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:
$$P(X=l) = \dfrac{\dbinom{N-k-1}{m-1} \cdot \dbinom{1}{1}}{\dbinom{N-k}{m}}$$

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:


*

*$P(X=1)=\dfrac{1}{N-k}$. This is immediate.

*$P(X=2)$. 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!)

*$P(X=3)$. 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.
