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 $Pr(\text{Red on}\;n\text{-th attempt}\; | \;1 \; \text{Red already})$ is $$ \frac{1}{N-(n-1)} $$ but I'm not sure for the other probability, or whether this one is correct for that matter.
Any help would be appreciated. 🙂
Best Answer
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:
$$\dbinom{2}{1}\dbinom{N-2}{n-2}$$
So, the probability that you have drawn exactly one red ball in the first $n-1$ draws is:
$$\dfrac{\dbinom{2}{1}\dbinom{N-2}{n-2}}{\dbinom{N}{n-1}}$$
Now, multiply by the probability that the last ball is red (you were correct):
$$\dfrac{1}{N-(n-1)}$$
Final probability:
$$\dfrac{2(n-1)}{N(N-1)}$$