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

Must be simple:
What Person A did is irrelevant to the question. You have Z balls where 1 has different color and you are looking for the probability of finding the different ball as the last at the m-th draw. 
So practically you are looking for the probability of having the different ball at the m-th place after making the Z balls in an order. You have 1 different ball so it is always 1/Z.
Z=N-k so the answer is 1/(N-k)
