 Hi, I'm Zor. Welcome to a new Zor education. I would like to present to you a couple of problems, which, well, I qualify as advanced probability problems. They're not really that difficult at all. But there is certain theory behind these problems, which I would like to share, and maybe that's why I put it in this particular partition. Now, this lecture is part of the course for teenagers, the course of advanced mathematics presented on Unizor.com. I think it's better if you watch this lecture from this website, because it contains notes, notes have the problems which I'm going to discuss, and the answers. So you can actually try to solve the problems yourself and check the answers. And solutions are also presented in the notes as well, basically what I'm talking about right now. All right, so the problem number one. Let's consider you have a bus in the city, which starts at a certain point, and then it goes along its route to the end point. And there are stops in between. Now, on the bus, on the first stop of the route, M people enter the bus. Well, I think it would be a little bit more understandable if I will say that M people enter the bus and you yourself also enter the bus as an observer, basically. So you see how many passengers are getting off on each of these stops. Now, let's assume that there are no additional passengers entering the bus. Only these M are supposed to exit on one of these stops. And there are S stops on the route, not counting the beginning one, obviously. So the possibility of any one of these M passengers to exit are basically equal. So any person with the same probability can exit on any stop. And that's where I would like actually to spend some time to talk about. Now, what do you mean the probability? Everybody knows where he wants to exit the bus, right? Everybody knows his stop. So what kind of a probability we're talking about? But again, let's go back to the beginning. I said you also enter the bus as an observer. You have absolutely no idea who are these people and what are these stops which they're supposed to exit. They know themselves. So each person knows about himself, but you as an observer, you have no information. And this is a very important philosophical aspect of theory of probabilities. Information and probability are very much related to each other. I spoke about this in the beginning of this chapter on probability. Basically, knowing about something about the subject is changing your evaluation of the probabilities of certain outcomes, of certain experiments with this particular object in mind. So in this particular case, considering you have absolutely no knowledge about who is entering, exiting on which stop, your only assumption can be basically to spread the probabilities evenly. So you assume that every person out of these M is having exactly the same probability to exit on any stop out of these S stops. That's the only reasonable assumption you can come up with. If you know something about these people, let's say some of the people are dressed in one way and you know that people who are dressed in that way mostly are located in that particular area of the city. So you can assume that the probability is greater that these people who are dressed in that particular way to exit in those particular stops. But not knowing anything like this. There's only reasonable assumption that all these probabilities are equal. And that's exactly what I mean when I'm saying that any person randomly can exit on any stop. It doesn't mean that he just doesn't know where he wants to go. It's you as an observer have no information about where this is supposed to happen. Now let's go to the problem. The problem number one. Okay, what's the probability of all passengers exit at different stops? So no two passengers are exiting on the same stop here. Now obviously I should assume that the number of stops greater than the number of passengers. Because otherwise the problem would have no solution. There is always the stop. If M is greater than S, there is always a stop where two passengers or more must exit. Otherwise it just doesn't work. But if M is less than S, and that's what I assume, actually even equal as well. Then this is possible. They can all exit at different stops without any problems. Now here is my question. What's the probability of all of these M people exit on different stops? Well, actually the problem is kind of easy if you consider that all you have to do is to just pick M stops out of S. And say, okay, these are the stops where people are exiting. And since the probability of these M people to exit on these M stops out of S doesn't really depend on what people are or what kind of stops we are choosing out of these S. The probability should actually be equal to the number of combinations from S by M. So I have to pick certain stops. And I have to divide it, obviously, by the number of all elementary events which I have in this particular case. So what are my elementary events? Well, any passenger, as I was saying, can exit on any stops. So there is an S possibilities for one passenger, S possibilities for another passenger, etc. So it should be S to the power of M. Now, that seems to be like an easy solution. But at the same time, there is certainly the satisfaction which I was just experiencing when I was coming up with this particular solution. And I decided to approach it a little bit differently and check if I will have exactly the same result. So let's just think about this. So we are counting the elementary events which are falling into this category, all supposed to be exit on different stops, right? So let's just think about it. The problem we will solve in the following way. First, I will number all my passengers, number one, number two, number three, etc., number M, all right? Okay, so I number them. Now, if I have already numbered them, then I can say that there are S different possibilities for the person number one to exit. And the person number two cannot exit on that particular stop where the first one entered, because they are supposed to be on different stops. So there is only S minus one possibility for number two. And they have to multiply, obviously, because it's all completely unrelated to each other, etc. And the last one would be S minus M plus one with the passenger number M. So that's number of cases, right? But let's just think about this. First, I have numbered them from one to M. Now, if I will number them differently, let's say whatever used to be number three will be number one, whatever used to be number five will be number four, etc. So if I will number them differently, I will basically do the permutation of people within these set of numbers from one to M. And do exactly the same thing. I will have exactly the same result. So one person will exit on one stop, another, another, etc., etc. So basically I have counted the same distribution of people among stops M factorial number of times. So I have to divide it by M factorial. And that's the true number of different cases of distribution of my people among stops with all having different stops. So this is a true number. But now if you look at this, this is exactly what C from S to M is. Because this is equal to S factorial divided by M factorial and S minus M factorial. So S factorial divided by S minus M factorial is exactly this one. So it's exactly the same answer, which is good. And that would be probably the end of this particular part of the problem. Now let's continue the same situation with M passengers, which are supposed to exit on these S stops. But I'm interested in a different event. I'm interested in an event when all