Urn Probability (need multiple ways) I need multiple ways to solve this question. Thank you! 
There are $A$ black balls, and $B$ white balls in an urn. You select balls one by one from this urn randomly without replacement. What is the probability that the $k$-th ball you selected is black? $(1\leq k \leq A+B)$, and $A,B$ are positive integers)
 A: HINT: think about it as having aligned $A+B$ persons labeled $1, 2, \ldots, A+B$ in a straight line, and some guy with a bag of $A+B$ balls wherein $A$ are black and $B$ are white. This funny looking guy with a long beard wraps all of the balls into red paper and hands them out, one by one, to each person in the queue. After all have gotten their present they unwrap simultaneously. What is the probability that the $k$-th person got a black ball? Does it matter if you are the first or second person? Or the $k$-th?
A: What M.B is saying is that the chance that "the first person gets a black ball" is exactly the same as "the $k$th person gets a black ball". As surprising as this sounds, it is completely true.
Imagine that after you draw each marble, you put it next to the previous ones, so you have a row of marbles. Once you've drawn the $k$th marble, you don't stop; you keep drawing until the urn is empty and you have a line of marbles $A+B$ long. (Drawing marbles after the $k$th can't possibly change the probability, right?) But you still know what the $k$th was, it's just the $k$th in the sequence.
But each possible sequence is equally likely! If you had numbered the balls, all you'd be doing is shuffling ${1, 2, 3, \ldots, A+B}$, and there's exactly one way to draw each permutation. It's exactly like shuffling a deck of cards. The chance that the Ace of Spades is at the top is exactly the same as the chance it's at the $41$st position. So the chance the first marble is black is equal to the chance the $k$th marble is black.
What is the chance of the first marble being black?
A: The Hard Way: You can calculate the probability directly. To motivate my calculation, let's first look at the case $k=3$. The probability 
that the 3rd ball is black depends on what happens in the first two draws. Taking all four cases into account we get
\begin{eqnarray*}\mathbb{P}(3^{\rm rd} \mbox{ ball is black})
&=& {B\over A+B}\,{B-1\over A+B-1}\,{A\over A+B-2}+ {B\over A+B}\,{A\over A+B-1}\,{A-1\over A+B-2}\\
&+& {A\over A+B}\,{B\over A+B-1}\,{A-1\over A+B-2}+ {A\over A+B}\,{A-1\over A+B-1}\,{A-2\over A+B-2}\end{eqnarray*}
Extending this argument we find that 
$$\mathbb{P}(k^{\rm th} \mbox{ ball is black})=\sum_b {k-1\choose b-1}{A^{\underline{b}} \, B^{\underline{k-b}} \over (A+B)^{\underline{k}}},\tag 1$$
where the underlined power is an alternative to the Pochhammer symbol for "falling factorial". The variable $b$ refers the number of black balls drawn. 
By a well-known formula (see page 245 of Concrete Mathematics) we know that 
$$\sum_k {n\choose k}x^{\underline{k}}\,y^{\underline{n-k}}=(x+y)^{\underline{n}}.$$
It follows that 
\begin{eqnarray*}\sum_b {k-1\choose b-1} A^{\underline{b}} \, B^{\underline{k-b}} 
&=& A\sum_b {k-1\choose b-1} (A-1)^{\underline{b-1}} \, B^{\underline{k-b}}\\
&=& A(A+B-1)^{\underline{k-1}}\\
& =&{A\over A+B}(A+B)^{\underline{k}}.
\end{eqnarray*}
Dividing this by $(A+B)^{\underline{k}}$ and using (1)  we get the result 
$$\mathbb{P}(k^{\rm th} \mbox{ ball is black})={A\over A+B}.$$
Reference: Concrete Mathematics: A Foundation for Computer Science, by Ronald Graham, Donald Knuth, and Oren Patashnik.
