 Hi, I'm Zor. Welcome to Unisor Education. This lecture is part of the course of advanced mathematics for teenagers on Unisor.com, and that's where I recommend you to watch this lecture from, because it has notes. Notes are basically the complete lecture, which I'm basically presenting right now. In this particular case, we are talking about problems, so it would be actually quite beneficial if you would try to solve these problems just by yourself. So read notes, they have the problem, they have also a solution as well, but try not to look into a solution. Anyway, after you went through the problem as presented in the notes, then it makes sense actually to listen to the lecture. Alright, we are talking about problems related to conditional probabilities. Alright, so I'm going to present a certain number of problems, one by one, and with solutions actually, right? So the problem number one, one more detail, I will extensively use the set theory and the measure theory as a representation of the probabilities. So we have a sample space, which is actually a set, we have elementary events, which are elements of this set, we have events as basically subsets of our set, and the probability as a measure introduced on all the different events. We are talking about finite sample spaces, so finite number of elementary events. So every event is just a combination of certain elementary events, and in as much as every subset is just a set of elements which belong to this finite set. And the measure is an additive measure with the measure of an entire set equal to one, that's what makes it a probabilistic measure. And I will extensively use the geometric interpretation of this, basically representing event as area on the whiteboard with the size maybe corresponding to the probability, to the measure. Okay, problem number one, we have two events, and I will again graphically, this is my sample space, and there are two events, one and two. Events are subsets, for instance, this represents all the different results of my experiment. Now this is certain results which form one particular event, and this is another set of elements which is another event. Now, I am talking about the case when these two events, let me, one is this one, and another is this one. Let's call them A and B, and I am talking about A is independent of B. So we are talking about two independent events. Okay, now the first property which I would like to present as a theorem, if you wish, or a problem is the following. Now the area which represents the event which is a combination of A and B, A and B, is one area which I would like to consider. Then the area which is A except all those which are in B, which is this area, A minus B. So these are all elementary events which belong to both A and B. This is belong to A but not to B. Similarly, there is this one, belongs to B but not to A, which means it's this part of the B. And finally, I would like to consider elements which do not belong to either A or B, which is usually something like not A union B, something like this. Now, usually the whole space, the sample space is delineated with a Greek letter omega. So basically this is omega minus A union B, that's what it is. So, what's my theorem is about? My theorem is about the following, that the probability of the common part multiplied by the probability of part which does not belong to A or B, which is omega minus A or B, equals to the product of probabilities of A without B times probability of B without A. So, probability of this times this is equal to probability of this times that. Well, and I'm talking about independent events. It's not like any event, obviously. It's only for independent events. Now, how I'm going to solve this particular problem or prove this particular theorem? It's actually relatively simple thing. What I will do is the following. You see, it's kind of difficult to deal with these events with their intersection, their difference, compliments, etc. I would prefer to deal with events which are completely mutually exclusive, do not have any intersection between them. So, what I'm going to do is the following. I will call the elements which are inside of both, which is this one, and event w. Now, sorry, it's called x. To correspond to my notes, that's x. So, this is inside this area. Now, this would be y, which is A without B, which is this part. This part. This is y. Now, B without A is this part, would be z. And everything outside would be w, outside of both. Now, I would prefer to deal with x, y, z, and w instead of A and B, because it's just easier. They do not have any intersection. And what's interesting is that the probability is an additive measure, which means if you have two events which do not have anything in common, the probability of union of these events is some of their probabilities. That's why we usually can use the plus sign instead of the union sign when we are combining elements which do not have anything in common. So, that's just quite a convenient thing. Now, using this, I basically have to prove that x times w equals to y times z. Now, where lowercase x, w, y, and z are the corresponding probabilities. So, probability of event w is lowercase w, probability of event x is lowercase x, probability of event y is lowercase y, and probability of event z is lowercase z. So, that's how I would treat them. Now, this seems to be much easier, right? But still we have to prove it, right? It doesn't really matter in what language I express my problem. We still have to prove it. Now, what's given? The only thing which is given about this, above and beyond the notation which is not really given, it's just my choice, is this one. A is independent of B. So, let's try to express this independence in terms of x, y, z, and w. So, that's probably the key to everything, right? So, what does it mean that A is independent of B? Well, many times we were talking the conditional probability of A under condition of B is equal to the unconditional probability. That's what independence means. If there is a condition B or there is no condition B, the probability of AB is exactly the same. Or, if we are expressing the conditional probability of A under condition of B as something which we know as A intersection B divided by probability of B, that's what this is. We defined this conditional probability and basically derived this in case of a finite probabilistic space and it's actually a definition for other cases, which basically means that the probability of A under condition B means the relative. So, basically, let's go back to the probabilities. So, instead of distributed among all the elements of the space