How likely are various outcomes in a lottery with multiple prizes This question is (I think) simpler than most posed here, but it's beyond my ability to solve.  
I'm trying to calculate the probability of various outcomes for a charter school lottery.  There are two classes of entrants: students otherwise zoned for under-performing schools, and all other students. 
Let x=the number of students zoned for under-performing schools
Let y=the number of students not zoned for under-performing schools
Let z=the number of spaces available in the school
It's easy enough to determine the expected number of students admitted from each class, but how can I determine the probability of various other outcomes.  For instance, how likely is it that the number of students admitted from a class is half the expected number?
For what it's worth, I'm a math teacher who has never had a stat, probablity, or discrete math course, although I've taught myself the basics.  If you have any suggestions for further self-guided study in probability I'd appreciate them, but right now I just need to solve this practical problem.
 A: In a simple lottery, all $z$ element subsets of the box of tickets have equal chances of being drawn.  When there are $x$ tickets of type "X" and $y$ tickets of type "Y", the number of these subsets is written $\binom{x+y}{z}$ (read "$x$ plus $y$ choose $z$").  This binomial coefficient has a simple, well-known formula,
$$\binom{x+y}{z} = \frac{(x+y)(x+y-1)\cdots(x+y-z+1)}{z(z-1)\cdots(2)(1)}.$$
This is the $k^\text{th}$ entry (counting from $0$) in the $x+y^\text{th}$ row of Pascal's Triangle.
Each $k$ from $0$ through the smaller of $x$ and $z$ is a possible number of the "X" tickets drawn.  The number of ways to draw $k$ tickets from category "X" and the remaining $z-k$ tickets from category "Y" is the product $\binom{x}{k}\binom{y}{z-k}$.  Whence the chances of drawing exactly $k$ tickets from category "X" are the ratio of this number and the total number of $z$ element subsets:
$$\frac{\binom{x}{k}\binom{y}{z-k}}{\binom{x+y}{z}}.$$
This is known as "the" hypergeometric distribution.  (For every combination of $x$, $y$, and $z$ describing such a lottery, there is a hypergeometric distribution giving the chances for all possible numbers of tickets $k$ that can be drawn from category "X".)
For example, with $x=80$, $y=120$, and $z=20$, the chances for $k=0, 1, \ldots, 20$ are
0.000018, 0.000289, 0.0021, 0.0097, 0.0304, 0.0705, 0.1247, 0.1724, 0.1894, 0.1668, 0.1184, 0.0679, 0.0314, 0.0116, 0.0034, 0.00078, 0.000137, 0.000018, 1.57e-6, 8.62e-8, 2.2e-9

Here is a plot of these chances:

The exact chances are shown by the heights of the black dots.  For reference, the red curve plots the density function for the normal distribution with the same expectation ($8$) and same variance ($864/199$).  It can be used (with a so-called "continuity correction") to approximate the chances.  Those approximate values are indicated by the heights of the thin red lines.  A close look shows that the red lines are a tiny bit short to the left of $k=8$ and a tiny bit tall to the right--but the approximation looks quite good.  Using this normal approximation sometimes is easier than doing the calculations, but its chief virtue is in allowing us to use our experience with and intuition about normal distributions to understand this one.  (The normal approximation will be reasonable provided both $x$ and $y$ are $5$ or greater and neither is large compared to the other.)
With these chances in hand, the probability of any stipulated event can be computed by adding the chances of its components.  For instance, the chance that five or fewer of the tickets are drawn from the "X" pool is the chance that $k$ is one of $0, 1, 2, 3, 4,$ or $5$.  The sum of those component chances is 
$$0.000018+ 0.000289+ 0.0021+ 0.0097+ 0.0304+ 0.0705 = 0.1130,$$
or about $11$%.
A: Are you selecting separately from the zoned and non-zoned pools? It seems you are randomly selecting from the total pool of zoned and non-zoned students. If so, you can calculate the precise probability for any outcome using the binomial theorem. If the numbers get unwieldy, you can use the normal approximation to the binomial to calculate the probability of an outcome as extreme or more extreme than a given value. Both these approaches are in most introductory statistics textbooks. See, for example, http://jwilson.coe.uga.edu/EMAT6680Fa06/Crumley/Normal/Normal1.html 
