 Hi, I'm Zor. Welcome to Unizor Education. I would like to have some very simple problems to consider the basic properties and basic definitions of probabilities. It's just illustrative problems. They're very easy. The purpose is to get acquainted with some very basic fundamental characteristics of probability and events and sample space, etc. What we will do is we will do it in the form of very simple problems which I present as everything else actually as part of the advanced mathematics course on Unizor.com. I do suggest you actually to go to the website, this one, and try to consider problems yourself first because they're really very very simple. It's the definitions of probability problems number three. I will consider it right now and let me just start from probably the simplest one. Let's consider you have a certain sample space which contains certain elementary events and out from these elementary events we have composed two different events X and Y. Now what I'm making next is an event which is the union of these. It's OR basically. So the condition is event X or event Y happens. Now this event has a probability and I would like to prove that this probability is less than or equal to some of their probabilities. Actually if you remember the previous lecture which was about event arithmetic contained the material when this was actually exact equality. That's in the case when these events X and Y are mutually exclusive. So certain number of elementary events went into X and certain other elementary events completely different ones went to event Y. In which case the measure of the OR between them actually includes all the elements of here plus all the elements of there and there are no common elements. Since there was no common elements there was an equality because the measure of their union was equal to the sum of their measures. But in general case this is less than or equal. Now why? Again if you remember in this previous lecture I actually presented the following formula. So you have to subtract the common elements which if I'm counting these elements and these elements then the common elements I'm counting twice. And that's why I have to subtract it once. So intersection between X and Y is basically the common elements and since I have added them into this and into that I have to subtract this to get the real measure. Now this is the probability, it's measure, it's always non-negative which means that if I will not use this I will have a greater, I will have a greater value. So P of X plus P of Y together are greater than P of X plus P of Y minus something, right? Because this something is either zero or positive, it's a measure, it's a probability. That's why we have this particular inequality. Now this is a simple problem as I was saying and it's basically a direct consequence from the material which I presented in the previous lecture about events arithmetic. Now another very simple consequence of the same material is the following. If you take any event A and event which is not A, I will put the vertical bar above A. That means it's not A. So the sum of their probabilities is equal to 1 always. Why? For a very simple reason. This is the sample space and this is certain event A, it has certain elementary events, right? Event not A includes all elements outside of this, right? So if I will add the measure of this plus measure of all outside and they obviously are mutually exclusive, so that's why we have the following equality. Since they are mutually exclusive, A and not A obviously don't have any common elements. So the sum of their probabilities is the probability of their union and what is their union? Well, this is the probability of entire sample space which I call omega, entire sample space. And this is by definition equals to 1. Okay, another simple thing. As I was saying, these are just illustrative examples of the material which I have presented in the previous lectures. Now another one. X and Y are two events. Now what's known about them is the following. Probability of their intersection is equal to the probability of their union. That's kind of an unusual formula, right? Now if you have pictures, so this is one event X and this is event Y. So this is their intersection and this is their union. And I'm saying that the measure of these are the same. What does it mean? Well, it means that X and Y are almost the same because if X and Y is exactly the same thing, as you see their intersection is the same thing and their union is the same thing and that's why they are actually identical. So I have to prove that from this follows that X is equal to Y but I will put the P on the top. Now P stands for probability. Now what it means is that they are almost the same and if they are not the same, if there are certain elements, elementary events which are in this but not in that or in that but not in this, these elementary events have measure of zero. By the way, it's not some kind of a rule that all elementary events inside the space, inside the sample space have to have the same measure equal to basically one over number of elements. Well, this is how we are conducting certain experiments which we call random. Well, there are some cases, for instance, let me ask you the following question. What's the probability of rolling a dice and getting an event seven? Well, we know that there are only six numbers which can be the result of the rolling of the dice, right? So what's the probability of event that the number is equal to seven? Well, zero. There is zero probability. So if I will establish a new sample space, the numbers from one to ten, which can happen as a result of rolling a dice, then numbers from one to six will get the probability of one six but numbers seven, eight, nine and ten will get the probability of zero. So there is nothing wrong with having a sample space greater than the number of practically occurring situations. So this is exactly the case. So if they are not equal, then it means that they are different only by those elementary events which have the probability of zero. That's what it means. So anyway, I have to prove it somehow. So how can I prove it? Well, let's just start with something which we know, let's say. P of x and y, sorry, or y is equal to P of x plus P of y minus P of their intersection, right? Now, I know on the other hand that the union has the same probability as intersection. So this is the same as P of x, y. So what do we have here now? Well, P of this intersection is equal to this. So two P of x intersection, y is equal to P of x plus P of y. Now, let's just think about it. So instead of having two, I will have P of x plus y of x and y plus, the same thing, right? Now, let's just think about it. This event is definitely smaller as far as the number of elementary events which compose it than this one, right? P of x and this is x which is intersected with y, which is obviously less than x. It's only those elements of x which are also elements of y and this is only those elements of x. So I know that P of x is definitely greater than or equal to P of x and y, right? Because this is number of elementary events is equal or smaller. And the quality is only if they are, if x and x and y are the same. So the difference between x and x, y and x and y is equal to zero by probability, right? So only those elements from x which do not also belong to y are here and not there. And if I want an equality, I have to have these elements which are from