Hi and welcome back. In this video we'll continue the learning objectives for Module 1. In particular, we're going to calculate probabilities of events using the axioms of probability. First, we're going to have to figure out what the axioms of probability are. What do we want to do? We want to assign a number and we're going to call that P of A, the probability of event A. This is going to give us a precise measure of what we mean by the chance that they will occur. Now, in statistics, we draw a sample from a population and we get an estimate of event A. We're going to be able to understand statistics a lot better if we understand how to calculate those probabilities from a more mathematical standpoint. Recall from video 1, we first start with an experiment that generates outcomes. Secondly, we organized all of those outcomes into a sample space S. Then we let A be some event contained in S. We want to figure out what it means to say the probability of A, the probability that event A will occur. Here are the three axioms of probability. First, an axiom is not something we prove. It's something that we assume to be true. A lot of mathematics begins with axioms, something that we assume is true, and we move on from there. What do we expect to be true about probability of A? The 1st axiom is that for any event A, the probability of A is between zero and one. It certainly doesn't make sense to have a negative probability and I don't know what a probability of 1. 25 would even mean. Having a probability between zero and one, that makes sense. Axiom 2 is that something in the sample space will occur. The probability of S, the probability of the whole sample space is one. Again, that should make sense. We're running our experiment. Our sample space consists of every outcome from that experiment. Something occurs, so the probability of S is one. Our 3rd axiom is that if A_1 A_2 through A_n are a collection of n mutually exclusive events, then the probability of the union of those events should be the sum of the individual probabilities. Using our Venn diagram, suppose we have event A_1 here, maybe event A_2 here, and so on down to event A_n. They're all mutually exclusive. That means there's no intersection between any two of them. The probability of A_1 plus the probability of A_2 plus the probability of A_3, all the way down to A_n should equal the probability of the union. I'm thinking here that the sample space has probability one. If you want to think of the probability of the sample space as being the same as area, that can help you visualize what we're talking about when we say the probability of the union is the same as the sum of the probabilities. Now we can extend this axiom three prime. Suppose we have A_1, A_2, an infinite number of mutually exclusive events, then the probability of the union of that infinite number is the sum of the probabilities. Visually what's going on? Same thing. If I think of this as A_1, A_2, and so on, forever and ever. Here's my sample space S, this probability of the union should be the sum of the probabilities. Those are our three axioms. Now, let's do an example. For this example, I want to return to the example from the previous video. I want to flip a fair coin until the first tail appears. We already discussed the sample space. I want to let An represent the event of obtaining a tail on the end flip. Probability of A1, A2, A5 and An, can be found as one half of one fourth one over two to the five and one over two to the n. Notice, that in the event of obtaining a tail on the end flip, means that it doesn't occur on any other flip. A1, A2, A3, A4, these are all mutually exclusive events. When I calculate a probability, I want to observe that the probability of S is the union of all of the individual, A's sub case. Union from k was one to infinity of Ak, that's going to equal the sum of the individual probabilities. We already calculated that on the previous slide. We get the sum from k equals one to infinity of one over two to the k. This is a geometric sum and it equals one. All right. Suppose I want to calculate the probability of B, the event that it takes at least three flips to obtain a tail. The probability of B is the probability of zero zero one, zero zero zero one, and so on. Now, observe that big compliment is the event you obtain a tail on the first, your second flip. The compliment, remember, the compliment of an event is everything in the sample space except the event B. Everything in the sample space except all these elements from B, that's just going to be one and zero one. Those are mutually exclusive. I can get a half plus a quarter. Just add the probabilities and I get three quarters here. We also know that the probability of S is B union B compliment, probability of B plus the probability of B complement, that's equal to one. I can rewrite the probability of B as 1 minus the probability of B complement, so that's 1 minus 3/4. I get 1/4 here. Let's talk about this in a little bit more generality, the consequences of the axioms of probability. There's three that I want to talk about right here. If I have A, intersect A compliment, that's going to be equal to the empty set, and A union A compliment, is all of S. I can write, here would be my A. Here would be my A compliment, my sample space S the intersection of A and A compliment is the empty set. The Union of everything is the whole entire sample space. That's going to equal one. We can separate these because they're mutually exclusive. So we end up with the probability of A compliment, is one minus the probability of A. Which is what we used in the previous problem. Also, notice the probability of A if I have A intersect, B equal the empty set, then the probability of A Intersect B is zero, and that's the