 Dear students, let me present to you and discuss with you the PMF of the discrete uniform distribution. Well, in the simple case, when the random variable x assumes values 1, 2, 3, so on up to n, the PMF is very simply written as follows. p of x is equal to 1 over n, where x is equal to 1, 2, and so on up to n, and p of x is equal to 0 elsewhere. This is the perfect way of writing the PMF of a discrete uniform distribution in the simple case which I am presenting right now. Now, after a little while, we will go a little towards its general case, but now you concentrate on this. So, let me give you an example. Suppose that we toss one fair die only one time. Throw the dice of the dice one by one. Then how can we write the PMF of this distribution? According to what I have just presented, we can see that n is equal to 6, because the six faces on top of it, the number of dots on top of them, of course, they are 1, 2, 3, 4, 5, and 6. So, therefore, n is equal to 6. So, now you put it in your formula. What do you get? You get p of x is equal to 1 by 6, where x is equal to 1, 2, 3, 4, 5, 6, and p of x is equal to 0 elsewhere. So, this is the simple case. Now, I would like to present to you a slightly more general case. Consider the following. Let x denote a discrete random variable that can assume any one of n distinct equi-spaced values from A to B. And let us suppose that the interval between them is 1. So, let me say it again. We are assuming that x, the discrete random variable x, is able to acquire any one of n distinct equi-spaced values from A to B at an interval of 1. Then, my dear students, we can say that the random variable capital X follows the uniform AB, capital U A, B inside bracket. The uniform AB distribution, if the PMF of the random variable X is given by the following. The probability that capital X is equal to small x is equal to 1 by n, where x is equal to A, A plus 1, A plus 2, and so on. And after that, we will write B minus 1, B. I am going to say it again. P of x is equal to 1 by n. X is equal to A, A plus 1, A plus 2, and so on, B minus 1, B. And along with this, we have to write where n is equal to B minus A plus 1. Of course, P of x is equal to 0 elsewhere. If n is equal to B minus A plus 1, then it is obvious that we can write the PMF as follows as well, that P of x is equal to 1 over B minus A plus 1, where x is equal to A, A plus 1, A plus 2, so on, so on, B minus 1, and B. And of course, P of x is equal to 0 elsewhere. Now, whatever I have told you, we do not need to be so confused about this. Let us take an example by which we can understand it quite easily. The first example I want to give you is by which we will get back to the simple case which I presented first. I told you that if we do a die, then 1, 2, 3, 4, 5, 6. So, in this one, you tell me what is A? What is the first value of x? Is it not 1? Right. So, if it is 1, then please note that for this example, A is equal to 1. After that, tell me what is B? Is it not 6? It is going from 1 to 6, right? So, that is correct. You write 6 at the place of B. Now, what we have just written as PMF, and you are seeing it on the screen, you put its values in it. 1 over B minus A plus 1, that is, 1 over 6 minus 1 plus 1. So, how much is that? 1 over 6, exactly the same as before. And along with that, the values of x, A, A plus 1, and so on. So, what is A? 1 plus 1 plus 1 plus 1 plus 2. And how far do we have to go? What is the last one? B. So, that is 6. What will happen before that? B minus 1, that is, 5. So, you saw exactly the same as what you had before. And after that, another example in which A is not 1, and B is not equal to n, let us also consider that it is not an issue. So, let us consider the tossing of one fair coin. In this case, what are the possible values of x? If x represents the number of heads, I think you will see very quickly that x can be either 0 or 1, because if the tail comes, then the number of heads is 0, and if the head comes, then the number of heads is 1. So, if we want to match this one with this formula, then can we say that A is equal to 0 and B is equal to 1? Of course we can, because A has to be the first one and B has to be the last one. If this is acceptable to you, my dear students, then look at that formula that you have to write for the probability. It is 1 over B minus A plus 1. So, B is equal to 1, A is equal to 0, what do I get? 1 over 1 minus 0 plus 1. So, how much is that? 1 over 2. X is equal to A, meaning x is equal to 0, and after that, A plus 1, meaning 0 plus 1, meaning 1, and that itself is equal to B. So, that's it. X is equal to 0, 1, or probability 1 by 2, exactly what it should be. So, isn't it wonderful that we have a general formula for the uniform distribution in which the values are going from A to B, and they are equidistant, and the interval is equal to 1.