 Dear students, I would like to present to you an example of computation of probabilities that cannot be found through the marginal PDFs. So let us look at this interesting situation. Suppose that the joint PDF of two continuous random variables x1 and x2 is given by small f of x1, x2 is equal to 4 x1 into x2, where x1 is going from 0 to 1 and x2 is also going from 0 to 1. Now if we want to take out this probability that x1 lies between 0 and 1 by 2 and x2 lies between 1 by 4 and 1, it is quite simple. It is quite simple. You will simply put the limits of the double integral according to what you have and you will solve it step by step and you will get it. For example, if we write it as integral 1 by 4 to 1 and then the next integral 0 to half and then f of x1, x2, 4 x1, x2. After that, you have to write very carefully that which of the dx1 and dx2 you have to write first. Since you had written 1 by 4 to 1 in the first integral, those are the limits of your x2. Therefore, you will write the most at the end, dx2. And what you had written in the second, i.e. 0 to half, are the limits of your x1. Therefore, you will write dx2 at the end, but before that you will write dx1. Because you know from calculus that if there is a double integral, then first you solve the integral inside and then you solve the integral outside. So, after that, the various steps are in front of you on the screen. And of course, you can do them yourself and doing them. What is the final answer? It is 15 by 64 and that is the probability of this particular event. Suppose instead that we are wanting to find the probability that x1 is equal to x2. Well, the probability of x1 is equal to x2 can be rewritten as the probability that x1 minus x2 is equal to 0. Now, since the random variables x1 and x2 are continuous random variables, therefore their difference is also, my dear students, a continuous random variable. This difference, this is also a continuous random variable. But the point that you have to understand here is that if we are saying that we are wanting to compute the probability that this new random variable x1 minus x2, whether you name it y or w or any other name, that this new random variable is equal to 0, so the thing to think about is that 0 is a constant. And we know by the basic definitions and all those things which are attached with it that the probability of a continuous random variable being equal to a constant is always 0. Continuous random variable, it is not discrete. Continuous random variable, it is always in an interval. So therefore, the probability of x1 minus x2 is equal to 0. In other words, the probability of x1 is equal to x2. This probability is equal to 0. This was another interesting point and now, importantly, let's take out another probability. And you will see that here it is slightly different and it is not like the very simple case when we take out a very simple distribution. So this is this one. When we are wanting to find the probability that x1 is less than x2. If we want to take this out, then how do we proceed? Students, in this case, we will write it as follows. The first integral from 0 to 1 or end we are writing dx2. That is, the entire range of the x2 variable in other words, according to our domain x2, we are taking 0 to 1. However, for the second integral, we are writing 0 to x2 and you must realize that it is alright because we also had to take out the probability of x1 is less than x2. So now this second integral is according to x1 because it was the first one according to x2. So now if we are talking about x1, then we want it to be less than x2. Which means that the integral has to go from 0 from the starting point up to x2 only. And then after that, we write down inside 4x1x2 which is the expression of our joint PDF. Alright, now that we've got all this, now we can simply solve it. First you solve the inside one and keep in mind that if we are solving it according to x1, so x2 is acting as a constant and then you do all the steps as you can also see on the screen. And once you have got that, after that you solve the outside integral which is according to x2. And doing all those steps, you obtain this particular probability equal to 1 by 2. So this is the procedure by which you are able to solve sometimes apparently somewhat complicated situations also. You just have to be a little careful as to how you put the limits to your integrals and also then solve accordingly.