 Hi, I'm Zor. Welcome to Unizor Education. This lecture is part of the advanced mathematics course for teenagers presented on Unizor.com. And that's where I suggest you to listen to this lecture from, because the lecture contains notes. And notes might actually explain certain things slightly differently than what I'm talking about. Basically, it's about the same. But anyway, there might be some details. Or maybe I'm omitting something during my presentation now, but I include it into the notes. So it's important that you read the notes. And actually, this is the problem-solving lecture, which means that I do recommend you first go to the website, go to the notes and try to solve these problems just by yourself. There are answers. There are three problems and there are answers in the notes. So before you're reading the solutions or listening to this lecture, I think it's very important for you to try to solve these problems just by yourself. Okay, this is the third set of problems on conditional probabilities. And let me just start one by one. Okay, the problem number one. You have two dice and I'm interested in the event sum equals to seven of both dice. Sum of the numbers in both dice equals to seven. Under condition that the number one dice shows the number which is greater than three. So we are interested in the conditional probability of the sum of the two numbers and two dice to be equal to seven. Under condition that the first dice shows a number greater than three, which is four, five or six. Okay, it's very, very important in all these cases. Well, in probabilities in general, but conditional probabilities especially, it's important to understand the graphics of these probabilities. So basically we are looking at the theory of probabilities from the measure theory standpoint. So I have an experiment with two dice, each one of them from one to six. So I have 36 different elementary events. And as usually I put them graphically into the table of six by six, basically signifying this is number one and this is number two. So it's one, two, three, four, five, six, one, two, three, four, five, six. And intersection of the column and the row basically means the row for instance this one is four and that's the first dice and the column let's say it's five, that's the second dice. So now let's graphically represent the events which we are interested in. First of all, we are interested in the event that sum is equal to seven. So let's just forget about condition. We are interested only in the event and we'll call it a. So the sum is equal to seven. Now, what are these? Well, the first one is one and the second one is six, right? So it's this one, or two, five, this one, or three, four, or four, three, or five, two, or six, one. So these are events, elementary events, which comprise together an event that the sum is equal to seven. Okay, now what are the events when the first dice has greater than three? So the first dice is the row. So it's this, this, and this. So events are this, this, this, this, this, this, this, all these which have the row greater than three, which is four, five, and six. Well, as you see we have one, two, three on the intersection which belongs to both. But let's forget about this for a while. Let's talk again about, I don't want really to, to apply formula, with the formula it's really solved in one second. But I would like actually to, to go through some understanding of what conditional probability actually is. Now, initially all these squares have the same chances to occur, which means there is a probability of 136 in each one of those, right? Now, what happens if I'm saying that the condition of my experiment is that the first dice shows greater than three? It actually means that the probabilities from being evenly distributed among 36 different occurrences now are concentrated only in this particular event, which means wherever, wherever, wherever I have an X in this, in this table. So, instead of everything being equally distributed among these squares, 136 each, now I have all the probabilities distributed evenly among only these 18, right? Three by six, only 18. So, now the unconditional probabilities of all squares was 36 squares by 136 each, right? So, it's one. Now, after the condition is applied, my probability is 18 are zero plus 18 are one-eighteenths, also one. So, I'm shifting whatever is not really happening. I know that the condition is true, right? Which means that none of these is happening. So, I basically reassigned, I redistributed the probabilities from all of these events evenly distributed to only those events which I know are actually occurring. And these which do not occur have zero probabilities. So, I have 18 zeros and 18 one-eighteenths. So, now all the probabilities here. Okay, fine. Now, what's the probability of this event now? This is some of this, this, this, this, this and this. Now, initially, if I do not have any conditions, I have six zeros here, right? And the probability of each was 136. So, I had 36 probability of this occurrence, right? But not anymore since my condition is applied. Now, these elementary events, these three zeros, have probability of zero. And these three zeros have