probability of the empty set. Think about this A and B, and this is my sample space. Finally, I want to think about this third consequence, the probability of A Union B. This is for an arbitrary A an an arbitrary B. Let's put this A and let's say that's B. So if I take the probability of A and so let's think about that, that would be all of this space in here. That would be my probability of A. Then I take the probability of B and I want to know what the probability of A Union B is. The probability of all the shaded area there? I get the probability of A plus the probability of B, so that's all of A and all of B, but notice that you've counted this intersection area twice. You've counted it once in A and once in B. We have to subtract it, and that's what we do here. We're subtracting it once. We counted it once in A, once in B, and then we subtract it out once and we get the probability of A Union B. We're going to use this in the next problem. Let's return to the car example. Recall we had a randomly selected car, it was inspected for three defects and we already discussed our sample space. I want to consider these three events. A is going to be the event, the defect 1 is present. These are the elements from the sample space that are in A. Likewise, B is the event defect 2 is present. C, event defect 3 is present. We need some information about the probabilities of those events. Suppose over many days you collect data and you find that 20 percent of the cars have defect A. That tells us the probability of A is 0.2 and 25 percent have defect 2. That's the probability of B is 0.25. Probability of C is 0.3. Now, these numbers I made up. They're not realistic. In reality, for a true car factory, you would expect these numbers to be very, very tiny. However, because this is just an example, I made the numbers a little bit easier to work with. What else are we given here? Defects 1 and 2, so that would be A and B, that's going to be five percent. That's 0.05. Two and three, that would be B intersect C. That would be 0.075. One and three, that would be A intersect C and that's 0.06. Then finally, 1.5 percent have all three defects. That would be A intersect B, intersect C and that would be 0.015. Let's do some calculations with these events. Suppose we want to know the probability that defect 1 did not occur. The probability that defect 1 did not occur, that's going to be the probability of a compliment. We know from the consequences of the axioms that this is going to be 1 minus the probability of A. That's 1 minus 0.2, so we get 0.8 for the probability that defect one did not occur. What about the probability that at least one defect occurs? Probability at least one defect. That's going to be the probability of A union B, union C. We're going to extend our third consequence of the axioms and we're going to try and calculate this probability. We will start with the probability of A plus the probability of B plus the probability of C. I have a visual here, Venn diagram. Suppose that is our A, that's our B and that's our C, so we're taking all of A plus all of B, plus all of C. We have to subtract out all of the intersections so we subtract out the probability of A intersect B minus the probability of B intersect C minus the probability of A intersect C. Look what we've done. We've got all of A, all of B, and all of C, we've subtracted out A intersect B, B intersect C, and A intersect C. This area in here was counted once for A and once for B, and now we're subtracting it out once. The same for this intersection area and this intersection area. What about the intersection of A, B, and C? Notice that that was counted in three times here, here, and here and then subtracted out three times, once here, once here, and once there. We've lost it. We've added it in three times, we've subtracted it out three times. We need to add it back in one more time. We end up probability of A intersect B, intersect C. That's going to be the probability that at least one defect is present. We put in all the numbers from the previous slide, so that would be 0.2 plus 0.25, plus 0.3, then we start subtracting. Minus 0.05, minus 0.075, minus 0.06, then we add in 0.015. Our final answer, when we add all of that together is going to be 0.7. Third question, suppose I want to know the probability that no defect occurs. Let's look at the events that are in each of these little spaces in our Venn diagram. A occurring, is going to be 1, 0, 0, or A and B, that would be 1, 1, 0, B alone would be 0, 1, 0, B and C, C alone, and A and C, that would be 1, 0, 1. All three occurring would be 1, 1, 1, and everything else outside of A, B, and C is the event 0, 0, 0, so no defect occurs. The probability of no defect is going to be the probability of A union B, union C compliment. That's going to be 1 minus the probability of A union B, union C, and that will be 1 minus 0.7, we get 0.3. The probability of no defect occurring is 0.3. Finally, I want to look at the probability that defect 1 and 3 occur, but 2 does not. What this is asking for is the event 1, 0, 1. That would be the probability that 1 and 3 occurs, but 2 does not. From all the probabilities that we've calculated, you can observe, here in the Venn diagram, we want this piece of the Venn diagram. If I take all of intersect C, and subtract out A intersect B, intersect C, we'll get the probability of just the 1, 0, 1 event. A intersect C, that has probability 0.06. A intersect B, intersect C, is 0.015, and we end up with 0.045 for our probability. I hope this has given you some idea of what the axioms of probability are, the consequences of the axioms of probability, and how to use those to calculate probabilities for specific events that you might be interested in. We'll resume in the next video. Thank you.
