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: 
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 $\frac{9!20!}{30!}\times\frac{30!}{9!21!}=\frac{1}{21}$.
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 } $