 Now, we can compute expectations directly from the probabilities, but it will be convenient to have formulas for some common probability distributions. For example, suppose a fair six-sided die is rolled. Let x be the number rolled. Find the expectation of x. So the possible outcomes are 1, 2, 3, 4, 5, or 6. And since the die is fair, all outcomes have the same probability, 1, 6. And so the expectation is the sum of the products of the outcomes with their probabilities. Now, every term has a factor of 1, 6, so we'll remove that, and find our expectation. And the preceding result can be generalized very easily. Let x be a discrete random variable with equally likely outcomes x1, x2, and so on up to xn. Then the expectation of x is 1n the sum of the outcomes. It's important to remember that these are the actual outcomes, so suppose a fair die is labeled with numbers 1, 2, 4, 8, 16, and 32. If the die is rolled and the number recorded, what is the expected value? So here there are six equally likely outcomes, so we'll sum those outcomes, and then, since there are six, we'll multiply the result by 1, 6, and find our expectation. Or say a coin is flipped three times and the number of heads recorded, let's find the expected number of heads. And since the possible outcomes, the number of heads could be 0, 1, 2, or 3, we could compute the expected value as the sum of the outcomes. And since there are four outcomes, we'd multiply this by 1 fourth. While we could do this, we would be wrong. Because remember, most sample spaces do not consist of equally likely outcomes. And in fact, this is a binomial experiment. And so again, the expected value is still going to be the product of the outcomes, 0, 1, 2, and 3, times the probability of each outcome. But these are binomial probabilities, and so we need to compute, for example, the probability that we get 0 heads will be, and similarly for the others. And so the sum of the products of the outcomes with their probabilities will be, or suppose a fair coin is flipped 500 times, let x be the number of heads, and find the expected value of x. Now, since we flipped the coin 500 times, x could be 0, 1, 2, 3, all the way up to 500, we could find our expected value of x by computing the sum of the products of the outcomes with their probabilities. But that's a lot of work, so let's try and find an easier way. And as usual, the easier way exists because at some point we did the hard work of making it easy. Now, in a binomial experiment, x is the number of successes, so it will always take on the values 0, 1, 2, all the way up to n, the number of trials. So suppose our binomial experiment consists of n trials, and the probability of success on any trial is p. Then our expected value is the sum of the products of the outcomes with their probabilities. Now, since each term in the summation is multiplied by our index k, the k equals 0 term vanishes, and we could actually begin our summation at k equal to 1. Next, let's consider k with the product of this binomial coefficient. We note that k times n choose k. Well, let's write that out. That's k n factorial divided by k factorial times n minus k factorial. And if we write out our factorials, an interesting thing happens. And so we'll do two things. First of all, we can remove this common factor of k, and second we can remove this factor of n outside of our product. We'll see why that's useful in a moment. What's important is that if we do, so our numerator is n minus 1 factorial, and our denominator is k minus 1 factorial times n minus k factorial. And notice that n minus 1 minus k minus 1 is n minus k. And so this expression is n times n minus 1 choose k minus 1. And so we can rewrite our summation. Now, note that in the summation n is a constant. It's actually the number of trials. So we can remove it outside of our summation. And for reasons that will become apparent soon, we'll also remove a factor of p. Because again, p itself is also a constant. So we can remove a factor of np. And here's the important thing to recognize. Notice that the sum of the powers is n minus 1, which is the same as the number of items we're choosing from. And what that means is these terms seem like they come from the binomial expansion. So remember that binomial expansion tells us that a plus b to the nth can be rewritten as a sum of binomial coefficients times a product. And the power in our expansion is the sum of the exponents on the two factors. So if we find the n minus first power of p plus 1 minus p, we get. And this is not our expression, because this sum only goes up to n minus 1, whereas our expression goes up to n. Well actually we can reindex. So we'll reindex with i equal to k minus 1. So k is i plus 1 and our summation can be rewritten as, and this is the expression that we have. And so consequently our summation is the n minus first power of p plus 1 minus p. But that's just the n minus first power of 1, which is going to be 1. And so the expected value of our binomial experiment is going to be n times p. So for example let's say we flip a fair coin 500 times, let x be the number of heads, find the expected value of x. And so since this is a binomial experiment with 500 trials and probability of success in each trial one-half, then our theorem says the expected value is np, and so our expected value of x, 250. Thank you.