probability of one-eighteenths, right? So, in theory, now the probability of this event under a different distribution of probabilities is actually three-eighteenths. That's how it's supposed to be, like, logically presented. We can actually formalize it and say something differently. Because if you actually pay attention to this, what is the probability of these three events which are both x and 0 in this case? Well, the probability is a relative, well, measure of whatever event we are interested in, relatively to entire probability, right? So, entire probability is concentrated here. So, these are number of events which are really occurring. And we know that their number is 18. And these are events which are occurring among them. And that's why it's three-eighteenths. So, it's a relative measure of these out of those. And if we are talking about evenly distributed probabilities, it's obviously the same as... Now, what are these three? It's basically an intersection of this event with this event, right? So, that's why the formula of conditional probability looks like this. So, the conditional probability is the area or a measure of the intersection of a and b. a is this, b is this, which is these three things, divided by the area of the condition, which is this. So, basically that's an explanation of what conditional probability is all about. And let me just go formally and use this formula to derive this particular value, three-eighteenths or one-sixth. The probability of b, it's probability of my condition. Now, this is probability of everything which has all the different pairs where the first is four or five or six. And the second was anything. These are all pairs of dices which have the first greater than three. Now, there are six of these, six of these and six of these, right? Because this second dice can be anything from one to six. So, it's eighteen. So, that's why we have eighteen. So, the probability of b is equal to eighteen thirty-six. We're talking about initial probability, which is basically a reflection of the measure. Now, the probability of this particular event, which is a, intersected with b, is... So, if it's four and we need the sum to be equal to seven, then it's only four-three. If it's five, it's only five-two. And if it's six, it's only six-one. So, we have only three different elementary events, which are on intersection of both. And these are, you see, four-three, five-two and six-one. So, the probability of a intersected with b is equal to three-thirty-six. And finally, if you divide one by another, if you divide three-thirty-six over eighteen-thirty-six, you have three-eighteenths, which is one-six, the same as we used to have. So, I was trying to explain things graphically first and then using the formula. So, that's how you probably should think about this. I mean, obviously, you can go directly to a formula, but I still would like you to pay more attention to the graphical representation, because it actually represents what conditional probability is all about. It's a relative measure of whatever good things which we are interested in, which means basically both, relative to the condition, the measure of the condition, which is this one. All right. It's such a simple problem and I spent so much time, just because I would like you to understand completely what the conditional probability is all about. So, if you understand it quite well, then just skip all these graphical representation and you can just go to a formula and derive the results, no big deal. All right. Number two. Very similar. I have again two dice. Now, I would like to know the probability of the second dice to have greater than three value under condition that the first dice is not the same as the second. So, they are different. Okay? And again, let me go first to graphical representation of this. This is my first dice. This is my second dice. Okay. I need the second dice to be greater than three, which is four, five, or six. So, it's these and doesn't matter what's the first dice. That's an event I'm interested in. Now, what's the condition? The first not equal to the second. Well, basically, it's everything except when the first is equal to second, right? So, all these outside of the main diagonal are my... So, except the main diagonal when the first is equal to the second. So, everything. So, that's my condition. X is my condition. Zero is the event I'm interested in. Let's again try to think about this from the redistribution of the probabilities. So, before probability was equal one over 36 to each square. Now, after I apply the condition, my probability is shifted to all the axis, outside of the main diagonal. And there are 36 minus 6, there are 30 x's here. So, instead of being distributed among 36 squares equally, now my probability is distributed only among 30 squares. Now, right? Yeah, 30 squares marked with x. Now, out of these 30 squares marked with x, how many of them are actually satisfying this condition? Well, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, right? So, out of them 15 satisfy this condition. So, if my measure, which used to be distributed evenly among 36 squares, and now it's distributed evenly among 30 squares, and only 15 of them are basically satisfying my condition, then the probability must be 1 over 2, right? 