General Solution to a Specific Problem The Original Problem:
An urn contains 7 white balls and 3 red balls. Balls will be drawn one-by-one at random without replacement until a white ball is drawn. One possible outcome for this experiment is W (a white ball is selected on the first draw). Another possible outcome is RW (a red on the first draw and a white on the second draw).
a) List all of the possible outcomes. (Hint: There are 4 possible outcomes.)
b) Let $X = \text{number of draws until white ball is obtained}$. Find the probability mass function of $X$ and cumulative distribution function.
c) Find $E(X)$ and $SD(X)$

Updated Problem
After finding the solution to the original problem, we are tasked to solve the problem again using general terms (unknown amount of each color ball). That is:
$$ \begin{align}
m &= \text{ # of white balls}\\
n &= \text{ # of red balls}\\
R &= \text{red ball}\\
W &= \text{white ball}\\
n &\le m\\
\end{align}$$
I am unsure how to start the CMF, the expected value, and the standard deviation. Any help on where to begin would be much appreciated.

Solution So Far
a) $W, RW, RRW, ..., nRW$
b) $P(X = x) = \frac{m}{(n - (x - 1)) + m}$

Many thanks in advance!
 A: A problem of this nature is best derived from first principles and a tree-based analysis. The figure below shows the event space as a tree. The leaves correspond to the events $\{X=k\}$ for $k = 1, 2, \dots, n+1$. Each node in the graph represents an event, and the product of the conditional probabilities on the path from the root to any node is the probability of that event.

Using this analysis, it is now easy to derive a closed form for $P(X=K)$. Look at the pattern:
$$P(X=1) = \frac{m}{n+m}$$
$$P(X=2) = \frac{n}{n+m}\frac{m}{n+m-1}$$
$$P(X=2) = \frac{n}{n+m}\frac{n-1}{n+m-1}\frac{m}{n+m-2}$$
$$P(X=k) = \left(\prod_{i=0}^{k-2}\frac{n-i}{n+m-i}\right) \frac{m}{n+m-(k-1)}$$
This notation is suggestive of the idea that for the first white ball to arrive at draw $k$, we need to draw $k-1$ red balls in succession.
Now let us analyze the expectation $E(X)$. We will number the red balls $R_1, \dots, R_n$. Let the indicator random variable $I_j$ = 1 if ball $R_j$ was chosen before the first white ball, 0 otherwise. What is the probability $P(I_j = 1)$? This is simply $\frac{1}{m+1} (can you see why?).
It follows that 
$$X = 1 + \sum_{j=1}^{n}I_j$$
$$E(X) = E(1+\sum_{j=1}^{n}I_j) $$ 
$$E(X) = 1 + E(\sum_{j=1}^{n}I_j)) $$
$$E(X) = 1 + \sum_{j=1}^{n}E(I_j)) $$
$$E(X) = 1 + \frac{n}{m+1}$$
Hopefully this should be enough for you to complete the rest of the questions.
