Probabilities of selecting exactly $n$ white and $m$ black balls I stumbled upon a tricky problem that I don't know how to solve.
There $a$ white and $b$ black balls in an urn. We select one first ball at random (let's call its color $F$) and then return this ball back to the urn. We then add $c$ balls of the same color $F$.
From this urn we select $n + m$ balls without replacement ($n + m \leq a + b + c$). What is the probability that we end up with $n$ white and $m$ black balls?
And the second question is: how can we show that in the $k$-th step of selection the probability of the ball being white is $\dfrac{a}{a+b}$. It is quite clear intuitively, yet I would be interested to see a formal proof of this.
Any help would be greatly appreciated. I would also be interested in incomplete solutions.
 A: Part 1)
Let's say we have urn with $a$ white balls, $b$ black balls. After first draw, we return ball back to the urn and add $c$ balls of the same colour. We're asked about probability of event $E - $exactly $n$ white, $m$ black balls (We draw exactly $n+m$ balls ).
Let $X$ - be first draw, that is $X =$ white with probability $\frac{a}{a+b}$, and $X =$black with probability $\frac{b}{a+b}$, we'll condition on $X$.
$ \mathbb P(E) = \mathbb P(E|X=white)\mathbb P(X=white) + \mathbb P(E|X=black)\mathbb P(X=black) = \frac{{a+c \choose n} {b \choose m}}{{a+b+c \choose n+m}}\frac{a}{a+b} +\frac{{a \choose n} {b+c \choose m}}{{a+b+c \choose n+m}}\frac{b}{a+b}$.
Note that when both $a < n$ and $b < m$ this probability is simply $0$, due to that Newton Symbols. 
Part 2)
I'll assume you count $k$ after that first draw when you add either black or white $c$ balls.
You want formal proof, so we'll use induction. Let $E_k$ - at $k'$th draw we get white ball. For $E_1$ we have:
$\mathbb P(E_1) = \mathbb P(E_1|X = white)\mathbb P(X=white) + \mathbb P(E_1 | X=black) \mathbb P(X = black) = \frac{a+c}{a+b+c}\frac{a}{a+b} + \frac{a}{a+b+c}\frac{b}{a+b} = \frac{a(a+b+c)}{(a+b)(a+b+c)} = \frac{a}{a+b}$
So assume it works for $E_k$, and let $Y$ be one draw after $X$ (that is the first one we count). Then we have:
$\mathbb P(E_{k+1}) = \mathbb P(E_{k+1} | X =white)\frac{a}{a+b} + \mathbb P(E_{k+1} | X = black)\frac{b}{a+b}$
Now, we'll condition once more, but on $Y$. We have:
$ \mathbb P(E_{k+1} | X = white ) = \mathbb P(E_{k+1} | (X,Y) = (white,white)) \mathbb P(Y=white | X = white) + \mathbb P (E_{k+1} | (X,Y) = (white,black)) \mathbb P(Y = black | X = white) = \frac{a+c-1}{a+b+c-1}\frac{a+c}{a+b+c} + \frac{a+c}{a+b+c-1}\frac{b}{a+b+c} = \frac{(a+c)(a+c-1+b)}{(a+b+c)(a+b+c-1)} = \frac{a+c}{a+b+c}$
Similarly:
$\mathbb P(E_{k+1} | X = black) = \frac{a-1}{a+b+c-1}\frac{a}{a+b+c} + \frac{a}{a+b+c-1}\frac{b+c}{a+b+c} = \frac{a(a+b+c-1)}{(a+b+c)(a+b+c-1)} =\frac{a}{a+b+c}$
Here I used our induction claim, but with different number of balls we started with ( for example $a+c-1$ white, $b$ black, or so. All that matters, after we perform $Y$ - draw, then we just have $k$ draws left and we can use it.)
Plugging it:
$\mathbb P(E_k) = \frac{(a+c)a + ab}{(a+b)(a+b+c)} = \frac{a(a+b+c)}{(a+b)(a+b+c)} = \frac{a}{a+b}$
