# Probability for “drawing balls from urn”

I'm afraid I need a little help with the following:

In an urn there are $$N$$ balls, of which $$N-2$$ are red and the remaining are blue. Person $$A$$ draws $$k$$ balls, so that the first $$k-1$$ are red and the $$k$$ ball is blue. Now Person B draws m balls:

What's the probability for Person $$B$$ to draw the last/second blue ball after drawing $$(m-1)$$ red ones? Meant is, that the drawing stops as soon as the blue ball has been drawn. So the blue one must be the last one.

My ideas:

• there are only $$(N-k)$$ balls in the urn left, of which $$(N-2)-(k-1) = N-k-1$$ are red and $$1$$ is blue.
• for $$m$$ we have: $$m \in \{1, \dots, N-k\}$$.

Shouldn't that be solved with Hypergeometric Distribution? With this I like to calculate the probability for $$(m-1)=l$$ successes of red, so that the $$m$$-th ball is blue: $$P(X=l) = \dfrac{\dbinom{N-k-1}{m-1} \cdot \dbinom{1}{1}}{\dbinom{N-k}{m}}$$

• @JimmyR. Sorry, meant is that the blue is the last one, so that the drawing-process stops with the blue one being drawn. – Vazrael Mar 31 '16 at 10:00
• Ok, sorry to make it clear: You want the probability $P(X=m)$ for $m=1,2,\dots, N-k$ where $X$ is the number of trials until you get the blue (inclusive), right? – Jimmy R. Mar 31 '16 at 10:01
• @JimmyR. absolutely correct und much better expressed than I did :) – Vazrael Mar 31 '16 at 10:02
• Ok!! Thanks, let's give it a new try (I will edit my answer) – Jimmy R. Mar 31 '16 at 10:03
• Discrete uniform came out! – Jimmy R. Mar 31 '16 at 10:08

The hypergeometric distribution accounts for the change in probability of success, but counts number of success in finite predetermined sample, which is not what you want. The geometric distribution, which counts number of trials until the first success works in a different setting: constant probability of success and infinitely many draws. So, follow a direct approach instead:

1. $P(X=1)=\dfrac{1}{N-k}$. This is immediate.
2. $P(X=2)$. For this you need to draw a red and then the blue, hence $$P(X=2)=\frac{N-k-1}{N-k}\cdot\frac{1}{N-k-1}=\frac{1}{N-k}$$ (where by now the pattern starts to reveal!)
3. $P(X=3)$. For this you need to draw two reds and then the blue, hence $$P(X=3)=\frac{N-k-1}{N-k}\cdot\frac{N-k-2}{N-k-1}\cdot\frac{1}{N-k-2}=\frac{1}{N-k}$$

So, the answer is discrete uniform distribution on $\{1,2,\dots,N-k\}$ (you can continue up to $m=N-k$ to verify this), i.e. $$P(X=m)=\frac{1}{N-k}$$ for any $1\le m\le N-k$.

Another (equivalent) way to think of this process and reach this result is to think as follows: You will order the $N-k$ balls, and you want to know the probability that the blue ball will be in the $m-$ th position Indeed, the probability that it lands in any position from $1$ to $N-k$ is the same, hence the uniform distribution.

• @ JimmyR. That's great! Thank you very very much for giving this helpful and detailed answer - your "Another (equivalent) way" was also very helpful for understanding this rather simple problem in a very smooth way :)) I'm dealing with a second very similar problem now, but I'll start a new topic and would be glad for your help once again. – Vazrael Mar 31 '16 at 10:21
• @Vazrael You are welcome. As you see, the other (short) way, only came to mind after putting down to paper a direct approach. – Jimmy R. Mar 31 '16 at 10:25

Try to look at this from the perspective of the blue ball.

He is on his own and in the company of $N-k-1$ red balls.

There will be $m$ picks.

$$P\left\{ \text{I am picked as last}\right\} =$$$$P\left\{ \text{I am picked as last}\mid\text{I am picked}\right\} P\left\{ \text{I am picked}\right\} =$$$$\frac{1}{m}\frac{m}{N-k}=\frac{1}{N-k}$$

More shortly you can think of the picking as putting the $N-k$ balls in a row.

What is the probability that the blue ball will receive spot $m$?

edit:

• Welcome to Math.SE. A better formatted answer would get more upvotes. So, take your time and learn formatting. Mathematical expressions and equations can be formatted using $LaTeX$ syntax – Shailesh Mar 31 '16 at 10:49
Person A took $k$ balls, one of them was blue. So when B starts picking balls, the urn contains $N-k$ balls, one of them blue. Suppose B takes all balls from the urn, one by one. The probability that the blue one appears at any pre-selected $m$-th position is the same for each $m \le N-k$ and it equals: $$1/(N-k)$$