Reversing Combinations to find probabbilities of variable two state systems First I'm going to lay out the problem and do an insanity check to make sure I've planned out properly so far. The first question is probably trivial, but the second one is definitely a bit more challenging and I don't think I'm duplicating another question I would easily find. There is a bonus question at the end taht is a bit of a curiosity but doesn't need to be answered for any particular reason.
I have an array with 30,000 components (lets say cards) that can be (randomly) in either a 0(ZERO) or 1(ONE) state. If I were to then split this array(deck) into 15 equally sized arrays and just look at the first element this is probabilistically equivalent to just selecting 15 elements (right?). 
From here I am calculating the probability of getting 3 1's in the 15 selections, depending on the number of 1s present. I have that the probability of drawing all 3 ones first is 1 in 4.499^e12. 
Questions


*

*What is (and hopefully the general form) of the probability of drawing 15 cards and getting 3 ONEs in a stack with only 3 ONEs in the entire deck? We can assume that the deck stay 30,000 unless you can also easily represent that in the general form.

*How would I reverse the operation to determine how many ONEs it would take to create a 1% (or x%) chance of drawing 3 ONEs in a stack of 30,000 under the same conditions (15 draws, Y many ONEs)? Completely general form for this too would be appreciated but the 30,000 case is good enough. 
Bonus: What about if we extended the number of possible states arbitrarily. How does that alter the probability equations? (I am guessing series and/or "product series" enter the mix)
 A: $(1)$ If I understand your question correctly, the probability distribution is hypergeometric. Using the notation at that link, we define:
\begin{eqnarray*}
N &=& \mbox{Population size (size of deck: 30000)} \\
n &=& \mbox{Sample size (15)} \\
K &=& \mbox{Number of "successes" in population (number of 1's in deck)} \\
k &=& \mbox{Number of "successes" in sample (number of 1's in selection).} \\
\end{eqnarray*}
If random variable $X$ is the number of $1$'s in our selection then
$$P(X=k) = \dfrac{\binom{K}{k} \binom{N-K}{n-k}}{\binom{N}{n}}.$$
$(2)$ We want to solve the following equation for $K$ given $N,n,k,p$, where $p$ is the required probability of getting $k$ $1$'s in the selection, E.g. $p = 0.01$ for $1$%.
$$p = \dfrac{\binom{K}{k} \binom{N-K}{n-k}}{\binom{N}{n}}.$$
This isn't easily solvable. So a practical method of solution is to try various values for $K$ and try to get the resulting $p$ close to what we want. For example, Excel's HYPGEOM.DIST() function, produces these results (Excel formula: =HYPGEOM.DIST(3,15,A2,30000,FALSE)):
\begin{eqnarray*}
K && p \\
500 && 0.001713583 \\
600 && 0.002845673 \\
700 && 0.004340868 \\
800 && 0.006222501 \\
900 && 0.008505936 \\
958 && 0.010017961 \\
1000 && 0.011199441 \\
2000 && 0.058887575 \\
3000 && 0.128512571 \\
4000 && 0.193706685 \\
5000 && 0.236317365.
\end{eqnarray*}
(Bonus) If you are still only interested in the probability of getting $k$ $1$'s for some $k$, then it doesn't change the formula in $(1)$ at all. You are still regarding $1$'s as "successes" and all other states as "failures".
However, if you want to know, more specifically, the probability of getting $k_0$ $0$'s, $k_1$ $1$'s, $\ldots$ and $k_c$ $C$'s for some integer $C$ then we have a multivariate hypergeometric distribution, which is a generalisation of the hypergeometric distribution. Here (as in the link), if we have in the total population (i.e. deck) $K_0$ $0$'s, $K_1$ $1$'s, $\ldots$ and $K_c$ $C$'s, then the probability of getting $k_0$ $0$'s, $k_1$ $1$'s,$\ldots$ and $k_c$ $C$'s in the sample is:
$$\dfrac{\prod\limits_{j=1}^{c}{\binom{K_j}{k_j}}}{\binom{N}{n}}.$$
