls in two draws is 
$P(bb)=P(X=2)=\frac{2}{5}\cdot \frac{1}{4}=\frac{1}{10}$
You have two ways to draw three balls where the last ball is black: $wbb, bwb$. Therefore $P((bw),b)=P(X=3)=\frac{2}{5}\cdot \frac{3}{4}\cdot \frac{1}{2}\cdot 2=\frac{1}{5}$
Here the sequence $(bw)$ (with the brackets) indicates that we have to regard all possible orders.
Similar calculation for $x=4$ and $x=5$
$P((bww),b)=P(X=4)=\frac{2}{5}\cdot \frac{3}{4}\cdot \frac23 \cdot  \frac{1}{2}\cdot 3=\frac{3}{10}$
$P((bwww),b)=P(X=5)=\frac{2}{5}\cdot \frac{3}{4}\cdot \frac23\cdot  \frac{1}{2} \cdot 4=\frac{2}{5}$
For sanity check you can always sum up all probabilities. The sum must be $1$.
For the pdf you write all  the probabilities  of $x=k$ ($2\leq k\leq 5$)  and additional the case $=0, \  \text{elsewhere}$
A: 
So, I think there must be 9 possible events that can happen. 

How did you arrive at that ?
You are drawing five balls from the urn, one-by-one without replacement, but stopping when the second from the two black balls is drawn.
To count these (equally probable) outcomes, you need to count ways to arrange the 2 black and 3 white balls in a line.
We can partition these ten outcomes into four events, enumerated by the random variable $X$, the count for the draw on which the second from the two black balls is extracted.   This count will be from $2$ to $5$.
The event that this ball is extracted on draw $k$ is the event that the first from the black balls is among the first $k-1$ draws .

 $$\mathsf P(X{=}k)=\dfrac{\binom{k-1}1}{\binom 52}\mathbf 1_{k\in\{2,3,4,5\}}\\=\dfrac{k-1}{10}\mathbf 1_{k\in\{2,3,4,5\}}$$

However, this is the probability mass function (pmf). 
For the cumulative distribution function (CDF) you want to measure $X\leq k$, the event that the two black balls are among the first $k$ balls drawn.

 $$\mathsf P(X{\leqslant} k)~=~\qquad$$
$$~$$

