General probability of choosing marbles There are $x$ white marbles and $y$ black marbles in a bag of $n$ marbles where $x+y=n$ and $0 \leq x,y≤n$. I will now choose k marbles to remove from the bag, where $0 \leq k \leq n$. 


*

*If there is $n-1$ white marbles and 1 black marbles in the bag, how would the probability for black marble to be removed equate to $p=\frac{k}{n}$? My argument would be: 
$$\frac{C_{k-1}^{n-1}\cdot C_{1}^{1}}{C_{k}^{n}}=\frac{k}{n}$$
which involves choosing k-1 white from n-1 marbles after one black marble, which contain the thinking that all white marbles are different, so I'm not sure if this is valid.
$$$$

*If there are $x$ white and $y$ black marbles, how can I calculate the general probability for each combination of possible white and black marbles removed? (e.g. if there is 5 black and 4 white marbles, and I plan to remove 5 marbles, how do I calculate the probability that 3 white and  2 black marbles are removed, or similarly the probability that 1 white and 4 black marbles are removed? Here I am also thinking to use:
$$\frac{C_{a}^{x}\cdot C_{k-a}^{y}}{C_{k}^{n}}$$
to solve the probability that $a$ white marbles is selected from $n$ marbles containing $x$ white and $y$ black marble, which I'm again not sure if it is the correct approach.
Note: If my method turns out to be valid, no answers to this question will be necessary.
 A: If a bag contains 5 Black marbles and 4 White marbles, and you withdraw 5 marbles at random without replacement, then then number $X$ of White marbles among the 5 withdrawn has a hypergeometric distribution with
$$P(X = k) = \frac{{4 \choose k}{5 \choose 5-k}}{{9 \choose 5}},$$
for $k =0, 1, 2,3,4.$
In particular, $$P(X = 3) =
 \frac{{4 \choose 3}{5 \choose 2}}{9\choose 5}
= \frac {4\cdot10}{126} =  0.3174603.$$ 
In R statistical software, where dhyper is a hypergeometric PDF, you can make a probability table for the distribution as shown below. [You can ignore row numbers in brackets.]
k = 0:5;  PDF = dhyper(k, 4,5, 5)
cbind(k, PDF)
     k         PDF
[1,] 0 0.007936508
[2,] 1 0.158730159
[3,] 2 0.476190476
[4,] 3 0.317460317   # Shown above
[5,] 4 0.039682540
[6,] 5 0.000000000   # Impossible to get 5 White

Notice that I tried to find the probability of getting 5 white marbles, which is impossible because there are only 4 white marbles in the bag. In writing the PDF you can either (i) be careful to restrict $k$ only to possible values or (ii) use the convention that the binomial coefficient ${a \choose b} = 0,$ if integer $b$ exceeds integer $a.$ If your text includes a formal statement of the hypergeometric PDF, you should notice which method is used.
Here is a plot of the specific hypergeometric distribution mentioned in your problem.

Computations of $\mu = E(X) = \sum_{k=0}^5 k*p(k) = 5(4/9) = 2.2222$
and 
$$\sigma^2 = Var(X) = \sum_{k=0}^5 (k-\mu)^2p(k)\\
= \sum_{k=0}^5 k^2p(k) - \mu^2 = 0.6173$$ are shown below.
You may find formulas for these in your text.
mu = sum(k*PDF);  mu
[1] 2.222222

vr = sum((k-mu)^2*PDF);  vr
[1] 0.617284
sum(k^2*PDF) - mu^2
[1] 0.617284

A: Your method is indeed valid.
For the first question, note that all marbles are equally likely to be chosen, and the probabilities of marbles being chosen must add up to $k$, so the answer is trivially $\frac{k}{n}$.
