 Hi, I'm Zor. Welcome to Unisor Education. Continuing the theory of probabilities, we are talking about conditional probabilities and in particular about independent events. Now this lecture is part of advanced mathematics for teenagers presented on Unisor.com website and that's where I actually suggest you to watch this lecture from because there are notes for the lecture which are very useful and I would recommend you actually to read the notes maybe before you listen to the lecture and definitely after that just to make sure that you understand all the concepts. Alright, so about independence, independent events. Well intuitively we kind of feel what independent events actually mean. However if we are talking about theory of probabilities and in particular about conditional probabilities, we can actually specify quite precisely what independent events actually mean. So in my particular case I would start with the definition of the conditional probability and we know this from the previous lectures. Conditional probability of event A under condition that event B occurs is equal to probability of their intersection, the event we are interested in and event which is a condition, the intersection of these two events which means the probability of them happening at the same time divided by the probability of the event B which is a condition. Now graphically it looks like basically this. So if you have certain area which is our sample space and this area represents a condition B and this area represents an event which we are interested in. So basically the probability of A by itself is a relative area of this figure relative to the whole sample space. So if sample space has a probability of 1 then the fraction which is taken by elementary events concentrated in this area is its probability. Same thing with event B. The probability of B is actually the fraction of all the elementary events which fall into this figure relative to all the elementary events in the entire sample. So basically what this says is that the conditional probability of A under condition of B occurs is basically a fraction of this area which is intersection of A and B relative to B. So we are kind of narrowing down the entire sample space to only those elementary events which are concentrated in the event B and so we have to disregard everything outside of B because the probabilities are actually shrinking towards this particular area making everything outside of the probability of 0 and making everything inside of B correspondingly greater as if this is a new area of equal to 1. So this is our new, if you wish, sample space and this is the fraction which is taken by all elementary events which are falling into A because all these elements outside of B have a probability of 0. So this area relative to this area is a conditional probability. Okay, that's just a deviation back to the definition of what conditional probability actually means. But now let's talk about what events we can call independent. Well, from just the meaning of the word independent if I would like that A to be independent of B, it means that the chances to occur for the event A should not really depend on whether event B occurs or it doesn't occur. So basically the definition of independence is this. So if probability of the conditional occurrence of the event A under condition that B occurs is the same as unconditional probability of the event A regardless of whether B occurs or it doesn't occur, then A and B can be called independent. Alright, so let's examine a concrete example. Same example as I was using in the previous lecture, we have two dice, but in this case we are making our event independent. I mean it seems to be that if event A is only dependent on what's on dice one. Let's say dice one is even. And event B is dependent only on the results of the dice two out of this pair. Let's say it's equal to five for instance. So the probability, so the first event A is that the dice one falls on the even number and the event number B is dice number two falls on the number five. They seem to be independent because these are different dice, right? Now the example which I was using in the previous lecture, my condition was that the sum of two numbers equals to six. Now when I'm talking about sum that makes them dependent, right? If one is going up then another should go down. So we are actually changing the whole structure of this experiment. But in this case when every dice is considered separately it seems to be like these events supposed to be independent, right? So let's check it out. So let's check out this probability, this equation whether it's held or not. Okay, so in this particular case since this is a definition of the conditional probability I have to check that this particular equality is held. So let's check it out. So we need three different probabilities, A, B and A intersection with B. Alright, the probability of A, well dice one is even. Well that means it's two or four or six. Now considering our sample space is still the same six by six square. One, two, three, four, five, six. One, two, three, four, five, six. So what does it mean that the first dice is even? So the first dice is even means these and these and these events, elementary events. Three times by six is eighteen. So the probability is eighteen thirty six. Now what's the probability of dice number two is five. Now dice number two five is the fifth column. So it's this one. That's six different elementary events, right? So this is one half, this is one six. And finally what's the probability of combined A and B? So it's an intersection this and this. So we have three cases. We have this one, two five, four five, and six five. So that's three thirty six. Well and obviously you can check that if you divide this one twelve over one six you would get one half, right? One twelve divided by one six is equal to six twelve which is one half. So this particular equality is held and since this is a basically definition of this the conditional probability is equal to unconditional probability. So we were just checking that our intuitive understanding of the independence, when one event depends on the one dice and another depends on another dice, well that confirms actually that they are really independent. Okay. So we have defined what's the independence actually is. We checked on a concrete example and let's just examine a couple of properties of independent events. The property number one, the independent events are symmetrical in some ways which means that if A is independent of B which means that conditional probability of A under conditional B is the same as unconditional probability. Then what follows from here that event B is independent of A. Well again intuitively it seems to be kind of true, right? So if I do not depend on you then probably you don't depend on me. But you know it's not always like that and sometimes there is a dependency which is kind of differently formulated. But in case of theory of probabilities and in case of dependency as this particular property that is true. So symmetrical property of the independence is actually the very true property of the theory of probabilities. So you can say that two events A and B are independent which means A is independent of B and B is independent of A. They are mutually independent. So you don't really have to specify which depends which is independent of what. All right now how can it be proven? Well that's actually trivial. I have these couple of theorems and they call them mini theorems or even micro theorems because the only thing the proof is actually in one line. So let's talk about this. Now this is given. So I have to prove this. All right so probability of B under condition of A again by definition is B intersections A divided by PA. Right? Now what do I know about this? I know this. So that is given and I have to prove that this is equal to this. Now obviously from the set theory you know that the intersection is a symmetrical commutative operation actually. Intersection of A and B is the same as intersection of B and A. So these two the numerators are the same right? So from here I can actually derive that