Etydhv

A bag contains $N$ balls, $2$ of which are red. Balls are removed, one by one, (without replacement), stopping when both red balls have emerged. Find the probability that exactly $n$ balls are removed.

I'm honestly not sure where to even start. I know it's an intersection of ($1$ Red ball in $n-1$ attempts) and (red on $n$-th attempt).

I think the $Pr(textRed on;ntext-th attempt; | ;1 ; textRed already)$ is $$ frac1N-(n-1) $$ but I'm not sure for the other probability, or whether this one is correct for that matter.

Any help would be appreciated. :)

You are definitely on track. In the first probability, order does not matter. You have drawn $n-1$ balls, and one of them is red. The total number of outcomes is the total number of ways to draw $n-1$ balls from $N$. The number of outcomes you are interested in are given by:

$$dbinom21dbinomN-2n-2$$

So, the probability that you have drawn exactly one red ball in the first $n-1$ draws is:

$$dfracdbinom21dbinomN-2n-2dbinomNn-1$$

Now, multiply by the probability that the last ball is red (you were correct):

$$dfrac1N-(n-1)$$

Final probability:

$$dfrac2(n-1)N(N-1)$$

Here's another approach, depicted graphically below for $N = 8$.

We can construct a grid, where the equiprobable slots are represented by squares, indexed by the sequence position of the two red balls. The X's down the diagonal indicate that the two red balls cannot occupy the same sequence position.

The result of the experiment is the maximum of the two positions; for instance, the cases where $n = 5$ are marked by a red $5$. From this it can be seen that there are $2n-2$ different squares where the number of balls drawn is $n$, out of a total of $N^2-N$ possible squares. Thus, the desired probability is

$$
frac2n-2N^2-N = frac2(n-1)N(N-1)
$$

as in InterstellarProbe's answer.

This is a problem where it's easier to condition the other way: compute
$P(Acap B)=P(B)P(Amid B)$, where $B$ is the event that a red ball appears on draw $n$, and $A$ is the event that a red ball appears exactly once in the first $n-1$ draws.

The probability $P(B)$ equals $2/N$, since at any draw there are $2$ red balls out of a total of $N$ balls that could be picked.

The conditional probability $P(Amid B)$ equals $(n-1)/(N-1)$, since this is the same as the unconditional probability that when you line up $N-1$ balls, only one of which is red, the red appears somewhere in the first $n-1$ positions.