# Probability of taking balls without replacement from a bag question

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. :)

• Please provide some context for this question. What have you tried? Where are you stuck? Just hit the 'edit' button under the problem and add some details that will let us know what sort of response you are looking for. – InterstellarProbe May 8 at 20:15
• Thank you, will do – mathshatesme123 May 8 at 20:19
• I guess you are thinking that before the $n$th draw, $n-1$ balls have already been drawn, leaving $N - (n-1)$ in the urn, if if one of the balls drawn was red, exactly one of the balls still in the urn is red. That seems correct to me. – David K May 8 at 20:30

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:

$$\dfrac{2(n-1)}{N(N-1)}$$

• Thank you! Can I just ask, why does order matter? Also, how have you formed the fraction? Is there like a particular formula for doing so? – mathshatesme123 May 8 at 20:50
• Order does not matter in the first probability because the result you are looking for is one red ball in $n-1$ picks. You do not care what position it is drawn in, just that it shows up. In general, the probability of an occurrence is given by $$\dfrac{\text{# of outcomes with desired result}}{\text{# of possible outcomes}}$$ To get the final probability, it was just multiplying out: $$\dfrac{\dbinom{2}{1}\dbinom{N-2}{n-2}}{\dbinom{N}{n-1}}\cdot \dfrac{1}{N-(n-1)} = \dfrac{2(n-1)}{N(N-1)}$$ I am not going through the steps to simplify, but that's the easy part :) – InterstellarProbe May 8 at 21:08
• Thanks again! :) – mathshatesme123 May 8 at 22:14
• Alternatively, you want the probability for placing one red ball among the first $n-1$ positions and one in the next position, when selecting any two positions among $N$ to place the red balls. $$\dfrac{\dbinom {n-1}1\dbinom 11}{\dbinom N2}=\dfrac{2(n-1)}{N(N-1)}$$ – Graham Kemp May 9 at 0:02

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

$$\frac{2n-2}{N^2-N} = \frac{2(n-1)}{N(N-1)}$$

This is a problem where it's easier to condition the other way: compute $$P(A\cap B)=P(B)P(A\mid B)$$, 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 $$P(A\mid B)$$ 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.