PA intersect B is equal to PA times PB right? So if I substitute this into this I will have the P of B over A is equal to instead of probability of intersection I will use this. So it's P of A times P of B divided by P of B is equal to P of A. Now I kind of easily reduced by P of B probability of B. Obviously you know from algebra that it's not such a obvious kind of simplification because you really have to check if what you are cutting off from this from this fraction is not equal to zero right? So basically all the cases when the probability of the event B are equal to zero should actually be excluded and considered separately. But well in any case we probably do not really have to pay too much attention because we really don't consider events with zero probabilities just because they don't occur. So let's consider that everything whatever I'm talking about assumes that probability of events is not really equal to zero whenever it's really important like in this particular case. But in any case wait a moment I think I made a small mistake yes I made a small mistake I really have to put it the denominator is P of A right? So it's P of A which is actually supposed to be reduced which is equal to P of B probability of B and this is exactly what's necessary to prove. Alright so this is my miniserium one let's put it this way. So whenever event A is independent from event B event B is independent from event A and they are mutually independent or just plainly independent events A and B. Okay now the miniserium two is something which I have already basically proven here. Remember I was talking about independent events and definition of the conditional probability as this from which I derived that the probability of simultaneously happening two independent events equals to the product of their probabilities. It's right from here. So if they are independent this is definition of independence and this is equal to this basically by definition of the conditional probability. So they are equal and from this equality I derive basically this property P of B I multiply by P A to get to get the intersection. So again the probability of simultaneous the happening two events A and B is equal to their product if they are independent. Now the reverse and this is the next theory is also true because if I have this as given I can from here get this right. So from here I'm getting this and this is a definition of conditional probability or I can derive this and this is also definition of conditional probability but in reverse order. So that's just another way to prove that they are both equivalent to each other. So for independent events the probability of simultaneous occurrence of both events is equal to the product of their probabilities. Now let's go back to example which I was using before with two dice and let's check if this is true. So we have two dice right one is dice number one is even that's my A event my B event dice number two is five. So the probability of this is equal to 1836 which is one half probability of this is equal to 16 because I have already the second one is 5 and the first one can be 1 or 2 or 3 or 4 or 5 or 5 or 6 so it's six different combinations and finally A and B so I have even first dice and 5 the second dice so it's 2 5 4 5 and 6 5 so its probability is equal to 336 which is 112 and obviously the multiplication of this by this gives you this. So with my point right now is that this property that the probability of the intersection of events sometimes by the way intersection of events especially independent events sometimes it's called a product of events but that's just terminology it's basically intersection in the set theory sense. So the probability of the intersection of two events is equal to the product of their probability so this is a characteristic property of independent events characteristic in a way that if events are independent then this is true and if this is true then events are independent as we have proven in both cases and this is just a checking for the case which we were already kind of considering before. Here is an interesting consideration which I would like actually to end with this lecture the graphical interpretation of this. I would like to graphically show the mutual independence of the events if one is independent from another then another is independent on the first one but I would like to present it graphically. Here is my point let's consider this to be my sample space. Now and I have two events A and B and I will use letters for different areas so the whole area would be x this area would be y this is z and this is let's say u okay now I don't like it this way let's use A, B, C and D or something like this this would be A this would be B this would be C and this would be D okay I like it better all right so what does it mean that event A which is A plus B right A plus B would give me the A event B is A and C and an entire space sample space is A plus B plus C plus D okay so all these areas A, B and C are not intersecting so D is only whatever is outside of this picture C is only this piece A is the intersection of A and B events and B is only this piece right so A is A plus B B is A plus C and sample space is this now we were talking many times that the real probability of event A is a fraction of the entire area which is taken by this particular event which means it's actually graphically A plus B divided by A plus B plus C plus B so the whole area is A plus B plus C plus B the A area is A plus B so the ratio is basically a model of the probability now same thing with B this is A plus C divided by A plus B plus C plus B okay now what is A conditioned on B well we know this is basically the area of intersection which is A divided by A plus C the condition and what do I know I know that these A and B are independent which means this is equal to this what I have to prove is that this is equal to B over A is equal to A over AB so I have to prove that this is equal to this so this is given this I have to prove all right okay so let's concentrate on this you don't really need this and let's do it so if this is equal to this then I can actually use so A plus B divided by this is equal to A divided by this right so A square plus AB plus AC plus BC A square plus AB plus AC plus BC is equal to multiplication this denominator and this numerator A square plus AB plus AC plus so that's what's given or BC is equal to AD now what do I have to prove I have to prove this equality now this equality is equivalent to again I will use numerator times denominator which is A square plus AB plus BC plus A square AB AC and BC and AC it should be equal to this times A which is the same A square plus AB plus AC plus AD right and again and I have exactly the same BC is equal to AD so if BC is equal AD as follows from the first one then this equation is true then this equation is also true so this is kind of a geometrical approach to theory of probabilities and again from from this purely geometrical standpoint let me just redraw this particular picture so what did I have I had this plus ABCD so what do I have here let's do it differently A divided by B is equal to C divided by D right so A divided by B is equal to C divided by D that's what's an interesting kind of geometrical interpretation of probabilities using the areas so the intersection two words probability of A minus the intersection actually it's A minus B if you wish in the theory of probabilities A minus B is actually this because this piece B it's equal to C which is B minus A divided by D which is the entire space omega minus A minus B so entire space I just use the Greek letter omega well you can approach these probabilities from this particular geometrical standpoint okay I recommend you to go to Unizor.com and read notes for this lecture again it's like book basically it's like reading the text book I think after the lecture it would be very beneficial for you and I always encourage people to to register with their supervisors which will allow them to enroll in the courses with ability to take exams which I believe is very important for self-imparation thanks very much that's it for today good luck