Hello again, everyone, and welcome back. In this lesson, we're going to build on our previous discussion of set theory to introduce the basic concepts of probability theory. Again, we're going to keep it relatively simple and just introduce a few fundamentals. As I mentioned, we're going to build on the concepts from set theory. If those concepts, specifically the definitions of outcomes, events, intersections, and unions are not clear to you, I recommend going back and reviewing the video on set theory once more before we begin. Otherwise, let's jump right in. Probability theory is built on three fundamental axioms related to the associated probabilities of events. To begin, let's consider an event, E. For any such event, there's an associated probability, and this probability is bounded between zero and one, where zero probability means that the event can never happen, probability of one means that the event is guaranteed to happen, and a probability in-between means that there's some chance that it occurs. For example, probability equals to 0.5 means that there's a 50 percent chance of the event occurring. Next, remember the definition of the sample space. The sample space is the set of all possible outcomes, and we denoted it with a Greek letter Omega. Our second axiom says that the probability of the sample space is equal to one. This is intuitive to understand because since the sample space is the collection of all possible outcomes and the probability equals one means that the event is guaranteed to happen, then the second axiom says that one of the possible outcomes is guaranteed to happen. Our final axiom relates to the summation of probabilities. Before we get into it, let's first define an important term relating to events. Two events are said to be mutually exclusive if their intersection is empty. That is, they are non-overlapping events or have no outcomes in common. The third axiom then says that if two events are mutually exclusive, then their probabilities will add together. Consider two events shown here, A and B. If the intersection of A and B is empty, then the probability of the union of those events, that is, the probability of either A or B occurring, is equal to the probability of A plus the probability of B. Now, let's see these axioms at work in our simple example of a dice roll. Recall that our die has six possible outcomes. Assuming a fair or non-weighted die, each of these outcomes is equally probable. That is, they all have an equal chance of occurring. So our first and second axioms together tell us that the probability of each outcome, 1, 2, 3, 4, 5, and 6, are all equal to 1/6. That is, they all have a one in six chance of occurring. Clearly, if we add these probabilities together, we get one, which according to our second axiom says that if I roll the die, I must get a number between one and six. Next, let's look at a slightly more complicated event, one that involves more than one possible outcome. Here, our event is the even numbers, which we can view as the union of outcomes 2, 4, and 6. Because these outcomes are mutually exclusive, the probability of an even number or the probability of the union of these events is simply the sum of their probabilities, according to our third axiom. This means the probability of rolling an even number is simply 1/6 plus 1/6 plus 1/6, which is equal to 1/2. Still with me? Great. Let's step things up a little bit by introducing the idea of conditional probability. If we consider two events, say E_1 and E_2, the conditional probability of event E_1 given event E_2 is the probability of E_1 given that event E_2 has occurred. That is, we've previously observed event E_2, and we now want to know what is the probability of event E_1. This is denoted by the vertical line you see in the probability, which reads probability of event E_1 given event E_2. The conditional probability E_1 given E_2 is equal to the probability of the intersection of these events divided by the probability of observing event E_2. In the following example, we'll try to give an intuitive understanding for why this relation holds. Now, let's consider our conditional probability formula. Here we're looking at the probability of E_1 given E_2. That is, the probability that we roll an even number given that the role is less than four. First, let's look at the numerator. The probability of the intersection of these events, as shown on the left, is equal to the probability of rolling a two, as we previously saw, which is equal to 1/6. Meanwhile, the probability of event E_2, shown on the right, is equal to 3/6 or 1/2. Notice that event E_2 has reduced the sample space to outcomes 1, 2, or 3. So the conditional probability is equal to the probability of rolling a two, given that we know we have rolled either a 1, a 2, or a 3. Therefore, the probability of E_1 given E_2 is the probability of rolling a two, which is 1/6, divided by the probability of rolling a 1, 2, or 3, which is 1/2. So our conditional probability is 1/6 divided by 1/2, which is equal to 1/3. This makes sense, given that we know we've either rolled a 1, 2, or 3, there is a one in three chance that we will roll a two. This concludes our introductory discussion of probability theory. In the lessons that follow, we'll continue to build on these concepts.
