Question: > 
> 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.

Answer: 
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.
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 } $