Suppose in an urn there are $20$ black balls labelled $1,2, \ldots , 20$ and $10$ white balls labelled $1,2, \ldots ,10$. Balls are drawn one by one without replacement. Find the probability that the white ball labelled $1$ is drawn before all the black balls.
My attempt $:$
If we want to draw the first white ball before all the black balls then I have to draw the first white ball in one of first $10$ steps.
Suppose I draw the first white ball in $k$-th step. Then in order to fulfil my requirement I have to draw white balls in first $k-1$ steps. That can be done in $\binom 9 {k-1} (k-1)!$ ways. For each of these ways remaining $30-k$ balls can be drawn in $(30-k)!$ ways. This $k$ can run from $1$ to $10$.
So the total number of ways to draw the first white ball before all the black balls is $$\sum\limits_{k=1}^{10} \binom 9 {k-1}(k-1)! (30-k)!$$ So the required probability is $$\frac1{30!}{\sum\limits_{k=1}^{10} \binom 9 {k-1}(k-1)! (30-k)!} =\sum\limits_{k=1}^{10} {\frac {9!(30-k)!} {30!(10-k)!}}$$
Now my instructor has given it's answer which is $\frac {1} {21}$. Does the above sum evaluate to $\frac {1} {21}$? Is there any other simpler way to do this? Please help me in this regard. Thank you very much.
Best Answer
The question is rather badly posed. "The first white ball" sounds as if it refers to the first white ball drawn, but apparently it's intended to refer to the white ball with the number $1$.
There's indeed a much simpler way to show that the probability of drawing the white ball with the number $1$ before all black balls is $\frac1{21}$. Consider the $21$ balls comprising the white ball with the number $1$ and the $20$ black balls. All orders of these $21$ balls are equally likely; in particular, the white ball with the number $1$ is equally likely to be in any of the $21$ positions in this order. Thus the probability for each of the positions, including the first position, is $\frac1{21}$.
On how to evaluate your sum, see the hockey-stick identity.