# Probability question

#### winds

##### New Member
I read this as a solution to a problem, but I'm not sure how to get it:

Say we have k balls, of which 1 is red and k-1 are black, and k is odd. If we split them into two "half" piles of m and n balls, with m = n + 1, then the probability that pile "m" will have the red ball is m / (m+n).

I understand the other part, when k is even, then probability is 0.5, because it's equally likely to end up in either pile. And I understand, in the odd case, that the probability will be greater for the bigger "m" pile. But why will it be exactly the ratio given above?

Any insight much appreciated. Thanks.

#### BGM

##### TS Contributor
You have a total of $$m + n$$ balls
You split them into two piles, one pile with $$m$$ balls
and another pile with $$n$$ balls
So I think the answer $$\frac {m} {m+n}$$ is quite intuitive.
The probability is proportional to the number of balls in the pile

$$\frac {{m \choose 1} {n \choose 0}} {{m+n \choose 1}} = \frac {(m)(1)} {m+n} = \frac {m} {m+n}$$