# $n$-th ball is white if all previous blacks are black

Suppose we have an urn containing $a+b$ balls, $a$ of them are black, $b$ of them are white. We start to take out balls without replacement. After $n$ steps we notice that the first $n-1$ are black, while the $n$-th one is white. Calculate the probability of the $n+1$-th step.

I know that if there are no conditionals then the probability that the $n$-th ball is white/black is $\frac{a}{a+b}$ or $\frac{b}{b+a}$, but what I don't understand is how will this conditional affect our overall result? Will it be simply $\frac{b-1}{a+b-n}$ (since we already took out $1$ white ball) or should I take some more things into consideration?

• Please, if you are ok, you can accept the answer and set it as solved. Thanks! – user Jan 10 '18 at 10:05

It will be simply $$\frac{b-1}{a+b-n}$$
There are $a+b-n$ balls left and $b-1$ are white, so therefore the chance that the $(n+1)$-th ball is: $$\frac{b-1}{a+b-n}$$ Here $n<a+b$ must hold.
There is nothing else to take into consideration. The probabilities of picking a certain color ball depend only on the balls that are in the urn. There are $a + b - n$ balls left, and $b - 1$ of them are white. So the probability of selecting a white ball is as you stated.