Find the probability that the white ball labelled $1$ is drawn before all the black balls. 
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.
 A: To evaluate $\sum_{k=1}^{10}\frac{9!}{30!}\frac{(30-k)!}{(10-k)!}$:
\begin{align}
\sum_{k=1}^{10}\frac{9!}{30!}\frac{(30-k)!}{(10-k)!}&=\frac{9!\cdot 20!}{30!}\sum_{k=1}^{10}\frac{(30-k)!}{20!(10-k)!} \\
&=\frac{9!\cdot 20!}{30!}\sum_{k=1}^{10}\binom{30-k}{20} \\
&=\frac{9!\cdot 20!}{30!}\left[\binom{20}{20}+\binom{21}{20}+\binom{22}{20}+\binom{23}{20}+\ldots+\binom{29}{20}\right] \\
&=\frac{9!\cdot 20!}{30!}\left[\binom{21}{21}+\binom{21}{20}+\binom{22}{20}+\binom{23}{20}+\ldots+\binom{29}{20}\right] \\
&=\frac{9!\cdot 20!}{30!}\left[\binom{22}{21}+\binom{22}{20}+\binom{23}{20}+\ldots+\binom{29}{20}\right] \\
&=\frac{9!\cdot 20!}{30!}\left[\binom{23}{21}+\binom{23}{20}+\ldots+\binom{29}{20}\right] \\
&=\cdots \\
&=\frac{9!\cdot 20!}{30!}\binom{30}{21} \\
&=\frac{9!\cdot 20!}{30!}\frac{30!}{21!\cdot 9!} \\
&=\frac{1}{21}
\end{align}
When handling the sum, the relations $\binom{n}{n}=\binom{n+1}{n+1}$ and $\binom{n}{r}+\binom{n}{r+1}=\binom{n+1}{r+1}$ are used.
A: Your expression $\sum_{k=1}^{10}\frac{9!(30-k)!}{30!(10-k)!}$ can be rewritten as $\frac{9!20!}{30!}\sum_{k=1}^{10}\binom{30-k}{20}$. Now $\binom{30-k}{20}$ is the number of ways to choose $21$ numbers from $\{1,\ldots,30\}$ with the first being $k$ (because the remaining $20$ numbers must be chosen from the $30-k$ numbers after $k$), so summing this over all possible values of the first number chosen (which must be in the range 1-10) gives all ways to choose $21$ numbers from $30$. Thus $\sum_{k=1}^{10}\binom{30-k}{20}=\binom{30}{21}=\frac{30!}{9!21!}$, and so the whole expression becomes $\frac{9!20!}{30!}\times\frac{30!}{9!21!}=\frac{1}{21}$.
A: 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.
A: If Balls are drawn one by one out of an Urn containing w White balls and b Black balls. Then the number of Black balls, before the 1st White ball is given by :$ \frac{1}{Total White Balls + B_{special}  }(Total Black balls)$. $B_{special}$ which will break the chain of Black balls. Then we can say, The expected number of Black Balls preceding the 1st White ball is given by $\frac{b}{w + 1 }$.  Here, w +  $B_{special}$ = [w + one more ball] = [w +1 ]. So, there are  [w +1 ] equally likely places where $B_{special}$  can sit to break the chain. Therefore the probability that a Black ball can be placed is $\frac{1}{w + 1 }$ . If you can understand this problem, then you will be able to guess, that the answer for the problem given will be : $\frac{1}{b + W_{specialWhoCanBreakTheChain} } = \frac{1}{b + W_{1} } = \frac{1}{20 + 1 } = \frac{1}{21  } $
