 Any questions on sample space, events, a probability or what we called a mutually exclusive, anything like these are the basic operations, you should be clear about how to do them, ok. Now let us move on. So in axioms of this probability, just assume these three basic things that probability of the event should be non-negative, probability of the entire sample space should be 1 and there should be if I take mutually exclusive event, finite number of mutually exclusive event and if I take their probability that should be nothing but probability of this individual event. And this can be extended also to uncountably many, we will come back to that, but for time being let us focus on the finite activity. These three axioms itself, they say lot of other they imply lot of other properties, ok. The first property is going to be, so since I said p of is equals to greater than 0 and p of omega, what do you expect? Can any, let us take another f which is a subset of omega, can f p of f what do you expect this to be in terms of the value, in what value it should be? It has to be expected to be 0 to 1, but I did not assume this, I did not assume this here, but this three assumption itself imply that this is indeed true and you can work out that, ok. I will just, so only based on our non-negativity, normalization and finite identity these three properties you can argue that probability of any event is going to be less than or equals to 1, ok. So, for that maybe just let us quickly go through that, let us take, let us first take any two events, one is b and I have another event a, here is event a is a subset of b, it is right. Now, if this is the case naturally I expect probability of a to be less than or equals to probability of b, right this is intuitive, but I did not in the axiom I did not assume this, but what we are now going to assume that the three axioms already also imply this, so how is that? Now, what we can do is this event b, I can write it in two parts, ok let me call this the one with horizontal lines. So, the one in the vertical lines, right that portion, that portion can I write it as b minus a, so this is what this one in the vertical I am going to call it as b minus a, that is I remove all the elements of a from b and whatever remains I am going to denote it as b minus a. So, now b can be written as union of these two things, a union b subtract a, let us call this simply c that is a union c, so this portion is what I am calling it as c, so is a and c mutually exclusive? Yes, right there is no overlap between them between now, but then from my now can I say that probability of b is nothing, but probability of a plus probability of b minus a, why is that? The third property mutually exclusive property that I have applied and if that is the case, I am done right, I am I am saying that this I have already proved, why is that? This quantity here is another event b minus a is also subset of omega, right. Now, we know that from the first axiom this has to be greater than or equals to 0, if this has to be greater than or equals to 0 then it must be the case that pb has to be greater than or equals to e of a, ok, so fine. Now, the axiom said how the probability of each event should look, now does this using that? So, if you again look into this carefully, I can treat probability as a function on my events, like I said f is subset all subsets and now p is a function on this which is going to give a value of 0 1 between them. So, probability is actually function on the events, for every possible event it is assigning a value in the interval 0 1, but now if you have events, you may take union of events let us say if you have two events and you have union of events e union f and you want to see that how this probability of the union is related to the probability of the individual events e and f, ok. So, what we are trying to do here is we have one event and another event and let us call this as left as a and this as b and now intersection is the entire thing. Now, probability of e union f 1, so what when you want to compute probability of e union of what you are going to do is, ok, you want to take this much and also the likelihood of this much, so when you did this you have double counted this region here. So, in this naturally gives me intuition that the union of when I take probability of the union of this it should be equals to probability of e probability of f minus this portion which I double counted, ok, this is the relation and in fact, this relation is actually implied by our three axioms, it is not that I am making this heuristic argument here like the probability of e union f is equals to this intuitively it should be like this, but actually this is what our three axioms also said and these are the steps based on which we can use the axioms and find it, ok, I will not go into the details, but yeah just using this axioms 1, 2, 3 you will be able to derive this and you can extend this to more than two subsets, when you have let us say three subsets, let us say when you have this three subsets and they can be arbitrary, there can be overlap in any way, I mean this the three circles they may not overlap at all or maybe only two overlap and they and the third may not overlap with the other two, any possibilities there, but let us consider this a case where all of them are overlapping with each other, where this is the common region among all. You can again verify that this probability of this intersect union of these three can be written as probability of these three each of the circles, from that you are going to take off that is common between e and f and that is common between e and g and that is common between f and g, but when you are taking out you would have taken common between e of g one extra time, so you need to bring that back and you will get this relation and again this is actually implied by our three axioms, like to prove this you will you will first prove this property and then extend it to the three subset case, ok. Now, I was talking about finite additivity right why we need to worry about the finite additivity case, like when I talked about these three axioms, the last axiom I restricted to finite additivity, but in practice I need to extend this to include countable events also. And now to look into that let us see what you want to do is, let us first see that where I may need a countable additivity. Let us take omega my sample space where omega is any possible integer 1, 2, 3 like that and I am assigning the probability that ith number happening is 1 upon 2 to the power i and this is true for all i belongs to omega. Now, I am interested in finding what is the probability that even number happens. So, notice that here I have given you probabilities of each of the outcomes. Let us see what I am saying is p of i is equals to 1 upon 2 i for i 1, 2, 3 like this. Now, what I am asking is I am interested in event e which is even numbers which is nothing but 2, 4, 6 like this. Now, I want to find the probability of e. How I will find probability of e? Probability of e can be thought as probability of 2 happening, 4 happening, 6 happening, but event 2 happening and 4 happening if I am going to treat them as separately, event 2 happening and event 4 happening. Are they mutually exclusive? They are right. Then I can treat each of this break this event as mutually exclusive event and then I can use those probabilities to add up to get this. But now, how many times I need to add? Let us go back to this. So, first of all whenever I say something is a probability you need to verify that like that axioms at least it follows the 3 axioms. The axioms are saying that it should be for every i it should be greater than or equals to 0 which is absolutely obvious here. And if I am going to take p of omega. So, the second axiom said this should be equals to 1 and is it 1 in this case? Like omega is nothing, but 1, 2, 3, 4 all the up to infinity right. You just add all these probabilities. You will get 1 here. And now, I am interested in this event and this event could be thought of 2, 4 I can write it as union of events. Let me write it as event 1. Let me write it, it will have to write. Event I am going to write it as event 1 is 2, event 2 is 4 like this I will write ok. Now, E is nothing, but union of these events. And now, I know that Eis are mutually exclusive. So, Eis are mutually exclusive. Now, let us say if mutually exclusive and then what I need is probability of E, I know that when if this union is over finite not a finite union. This is a countably many terms are there. So, if I have to do that I need to do this. I need to have to do this or like P of E is nothing, but P of union of EI. And if this P of this has to be equals to this where I need to add in countably many. So, I need to allow myself flexibility not just add finite mutually exclusive in it even add countably many mutually exclusive event and that is where I need to extend my x third axiom. And now, we are going to say that if you have a sequence of a mutual events E1, E1, E2 like this then they are all defined on the same sample space omega. Then probability of the union of those mutually exclusive events is nothing, but sum of all of them. So, that is nothing, but summation of E of I. So, this is what this is how we are extending the finite additivity of the mutually exclusive event to the countably many case. But now, the question is this clear to all of you, but now finite count a finite additivity and uncountably additivity is fine, but what about the uncountable case?