Alternative interpretation of ball and urns problem Suppose an urn has r red balls and b black balls. They are withdrawn one at a time at random until a total of k, k $\leq$ r, red balls have been withdrawn. Find the probability that a total of n balls are withdrawn.

A first approach would be to consider the probability of drawing k-1 red balls in the first n-1 withdrawals, then multiply it by the probability of the final ball drawn being red. This gives

$$\frac{\binom{r}{k-1}\binom{b}{n-k}}{\binom{b+r}{n-1}}\cdot\frac{r-k+1}{b+r-n+1}\;.$$

All this is well. However, I wanted to try solving the problem not by considering probabilities in each step, but by counting outcomes as a whole. This is where I run into problems and wish that someone can enlighten me.

This is one of the futile attempts that I made:

$$P = \frac{\text{#ways to draw k red in n attempts}}{\text{#ways to draw k red in any number of attempts}} = \frac{\text{#ways for red ball to be nth position and exactly k-1 red balls preceding it in any of the n-1 slots}}{\text{#number of ways for any number to choose k red balls and insert any number of blue balls in their way}} = \frac{\binom{r}{k-1}\binom{b}{n-k}(r-k+1)}{\binom{r}{k}2^b}\;.$$

Clearly, the numerator is the same. Since in my second method i'm not involving the variable in the denominator, this already hints to disaster as the former answer does have an n in the answer. After a few more tries I'm no longer sure what objects I'm supposed be counting exactly. Please shed some light
You can only obtain a probability as the quotient of the number of favourable outcomes over the total number of outcomes if all outcomes are equiprobable. The way you're trying to count, there's no reason to expect the outcomes to be equiprobable, as they involve different numbers of draws.

The equiprobable outcomes that you'd need in order to do it this way are all possible outcomes of $b+k$ draws (including the ones with more than $k$ red balls drawn). But counting those is a bit complicated; it's easier if you go all the way and consider all possible outcomes of $b+r$ draws. That yields

$$
\frac{\binom{n-1}{k-1}\binom11\binom{b+r-n}{r-k}}{\binom{b+r}r}\;,
$$

which coincides with your first result.
I think the question should be - Find the probability that a total of n "red" balls are withdrawn.

Then let p = r/(r+b) and q = 1 - p = b/(r+b), and let X be the number of red ball. So X has a binomial distribution.

Thus

\begin{align}
\ P(X=n)= \binom{r+b}{n} p^{n}q^{r+b-n}
\\ =\frac{(r+b)!}{n!(r+b-n)!} p^{n}q^{r+b-n}
\end{align}