 Hi, this is Dr. Don. I want to talk about the geometric distribution. This is a form of the binomial that you run into occasionally, and I wanted you to be aware of it just in case it happened to show up on a quiz or an exam. In geometric distribution, it's something that we see a lot. And the common example is how many shots will a basketball player have to take until they hit a basket? So you're repeating a trial until you get a success. It is a discrete because they're all individual, countable items. We're looking at success or failure. The trials are independent. That each shot, for example, is independent. And the probability of success P is constant on each shot. We're not really going to get into the math behind the geometric, but just remember that we're looking at X, where X is the trial where we have our first success. P is constant, that's the probability of any given shot, any given trial. And Q is 1 minus P again. Probability of X where X is the trial of the first success is equal to P times Q to the X minus 1. If we use that formula, in a case where X is equal to 3, the basketball player makes his first basket on the third shot. P is 0.3, he's got a 30% chance each time. And we just do this math. We get the probability of X equal to 3, P times Q to the 3 minus 1 is equal to 0.3 times 0.7 times 0.7, since Q would be squared gives you 0.017. Okay, I have stat crunch here and we go to stat, calculators, and geometric. And we bring up what looks like our typical calculator that we've been doing with binomial and the normal. Here we need to enter our P which is 0.3. We want to look at the probability where X equal to 3 where the success occurs on the third shot. We put that in, click compute, and we get a probability of 0.109. But wait, that's not what we got. No, we got a probability of success on the third shot of 0.147. So what's going on? After doing a lot of digging, I found out that there's an alternate form of the geometric distribution formula. And in that alternate form, we're looking at the number of failures until the first success where Y is equal to X minus 1. If X is equal to 3, the first success occurs on trial 3, then Y is equal to 2. And the probability of Y is equal to P times Q to the Y. And that makes sense. You need to enter not the number of the trial with the first success, but the number of failures before you hit success. So if I put P equal to 2 in there, and it compute, I get the 0.147 that we're looking for.