 A particularly important type of combinatorial probability is a hypergeometric probability. In a hypergeometric probability, the sample space consists of equally likely outcomes, the outcomes are permutations, and we can describe the event as selecting groups from a group. For example, we might select a set of five cards from a deck of 52 cards, or maybe a committee of three persons from a group of twenty persons, or a sample of ten items from a collection of one thousand. For example, suppose we're selecting a committee of three persons from a group of twelve men and eight women. If the committee members are randomly selected, what is the probability that all persons on the committee are men? Now, randomly usually means we have a sample space of equally likely outcomes. Since our outcomes are permutations, the actual committee members, with equally likely outcomes, we have a combinatorial probability. And further, since we're selecting groups from a group, men from men, women from women, this is in fact a hypergeometric probability. So first, we'll determine the size of our sample space. Now, we're forming this committee of three persons, and so we're choosing three people out of twelve men and eight women, twenty people. And we can do this in twenty combined three ways, and so our formula gives us. And here's a useful thing to keep in mind. After simplifying, the numerator and denominator of our combination formula should have the same number of factors. And in fact, if you look a little closer, you can see patterns that emerge over not only how many factors we have, but where those factors actually are. In any case, we find that our sample space has twenty combined three 1,140 outcomes. Now, to select three men, we need to select them from the twelve men that we have available. And so here we'll use our different notation for expressing the number of combinations. Twelve choose three, and our formula gives us 220. And so of our 1,140 outcomes of these twelve combined three 220 correspond to committees of three men. And now we're ready to state our probability. Provided the outcomes are equally likely, the probability of the committee consisting of three men will be, or about one in five. Now, while it's not absolutely necessary to do this, it's actually good practice to always state the assumption that the outcomes are equally likely. And part of this is preparation for statistics. Because in statistics, one of the questions we ask is, are the outcomes equally likely? And so here it's important to understand that this probability we get is based on the assumptions that they are, and that may have implications further on down the line. So suppose we have a bag with three red balls, five white balls, and two green balls. If we randomly select three balls without replacement, what is the probability that all three are green? So there's three plus five plus two equals ten balls in the bag. And if we're selecting three without replacement, well, that's ten choose three ways to select them. And we'll make our assumption clear. Assuming we have equally likely outcomes, there are 120 ways to select three balls. We want to choose three green balls from a set of two. Well, that's obviously three choose two. We calculate that and there's three ways of doing that. And so our probability is, wait a minute. How can I select three green balls if there's only two green balls in the bag? And here's an important idea. Using a formula you don't understand is a guaranteed path to disaster. So remember the question we're really answering with this formula is, how many ways can we choose three green balls from a set of two without replacement? And so maybe we can set this up. Our permutations are, for the first ball, we have two choices. For the second, we have one choice. And for that third ball, there are zero choices left. There are no green balls left to choose from. And so there are zero ways we could make this selection. And so in fact, zero of the outcomes are sets of three green balls. And so the probability is zero out of 120 or zero. Well, how about something that doesn't have a zero probability? Suppose that we select three balls without replacement and we want to find the probability that two are red and one is green. Well, let's think about this. We want a set of two red and one green. And we can get this by picking the two red and then picking the one green. And since our choices for the two red can't be our choice for the one green, this is a permutation. So we'll pick the two red. There's three red balls to choose from. And we're going to select two of them. So there are three combined two or three ways to do this. We'll pick the one green. Since there's two green balls, there's only two ways to pick one green. And so there will be three times two six ways of selecting two red and one green. And previously we determined there's 120 ways of selecting three balls. And again, making clear our assumption, assuming the sets of balls are equally likely, the probability of selecting two red and one green will be six out of 120 or 0.05 or about one in 20.