x but not from y, they are supposed to be, to have the probability of zero. Now same thing here, probability of y again is greater than the probability of x and y. This is greater than this. So this is greater or equal to this and this is greater or equal to this. So the whole thing is greater than or equal and the only quality can be achieved if in both cases elements of x which are not in their intersection are having the probability of zero and elements of y which are not in common with x supposed to have the measurement of zero, the probability of zero. So basically we have that the whole probability is concentrated in their common place. This is x and y. And whatever is outside of this common place but in x, let's say this is x and this is y. So these are supposed to be, to have the probability of zero and these are supposed to have a probability of zero. Only in this case we will have equality here and equality here and equality here. That's the only way. So that's how we basically came up with this idea that if there is a difference between x and x and y, so if there are any elements of x which are not common with y, they have to have the probability of zero. And if there are any elements of y which are not common with x, they also should have the probability of zero. That's what I meant that x and y are almost identical. They are identical with certain precision of elements which are not measured to any significant probability which has the probability of zero. Okay. By the way, this is an important consideration. You do have in certain cases elements of probability zero. I was just giving you an example of rolling a dice but having the numbers from one to ten as possible events. But only from one to six have a real probability associated with them, equal to one-six each. And numbers from seven to ten have the probability of zero associated. It's impossible event, right? So these events x and y are identical to the precision of impossibility, if I can say so. Okay. Then I would like to have, like, exercise about what's elementary events, what's events, and what's an operation with events. The simplest possible experiment which I can conduct is flip the coins. Flip coins. Two times. Coin two times. So first of all I would like to characterize the elementary events. What are elementary events if I'm flipping two coins? Well, obviously it can be head-head, it can be head-tails, can be tail-head or tail-tail. So I have four different outcomes if I'm flipping two coins, or one coin two times, doesn't matter. Now, obviously if it's the symmetrical coin, I have four equally probable outcomes of this experiment. By the way, one coin is completely independent from another. That's why I can assign absolutely equal probabilities to all of them. And obviously these probabilities are one-quarter each. Now, next. Next I would like to characterize in this language the certain events and their probability. Now, event number one says no tails. Now, what elementary events are in this event, e1, which is no tails? Well, this, this and this, they're all having tails. So the only one event which has no tails is this one, it's only hh. Only one event and the probability is equal to one-fourth. Next, only one head, only one head. Okay, only one head means this is not good, this is good and this is good. This has two and this has zero heads, so only these two. So e2 is equal to ht and th. These two elementary events constitute an event e2, which is only one head. And the probability is one-quarter plus one-quarter, which is one-half. Right? Probability of any event is a sum of the probabilities of elementary events, which are included. e3. Okay. Number of heads not equal to number of tails. Okay. Well, number of heads is two, number of tails is zero. So this is good event, right? Now here it's one and one. They are equal, which is not good. Also one and one. And this is, again, zero heads and two tails. That corresponds to this condition. So these two events correspond to the two elementary events are together composing an event e3. Probability is one-quarter plus one-quarter, which is one-half. And the fourth event, which I'm considering, is number of heads equal to number of tails. Now, obviously, these are ht and th, and the probability is also one-half. So these are four events, which I am considering right now. Next. Next is what's the e1 or e2? Now, you remember that whenever we are talking about or condition, it means that all elementary events, which are either in this one or in that one, are supposed to be included, right? So this is, in this one is hh, and e2 is ht, th. That's the event, which is e1 or e2, which means either no tails or only one head. These events, which I'm interested in. The probability is obviously equal to one-three-quarters, right? Next. e3 or e4. e3 or e4. e3, e4. So it's basically hh, tt, ht, and th. And this is the probability is equal to one, because these are all the elementary events which are, in theory, possible, all four of them. By the way, e3 and e4 can be characterized in a different way. Basically, e3 is a negation of e4, or e4 is a negation of e3, because they're complementing each other to a complete set. Next. e2 and e4. Now, and means we're talking about common elementary events, right? e2 and e4. This is e2, this is e4. And as you see, they are identical. So that's quite interesting. The event, only one head, is equivalent, is identical, basically, to the event when the number of heads is equal to the number of tails. So the probability of this, well, first of all, what's included, ht and th, and the probability is one-half, quarter plus quarter. Next is e3 and, no, e1, sorry, e1 and e3. e1 is double h, e3 is hh and tt. hh is in common. So that's the only result of the end condition between e1 and e3. Common elements are only hh. So that's the result of this end condition and the probability is one-fourth. And the last couple of things. Not e2. The complement, the opposite to e2. Well, let's just think what is opposite if you have a total number of elementary events. And e2 is these two elementary events. And what's left from the total, if you subtract these, would be the negation. So the result would be hh and tt, right? Out of four elementary events, if we are trying to complement these two events, that's what will be there, right? And the probability is obviously one-half. So that's actually not e4. e4 is this. So the opposite is the same thing, hh and tt. And that's not a coincidence because e2 and e4 are exactly the same. I just expressed these events differently. e2 means only one head, e4 means number of heads and number of tails are the same. But it's exactly the same event. The same elementary events are going into it. So the opposite to e4, the same as opposite to e2, is this set of two elementary events with a probability of one-half. Well, these are all illustrative, very simple exercises with elementary events for one particular case. I mean, there are millions of different examples which are, in some way, analogous to this one, which we can present if we have time anyway. So what I suggest you to do right now is go to the website, go to theunisor.com to this particular lecture. It's problems number three among the formal definition of probabilities. And try to do it yourself again, just in a piece of paper, and see if you can do it. That would be a very good exercise. Other than that, that's it for today. Thank you very much and good luck.