1530s. That's the idea. Well, let's try and check if it works from the formula base. Again, we have this formula for conditional probability, intersections b divided by probability of b. Alright. So, let's just have both of them. A intersection and b. So, the second number is greater than 3, and they're not equal. Well, basically that's exactly what my x and 0 are. 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, out of 36, right? So, conditional probability, sorry, the intersection is equal to 1536. Now, what's the condition b? Well, that's all the x's, right? And there are 30 of them. We can count it. So, the probability of my condition is 3036. And if I divide one by another, which is 1536, by 3036, well, I'm getting 1530s, which is 1 half. So, this is just a pictorial representation of basically the same thing. Because all these squares where x and 0's are in the same square, this is basically the intersection, right? And the b is just x's everywhere. So, we basically divide the number of intersection, number of squares in the intersection by the number of squares, which corresponds to the entire condition. Okay, fine. Forget about dice. Let's talk about something which is much closer to students' life. Let's talk about exams. So, a student has 100 questions to be studied for an exam. On the exam, he will get only one question, and he should answer it correctly or incorrectly. Well, let's just assume that if the student studies the question, then he answers correctly. If he doesn't, he doesn't, right? So, we have 100 questions, but students doesn't have time to study all of them. So, he studies only 70. Okay. If you just don't know anything about anything else, your intuitive understanding of the probability of passing the exam is 70 over 100, which is 0.7, right? So, that's what we probably shouldn't get. This is something like unconditional probability, well, provided that certain things are relatively randomly distributed, etc. Okay. So, let's go to a concrete problem. A concrete problem is as follows. The professor decided to divide 100 questions into two groups and assign students to two different groups. So, the questions from number 1 to number 50 are at group 1, and group 2 would have questions from number 51 to number 100. So, now let's just think about it. Student has started only 71st. I probably didn't say the 71st questions from numbers from 1 to number 70. So, let's just think about it. If he happens to be in the first group, where the first 50 questions are assigned, are chosen from, basically, for the group, then he has perfect chances to pass an exam, because he has learned all 71st questions, and only the 51st questions are going to the first group. So, that's good. So, if he falls into the group number 1, that's great. If he falls into the group number 2, well, it has 50 questions, but only the first 20 of them, from number 51 to number 70, he has learned. So, it's easy. He might pass or might not, depending on what exactly happens. Okay. So, that's everything about the problem. Now, the question. Well, I actually have three questions. Question number 1, what's the probability of passing if he falls into group number 1? B, what's the probability of passing if he falls into group number 2? C, what is a total probability of passing? And we will talk about total probability. We did talk about this before, and we'll talk again. Okay. So, all right. So, let's just think about it. First of all, we assume that his selection into group number 1 or group number 2 is completely random. So, the probability of getting into group number 1 is the same as probability of getting into group number 2, which is 1 half. That's a reasonable assumption. Okay. So, how can I calculate this one? Well, that's easy. Again, let's just use this formula. So, the conditional probability of passing if he is in a group 1 is equal to a probability of passing considering he is already in the group 1. So, he is chosen into the group number 1. And then we can calculate the probability of passing. And this is obviously equal to 1, right? Because in the group 1 there are only first 50 questions. He knows first 70, so it's much more than needed. So, that probability is equal to 1. And that's probability we know it's equal to 1 half. Sorry, 1 half. So, all together it's supposed to be 1. So, the conditional probability of passing exam if he is chosen under condition that he is chosen into group number 1 is equal to 1, right? So, all questions are studied by him. So, that's why this probability is basically equal to probability of g1, which is 1, 1 half. Now, in this case, the situation is different. It's different. In this case, well, in the bottom you also have 1 half. But what is the probability of passing exam and to be selected in group number 2? Well, out of 100 questions, 50 are selected into group number 2, right? And out of these 50, he knows 20 and the other, he doesn't know, right? So, that's supposed to be PG2. Okay, so what's here? Again, if you will just divide the entire set of questions into these two groups, you have 20, right? 