omega, the probability is concentrated in B in this area. So, everything is concentrated in B in all elements. Let's consider it evenly distributed within the elements of B. So, the common part, which is this one, which I basically described as an event x, that's what the measure of probability in the numerator and denominator is the total thing. So, we are talking about basically a ratio or a fraction. What part of the B is taken by events which also belong to A? That's what conditional probability that A is happening in case B is happening. So, what part of the B is taken by this particular area which is a combination of A and B? So, that's what this formula is. That's a definition of conditional probabilities. Now, let's express it in our language. A intersections B is x. Now, what is B? B is this, right, which is x plus z. And I can use really B is equal to x plus z. I can use plus because x and z have no common elements and we have agreed that union in case of there is no common elements can be expressed as a plus sign because the probability of the union in this case is the sum of these two probabilities, right? So, we have probability of B equals probability of x plus probability of z. That's because x and z do not have anything in common, right? So, how can I express this formula? And by the way, probability of A is correspondingly x plus y. So, this formula basically means the following. Probability of A, that's x and y, right, which is x plus y equals probability of A intersections B which is x divided by probability of B which is x plus z. That's what's written. Okay, so this is given and this we have to prove. So, it becomes actually a plain algebraic exercise, how from one formula I can get to another. And let's not forget that sum of all these probabilities, x, y, z and w is equal to the probability of the entire sample space which is always one, right? So, there is also this condition, x plus y plus z plus w is equal to one. So, this condition plus this condition must result in this. Question is, can I derive it? Well, let's see if I can. Okay, so now we have a clean room to make a relatively simple calculations. Alright, so this is given and this is given. I have to prove this. Well, probably the best thing is, well, let's just convert this and that will be x is equal to x square plus xy plus xz plus yz, right? Or x factor out x plus y plus z plus yz. Or, now x plus y plus z is obviously one minus w, right? Multiply x minus xw plus yz. And what do I have? x is equal to this. x can go and I have 0 equals minus xw plus yz or xw is equal to yz which is exactly what's necessary to prove. As you see, the proof itself is very simple. What's really kind of maybe difficult or takes some time is to come to this purely algebraic exercise, very easy exercise, using certain other events which we have introduced. So, instead of a and b and an entire space omega, we have introduced x, y, z and w which do not have any intersection and we expressed original events, a and b in terms of x, y, z, etc. And we have expressed their independence a from b in this particular way which is one of the things which we have definitely used. So, that's the end of this particular problem. This is the proof and we will use this particular theorem. Let me draw again my graphical representation of this problem. So, this is omega, this was my a and this was my b. Now, this is my x, y, z and w. So, let me circle my original letters which have something like an intersection, etc. and x, y, z and w without the circle, these are independent parts. So, y is the part of the a which does not belong, which does not belong to b. b, z, sorry, is part of the b which does not belong to a. x is a common part and w is something which is outside of books. That's what it is. And this is the equation. So, the probability of the events which are common for a and b multiplied by the probability of events which do not anything from a nor b is equal to the product of probabilities of a without b and b without a. So, that's what this particular theorem is. x, w is equal to y, z. Alright. Finished with this problem and let's go next. Okay. We have to prove that if a is independent of b then not a also is independent. So, again, the condition is exactly the same. All I have to do is to prove that if a is independent which is basically what we were doing in the previous problem then not a is independent as well. Well, let's just think about it. Independence a from b we have actually written down as the probability of their intersection which is x divided by the probability of b which is x plus z and that's equal to, that's the conditional probability of a relative to b and unconditional is x plus y. So, this is the expression of a is independent. This is given. Now, what's necessary to prove Now, let's think about how can I express not a in the language of x, y, z and w. Well, not a means, okay, if a is this, not a is this piece which is z plus everything outside which is w, right? So, not a is equal to w plus z. Now, sometimes I'm using lower case and I mean the probabilities. Sometimes I'm using the upper case which looks exactly the same as lower case, sorry, but that means events, right? And obviously the same letter means the probability of that corresponding event. So, these are probabilities. These are events. So, event w which is everything outside plus this event which is b but not a, they constitute together everything which is outside of a. So, I have to prove that this particular event is independent of b, which means what? I have to prove that the probability of w plus z, intersections event b divided by probability of b. That's the conditional probability of w plus z relative to b, right? And that's equal to unconditional probability. If I prove that, that means that w plus z is independent of b, correct? All right, now, so let's talk about probability of w plus z is w plus z. Consider this is a lower case. Now, probability of w plus z, intersections b. Okay, w plus z, intersection with b, which is this. So, what's the common? Only the part z. And probability of b is x plus z, x plus z. So, I have to prove this. What's given? Well, basically, given this, or actually this, which is simpler, which we have proved before. So, I have to prove this, knowing that this is true and knowing that x plus y plus z plus w is equal to 1, all right? Okay, so let's do it. How can I prove it? Well, again, let's just multiply. So, z is equal to x w plus x z plus w z plus z square, which is equal to x w plus z times x plus w plus z, right? x plus w plus z equals x w, sorry, x w plus z times. Since x and y and w and z are