 Okay, let's talk a little bit more about Bayes' theorem. So again, we can get back to Bayes, and we're going to start off with our 2 by 2 contingency table, and again, we have various probabilities. At this time, we're going to consider the probability just of the A given that B has occurred. And so, again, we know that this is by the frequentist interpretation how many times A has occurred given that we know that B has occurred. So we know that B has occurred, we're in this A plus B set, and then A has occurred A of those times, and our frequentist interpretation tells us that this is the conditional probability. Now, once again, we would like to have probabilities on both sides, and so, let's consider this A. This is the number of times the event has occurred, and on the one hand, I can view this as the probability A over A plus B, but if I look at A in isolation as a probability, if I divide by the total number of times that our event has had to occur, then I get the following quotient. Now, let's consider the two pieces separately. The denominator here, A plus B over A plus B plus C plus D. Well, A plus B is the number of times that B has occurred. So this denominator here is the probability that B has occurred given that we know nothing else. We don't know whether or not A has occurred. So this is, it's occurred A plus B times at the total of A plus B plus C plus D times. So this denominator is just the probability that B has occurred. If I look at the numerator, this is the number of times that both A and B have occurred, but it's out of a total of A plus B plus C plus D times. So I don't know anything else other than both events have occurred, so I can express that as the probability of both A and B. And what this quotient gives us is another form of Bayes' theorem. Now, something that's probably useful to point out, this formula is not very useful in practice. It's a lot like the formula for the probability of non-mutually exclusive events. The number of times we'd actually use the formula as it's written are fairly uncommon because, quite frankly, the information we need is hard to come by. However, the formula itself, because it relates the conditional to the unconditional probabilities, is very useful from a theoretical point of view. And that particular importance comes when we start to look about dependent and independent events. So let's consider a pair of independent events. So first off, we have Bayes' theorem, which applies whether or not the events are dependent or independent. We have the relationship the probability of A given B is the same as the probability of both A and B divided by the probability of B. However, there's something useful that happens. If the events are in fact independent, then I know that the probability of A given B is the same as the probability of A by itself. In other words, knowing that B has occurred leads to no change in the estimate of our probability. So this probability of A given B is just the probability of A. And now, again, nobody likes fractions, so I'll multiply across. And that tells me the probability of A times the probability of B is the probability of A and B both occurring. And so what that tells us is that if I have independent events, then the probability that they both occur is just the product of the probabilities that each occurs individually. All right, so let's take a look at another problem. So suppose the probability of drawing an ace is one-thirteenth, and the probability of rolling an even number on a six-sided die is one-half. Again, we're not going to worry too much about where these probabilities come from, at least not yet, but assume that they are found, obtained, determined, somehow we get these probabilities. And so now I'm going to do two things. I'm going to draw a card and I'm going to roll a die. And so I want to know the probability that I both draw an ace and also roll an even number. So A is the event that I draw an ace and B is the event that I roll an even number. And the first thing we do want to check is to see whether or not A and B are independent. So let's think about that, the probability of A given B, the probability that I draw an ace, given that I know that I've rolled an even number. All right, well we know, I'll roll the die, it's an even number, and you probably don't change your confidence or not that I've drawn an ace, knowing that I've rolled an even number wouldn't seem to give you any useful information about whether or not I drew an ace. So given that the probability of A given B is the same as the probability of A, I can conclude that the events are at least reasonably independent. And so independence allows me to use the formula for the probability of independent events, the probability that both occurs is just the product of the two probabilities. And I know the one probability is 1 13th and the other probability is 1 1⁄2, so my probability is going to be the product 1 26th.