 Hi, I'm Zor. Welcome to a new Zor education. We continue doing what I consider to be the most important part of this whole course, which is solving problems. Now, this is the fifth set of problems on zero probabilities. These I call easy problems. There will be some more advanced ones, hopefully. This is part of the whole course of advanced mathematics for teenagers presented on Unizor.com. That's where I suggest you to watch this lecture, but before doing that, I strongly recommend you to go to the website and look at the notes to this lecture. Notes basically contain all these problems and solutions, but don't pay attention to solutions first, try to solve yourself all these problems. Then you can read the solution, you can listen to my lecture, where I will explain the solutions, how I understand it. So, basically, let's start. Okay, the problem number one. I have a box and in the box I have R red balls and B black balls randomly positioned inside the box. I don't know how it's all inside the box. Now, I am randomly choosing N balls from the box and randomly it means basically that the probability to pick any one of these balls is exactly the same. Now, I choose N balls. Now, I'm interested in the probability of having S red balls, where S is obviously from 0 to R. I cannot draw less than 0 and I cannot draw more than R red balls. So, for these S and for this N, N is obviously greater than 0, I have to determine the probability of taking among my N balls exactly S red balls. Alright, let's just think about this. What's the sample space? Sample space is all the different groups of N balls which I can pick from this box. Now, in the box I have R plus B balls and out of these R plus B balls I am just randomly taking N balls. So, many different groups of N exist, obviously number of combinations from R plus B by N. So, this is the total number of elementary events in my sample space and they are all equally probable which means that the probability of each one of them is 1 over this number. Now, all I have to determine now is how many good elementary events satisfy this particular condition. So, how many groups of N balls contain exactly S red balls? Well, let's just consider. We have N balls, right? S of them are red and correspondingly N minus S is the number of black balls, obviously, right? So, I have a freedom of choice to have S balls out of R. So, number of combinations from the R red balls by S is number of different groups of S red balls which can be part of my N balls set. Now, with each of them I can have a freedom of choice of the black balls, right? So, if N is total, so N minus S black balls are supposed to be picked from the B black balls. So, this is the number of different combinations of N minus S black balls which I can get from the B black balls. And obviously, with each of these combinations I have each one of those combinations, so that's why I'm multiplying them. And each one has the probability one over this, so the total probability is this. This is the answer. Now, obviously, my conditions should be that, well, S is supposed to be from 0 to R. Now, also, S cannot be greater than N. Obviously, I cannot, if I choose four balls, I cannot have seven red or something like this. So, this is all for all these conditions which basically are necessitated by the formula here. Okay, that's it. That's problem number one. Okay, problem number two. I have a standard deck of 52 cards. There are four different suits, as we know, and the cards have different ranks. Now, for each suit I have cards from 2 to 10, Jack, Queen, King and Ace. Now, I will call these cards of all different suits, doesn't really matter, numerics. And these cards I will call pictures. This is basically my way to divide the total of 52 cards in two groups, two categories. I have 36 numerics and 16 pictures, obviously, because I have four different suits. So, for each one I have all these cards. Alright, now, now I'm randomly picking two cards out of the deck. And I'm interested in all the different probabilities that I have two numerics, I have two pictures, or I have one plus one, one numeric and one picture. Now, I'm picking two cards, doesn't really matter the order. I'm just picking a pair, alright? So, there are three different events. I have both numeric, both pictures, or one numeric and one picture. And I need the probability of picking each one of these. Well, first of all, let me just point out that there are no other cases. So, these three are mutually exclusive events and together they basically spend the whole set of different choices. If I have two cards, I can have only these three choices, either two numerics or two pictures or one of them numeric and one picture. There are no other choices in this particular case. So, now let's talk about the probabilities. Now, again, the elementary events which comprise our sample space are all the different pairs which I can take from the 52 cards, which is number of combinations from 52 by 2. Now, let's think about the first case. So, two numerics, the probability of two numerics is equal. Now, obviously in the denominator, I have to have this. Now, what do I have to have in the numerator? Number of different pairs where both of them are numeric. There are only 36 numerics, so how many different pairs can I get? Well, obviously this number of combinations of two cards out of 36. Now, similarly, the probability to have two pictures would be equal to number of combinations of two out of 16 pictures divided by the same denominator. Because this is one over this is the probability of each pair. And finally, if I have numeric and picture in any order, I'm not really suggesting that this is first and this is second. It doesn't really matter. So, I have a group of 36 numeric and I'm picking only one of them which is number of combinations from 36 by 1 times number of combinations from 16 by 1 divided by the same denominator. Well, obviously this is 36 and this is 