passengers are exit on the same. So the first problem was all different. Now I'm talking about all the same. Well, this seems to be actually much easier. Because if I will say, okay, what's the probability of all of them exiting on stop number, let's say 3? Well, it means that the first one exits on stop number 3. And the probability is obviously 1 over S. And the second one will exit and stop number 3 will be 1 over S, etc. And the Mth one will be on stop number 3. So for any particular stop, this is a probability of all of them. This is an elementary event, basically. So this is the probability of all of them exiting on stop number 3. But since I'm not talking about stop number 3, I'm talking about any stop as long as all of these people will exit on this stop. I have to multiply it by the number of stops, which is S. So the result is 1 over S to the power of M minus 1. So that's the second problem. Now the third problem is also not very difficult. The probability of one passenger exits at stop number 1. One passenger exits at stop number 2, etc. One passenger exits at stop number M. And I need the probability of this. So I have sequential stops 1, 2, 3, 4, 5, M. And I need the probability of people exiting 1 on this, 1 on this, and 1 on this, etc. Now, so again, what kind of freedom of choice do we have here? We have fixed stops. What we don't have fixed is which passenger exits on which stop. So basically, as long as I will put them in order, all the passengers from 1 to M, I can say that passenger number 1 goes out at stop number 1, passenger number 2 goes on stop number 2, etc. So all I'm saying right now is that as long as I have ordered passengers in certain way, then there is only one solution. And there is only one event basically, elementary event. Because each passenger exits on one particular stop, and that's an elementary event. And the probability of this again is 1 over S to the power of M. So now what I basically have to take into consideration is that I don't really order passengers in any particular way, which means any order will be good, and there are M factorial orders. So all of these correspond to this particular event. No matter what order I put passengers in, the result will be exactly the same. So this is total probability of having one passenger, exactly one passenger exit on first stop, one passenger on second, etc., up to one passenger on Mth stop. And again, let me go back. These are probabilities from the point of view of observer, which has no information about where people live and which stop they're choosing, etc. And that's why all of these elementary events have exactly the same probability, which is 1 over S to the power of M. Okay, that's it for the first problem. Now, the second problem is related to the deck of 52 cards, standard deck. You have four suits, spades, heart, diamonds and clubs. And you have 13 ranks in each suit, from 2 to 10, Jack, Queen, King and Ace. So, now you are randomly pulling six cards out of the deck. My question is, what's the probability of all four suits to be represented among these six cards? So I should have at least one spade, at least one heart, at least one diamond and at least one club. It doesn't matter what rank of the cards I have. So at least one representative from each suit I should have. And I need the probability of this if I'm completely randomly choosing six cards. Well, completely randomly, what does it mean? Well, it means that any set of six cards pulled from the deck has exactly the same probability. And how many different elementary events I have, obviously. Number of combinations from 52 by 6. Right, whatever it is. 52 factorial divided by 6 factorial and divided by 46 factorial. So this is number of different sets of six cards which I can choose. Now, which of them are of my interest? So how many of these combinations have all four suits represented? So I have these four suits and I have one card of each suit for sure. Because these are all the elementary events which I'm interested have one card of each suit. In addition, I have six cards, now this is only four. So I have two additional cards. Now, two additional cards can be either in one particular suit or they can be in two different suits. There are no other combinations, basically. So I will consider these two cases separately because they are mutually exclusive. And I will evaluate number of cases when I have two additional cards belong to the same suit as one of these. And another case, this is case number one and this is case number two. When two cards are of different suits. Okay, now let's think about it. Let's consider the case number one. So the case number one is I have three cards in one suit and other suits, other three suits have only one card. How can I count the number of combinations? Well, the first freedom of choice I have is which suit has three. I have four suits, so number of combinations of four by one, which is actually four. I have four choices, right? Now, if I have chosen a particular suit, let's say it's hard in this case, where three cards are presented, the freedom of choice now I have is what's the rank of these cards? The rank of this card is I have to choose any one card out of 13 ranks, right? This one as well and this one as well. So I have C for 13 to 1 in the power of 3, right? 13 choices for this, 13 choices for this and 13 choices for this. And I'm picking only one card out of six. Now in this case, I pick three cards out of 13 of different ranks. So I have to multiply it by number of combinations from 13 by 3. And this is the number of choices I have for this particular case number one. Now, case number two. I have two indifferent suits, which means I have two suits with two cards and two suits with one card. The first freedom of choice I have is I have to pick which two suits I have two cards. Now, out of four suits, I have to pick any two. Now I've got these two. Now what choice do I have? In these two suits, I have one and one card out of 13. So I have 13 choices for this and 13 choices for this, right? I have to pick one card out of 13. So it's one card out of 13 square because there are two of them. Now in these cases, I have to pick two cards out of 13 and that's a square. So I pick two cards and there are two suits. So that's why I have freedom of choice two out of 13 multiplied by two out of 13. So this is the case number two. Now if I will add them together, that would be the number of combinations which satisfy my condition of all four suits to be represented. And I have to divide it by the number of elementary events of different six card combination out of 52. So the sum of these is obviously the number of combinations where four suits are represented. And on the bottom, I have the total number of combinations. Now obviously when I'm talking about something like this, it's four. This is 13 cube. Now 13, now this is 13 factorial divided by 3 factorial and 10 factorial plus. Now this is 4 factorial divided by 2 factorial and 2 factorial. This is 13 square and this is 13 over factorial over 2 factorial, 11 factorial square. So sum of these divided by this, which is 52 factorial, divided by 6 factorial and 46 factorial. So that's the total answer. Well, that's it. What I would suggest you to do right now is to go back to the Unisor.com website. Look at this lecture again and notes contain the problems. Try to solve them by yourself right now and see if you will get the same results using the logic whatever you have learned. It's just a good exercise of how you remember certain approaches to calculate certain probabilities. Well, that's it. Thank you very much and good luck.