20 out of 50, he knows. So, let me just think about it. That's supposed to be, let me just draw it again. So, if you have 100 questions, right? So, these are two groups. So, these are 50 questions and these are 50 questions. This is group 1, this is group 2, right? And he knows these, all these, and part of these. So, the intersection is this, right? So, it's 20 and this is 50. So, he knows 20 out of 50, right? So, it's 0.4, am I right? Hold on, one second. The area is equal to, the area of G2 is equal to, let's do it differently, let's do it 20 out of 50. That's how it's supposed to be, right? So, this is better to be specified as 50, 100 out of 50, 100. And this is 20, 100 over 50, 100. That's the probabilities. All of these, these are for the first case. All of these, there are 50 of them. So, the probability of the intersection is 50, 100, which is 1 half, and the probability of the G1 is exactly 50 of 100, and that's why we have 1. Now, in the second case, we have the probability of only those which he knows, which is 20 out of 100. So, it's 2100 over 5100. All right, so, we've got this as 1 and this is 0.4, right? Okay, that's my first two questions. So, if he goes into group number 1, he gets the probability 1 of passing, and if he goes to the second group, it's probability 0.4, which is understandably, right? So, you have, you know everything, that's why if you go into the group number 1, you know everything, so that's why it's probability of 1. And here, you know basically 20 out of 50, which is 0.4, which seems to be reasonable, right? Okay, but now I'm asking about the general probability. General probability, it feels supposed to be this, right? Because he knows 70 out of 100. But how can we calculate, based on whatever these calculations are? Well, let me just reiterate what the total probability is. So, if you have a sample space and somehow it's divided into two events, x and y, they are mutually exclusive and some of these two events cover the entire sample space. Now, in our case, g1 can serve as x and g2 can serve as y, right? Only two groups and he falls into one of these groups. They are mutually exclusive and the sum of these covers basically an entire random experiment. Okay, now then, any event a, whatever I'm interested in a, can be represented as a intersection with x and a intersection with y, right? A intersection with x is this piece, a intersection with y is this piece. Now, I'm using plus here instead of a union. Basically in the set theory, I should use the union, but I think we have agreed in some other lecture that if these events are mutually exclusive or using the set theory terminology, if these two subsets are non-intersecting, then I'll just use the plus here. And the reason for this is that the measure, you can consider measure as an area, for instance, of the a. It's sum of these two areas, right? And this is a real plus. So, this plus is some kind of invented plus. It's not an additional real. It's a union. I'm just using the plus, but then using the plus because of this. Because if I'm going into numbers with measures, then the measure of a is equal to sum of these measures. And the probability is a measure basically because it's just a relative area of this divided by the entire area of the sample space, which is, you know, assumed to be equal to one. That's why it's basically a measure. And now I can use this. It's equal to. Now, remember from the formula of the conditional probability, which we have many times wrote before. The area of the measure of intersection is equal to measure of the condition times measure of the event under condition of that condition, right? So, that's p of x times p of a conditional x plus p of y times probability of a under condition one. And now, we know all these components considering that g1 is x and g2 is y. So, the probability of g1 is equal to one-half, right? That's the probability of falling into the first group. So, it's one-half times. And this probability is equal to one plus another one-half, that's probability of y, times conditional probability of 0.4 equals two, zero-five plus zero-two equals zero-seven. And that's exactly what we expected. It's some kind of a checking, if you wish. All right? So, this is, by the way, a formula of total probability, which we talked about before. And I just repeated it again. It's a very simple one, and it's related to the fact that x and y are non-intersecting and covering an entire sample space. That's why we can divide every event A as a union of two non-intersecting events, all right? Well, that was my last problem. Thanks very much for listening to me. And I certainly recommend you to go to the notes for this lecture again and try to use this graphical representation, which I was trying to use a couple of times for every probability problem of that type, especially with conditional probabilities. It's very, very useful. So, again, try to solve these problems yourself. And by the way, if you're a registered student, then you can basically enroll in an entire educational process. So, that's why I recommend you to use Unisor.com for your studying rather than getting the lectures from YouTube or anything like that. Well, that's it. Thanks very much and good luck.