16. So, anyway, this is basically the answer to the problem. But as a prudent man, I would like to check my answer. In this particular case, how can I check the answer? Well, I should really add up these three probabilities and check if it's equal to 1, the result of this. So, in other words, I will just add all the numerators together and see if they, because they share the same denominator. So, I will add numerators and I will check if I will have the same denominator. So, I need to calculate C 52 2 which is 52 times 51 divided by 1 times 2. Okay, this is equal to, I have already calculated it somewhere, 13 26. Now, this one, 36 by 2 is 36 times 35 divided by 1 2 which is 16 2 which is 16 times 15 divided by 2. It is equal to 130 and finally this one which is times C 16 by 1 which is actually 36 times 16 is equal to 576. Now, you add these three numbers, 6 0 0 6 7 2 5 2 2 1 goes 6 13, yes. So, balance is out, so to speak, right? So, it actually gives me assurance that my calculations are correct. Now, the third one. Alright, the third one is a very simple one. There is a deadly game of Russian roulette. One takes the revolver, let's say we have seven, capacity of seven bullets in the revolver. Puts only one bullet, spins the cylinder and shoots. My question is what's the probability of shooting twice in a row and not firing both times? So, both times we are not getting the bullet in the proper place to make a shot, okay? So, let's just think about it. If I have only one bullet out of seven and hopefully revolver is very symmetrical, so it's basically a random when I'm spinning the revolver, the probability of getting the bullet in that specific place for the firing is equal to one cell, right? So, that's the probability of fire one bullet. Now, if I don't, the probability of not firing is obviously six-sevenths, because there are six different empty spaces, empty spots. So, I'm doing it once and then I'm doing it another time. Now, these both times are independent, which means that the probability of happening both is actually equal to the product of their probabilities, which means that I have to multiply six over seven by six over seven, and this is my probability of not firing two times in a row. Well, significantly greater than half, so it's good chances. For three, by the way, it would be, well, for three it would be what, thirty-six times two-sixteen, and this would be three-fifth times seven, three-forty-three, something like this. Well, it's still better than half. So, even three times in a row you have better than fifty percent chances of not really firing the revolver. Okay. Now, my last problem is, all right, there is a legend about King Arthur and the Knights of the Round Table. So, let's consider we have, by the way, there are many Knights of the Round Table. I checked the Wikipedia and there are names, names, names, there are many different names, but I do know two main kind of people in this set. One is King Arthur and another is Sir Lancelot. He is a very famous Knight. So, let's consider that all the Knights of the Round Table and there are, let's say, N Knights, N Knights of the Round Table, they are coming and they are taking randomly the places around the table. So, let's assume that the chairs are not assigned. So, whenever they come, everybody sits randomly, basically. So, the probability of any permutation is exactly the same. So, my question is, what's the probability these two, King Arthur and Sir Lancelot, to sit next to each other? I suggest two different solutions. Solution number one, let one of them take any place, doesn't really matter which one. Then, let's consider how many choices for the second guy exist. Well, if, let's say, King Arthur took one place, there are N-1 other choices. Now, only two of these N-1 choices, this is King Arthur, then only two choices are next. So, here is one choice, here is another choice where Sir Lancelot can take the spot to be next to King. So, all others are basically not satisfactory. So, we have two out of N-1 choices good in some way, right? So, the probability should be two over N-1. So, out of N-1 spots, only two spots are considered to be good. Well, if you have some doubts about this approach, because I basically say, okay, let King Arthur choose any place. Let's think about this more, I know, more theoretically, if you wish. So, I was saying that any permutation of knights around the table is equally probable and there are N factorial permutations, which means every one of them has the same probability one over N factorial. Now, let's see how many different combinations are considered to be good combinations. Well, King Arthur and Lancelot can sit, there are N-different spots for King and for each one of them, two spots for Sir Lancelot on both sides of the King and that's different permutations obviously, right? We are considering all the different permutations. So, it's two times N positions for these two. So, any of N positions of the King and for each one of them, two positions of the Sir Lancelot are around this, right? Now, all other N-2 knights can sit whatever way they want. So, the permutations are N-2 factorial. So, that's the answer basically. But if you think about it, what is N factorial? N factorial is N, the previous one N-1, previous one is N-2, etc., etc., up to 1, right? So, N-2 up to 1 is canceling this one, N canceling this one and we still have 2 divided by N-1, the same answer. Alright, that concludes this particular lecture. I do suggest you to go to unison.com to this particular lecture and again, go through the notes, try to solve all the problems yourself if you didn't do it before and I think it's just very beneficial for you to go through these problems and any other problems yourself. Because again, that's the purpose of this course. You would feel much more comfortable if you will solve all these problems yourself. That's the purpose actually. Okay, that's it for today. Thanks very much and good luck.