equal to 1, then this is equal to 1 minus y, right? Which is equal to x w plus z minus y z. So, this is equivalent to this. Now, we are excluding all these weird cases of the probabilities equal to 0 because it's like dividing by 0. We are excluding all these cases because when probability is equal to 0, it means events actually do not happen. So, let's just forget about all these extreme cases. So, this is equivalent to this. And this, by the way, I can reduce it by z and I have x w is equal to y z. So, this is equivalent to this. So, instead of proving this, I can prove this. But this has already been proven in the previous problem. So, from this, by transformations, we came to this. All transformations are reversible, again considering the probability is not equal to 0. So, from here, going backwards, I can get to my original equation, which I wanted to prove. So, again, if A is independent of B, then not A is independent of B as well. All we have to do is just express it in this language of x and y, and it becomes a plain problem. By the way, in the notes for this lecture, I spent some time to use it without this language of x, y and z using more probabilistic approach. Please read the notes on unizor.com and you will see basically a second solution, which seems to be exactly kind of as simple as this one. But still, it's kind of interesting to approach it slightly differently. Okay. My third problem is very much like this. So, if A is independent of B, then instead of not A is independent of B, now my theorem is A is independent of not B. And you just have to believe me. I don't want to spend some time. It's very much like it. All you have to do is to express this particular independence as an equation like this one and basically use a plain algebra like multiplication, whatever, and you will get exactly the same thing. So, it's a very easy thing to do. I don't want to do it because it's really easy and, again, it's presented in the notes. I do suggest you to read it. It's very simple. So, no time spent on the third problem. Now, problem number four is... Okay. So, we bypass problem number three and go to problem number four. Here it is. Now, we know that if I have any event A, I can always construct an event not A, which is a complementary event. Like, for instance, if A is this thing, not A is everything outside of this. So, geometric approach is very, very useful in all cases of these problems. And what's the measure of the sum of these? Well, the measure of the A and measure of not A is the measure of an entire sample space, which is always one. So, I can always say that probability of A plus probability of not A is equal to one. Not A, you can always write like omega minus A. That's exactly the same thing. Right? Omega minus A would be not A. So, the probability of two complementary events is equal to one. Now, my theory which I would like to prove, or the problem which I would like to save to solve is to prove that conditional probabilities also are equal to one, where B is any event actually. It doesn't really matter. Look at this from the geometric perspective. What is a conditional probability of A under condition B? It's the area which is common divided by the general area which is all B. Now, what is not A under condition B? Well, it's everything outside of the A. But again, the measure is all concentrated within B. Right? So, we can completely disregard everything which is outside of them both and concentrate only on this piece which is outside of A but still within the B. Right? So, I'm adding ratio of using this language X over X plus Z and not A which is basically Z over X plus Z. And I'm adding them together. Well, guess what? This is the common denominator and the numerator you get the same X plus Z. So, you have X plus Z over X plus Z which is always one. Right? So, as long as you express this equality in the language of X, Y and Z geometrically, well, I'm using the word geometrically in quotes actually because it's not really geometry. I'm representing my probabilities as areas and it's just a new illustration. So, as long as you do this, the problem becomes trivial. So, the conditional probabilities of two complementary events A and not A also give one in sum. Okay. Now, let me reverse the problem. Is this true? It looks like symmetrical. Remember, the previous two problems I was telling that they are really very similar whenever I was talking about independence. If A is independence from B, then not A is independent from B and also A is independent from not B. They're kind of symmetrical. But is there a symmetry here? If the sum of two complementary conditional probabilities of two complementary events is equal to one, we just proven this, how about sum of two conditional probabilities against two complementary events? Conditional probability of the same event A, but you have complementary conditions. Is this true? Well, the answer is not and it's very, very easy to prove. Think about this. Let's consider two independent events, A and B. Then conditional probability of A under condition B is the same as conditional probability of A, right? Because that's what independence means. Now, we have proven that if A is independent of B, then A is independent of not B as well. So, if they are independent, this is also T of A. And I have this, which means T of A is equal to one-half. Well, I mean, obviously I can choose any event A and obviously I can choose an event, the probability of which is not equal to one-half. And that would disprove this particular thing. So, it looks like if this is true, then this must be true. And since it's not, then forget about the original equation. Well, that's it. That's my last problem. That's a very simple one. So, what's the most important part of this? Probably the most important part is to use this geometric interpretation of the probabilities and representing events which do have something like intersection in terms of other events, X, Y, Z and W, which do not have any intersection among themselves. And you can always combine, like, X and Y to get A, X and Z to get B, X and Y and Z and W to get an entire space. So, everything is built from these non-intersecting components and that's very useful because the probabilities are edging. If you add two events, then the probability of the sum is the sum of the probabilities. Non-intersecting events are very useful in this particular case. Because the measure is additive. Well, that's it. Thank you very much. Don't forget to go to Unisor.com and to read again these problems and try again to solve them yourself. That would be very useful. All right? Thanks very much and good luck.