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.

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.
