 Hi, I'm Zor. Welcome to a new Zor education. This is the beginning of a couple of lectures dedicated to problems in theory of probabilities. Well, problems is my favorite part of this course. I think the whole purpose of the course is actually to present you with certain problems. Some of them easy, some of them challenging. So the course is presented as part of the advanced mass course for teenagers. It's presented on unizor.com and that's where I suggest you to listen to this lecture because it has notes, some hints maybe, and as far as the problems are related, I'm always encouraging you to try to solve these problems yourself. They are explained basically on this website, exactly in parallel to this lecture. But don't read the solution, don't listen to me before you start doing something yourself. So try to solve it yourself and then you can listen to whatever I'm suggesting. Maybe it's a different way to solve the same problems or whatever. Okay, so today I have four different easy problems in the theory of probabilities and let me just start from the first one. Okay, you have a standard deck of 52 cards. Now this deck contains four different suits, spades, hearts, diamonds and clubs. Each one has 13 different ranks of the cards. 2, 3, 4, 5, 6, 7, 8, 9, 10 Jack, Queen, King and Ace. Alright, so you have this standard deck. Now you randomly pick two cards from this deck and the problem is what's the probability of picking two spades? Probability of two spades. Okay, I suggest two different ways to solve this problem. The way number number one is the following. Well, to pick up two spades you have to pick the first spade and the second spade. As two completely independent random choices. Now, if you have 13 different spades in the deck of 52, the probability to pick the first spade is equal to, obviously, 1350 seconds because all the cards have the same probability to pick, which is 150 seconds and 13 of them which are all spades are actually good for you. So that's the probability to pick the card of the suit, spades for the first your pick. Now you have the second experiment which is completely independent of the first one. But the second experiment has slightly different numerical conditions. It doesn't have 52 cards anymore, it has only 51. And from this 51 cards you have to randomly pick one which is supposed to be a spade. Now, you have already picked one spade so there are only 12 left, which means that the probability to pick the spade on the second try is 1250 seconds. And now, since these are independent events, and the probability of the first one is this and the second one is that, the probability of both of them happening together is a product of probabilities as we know for independent events the probability of end events is the product of probabilities as long as they are independent. Okay, so that's the answer. Well, you can obviously reduce it slightly this is one fourth and this is three, one, so it's one seventeenth. So that's the answer. Now, on the other hand you can approach this problem slightly differently. Let's have the situation not as two sequential picks but as one pick of two cards. Now, how many different pairs of two cards you can pick out of 52 cards deck. Well, obviously it's this one, which is equal to 52, 51, 1, 2 which is 26 times 51. Now, that's the total number of pairs. Now, which are good for you? Well, the good pairs are those which are chosen from 13 spades and how many different combinations of two cards out of 13 we can pick. So, these are our good combinations. So, this is number of good pairs, this is number of all pairs and therefore the probability of two spades equals to this divided by that, which is two, three, one, seventeen, exactly the same answer to be expected. So, that's two different ways to approach the problem. Either you pick one and then another or you pick a pair. It doesn't really matter, the result is exactly the same. Okay, second problem. You have exactly the same 52 cards deck and you are looking for a probability to pick a card which has a rank greater than 5 which is 6, 7, 8, 9, 10, like Queen, King and Ace. So, this is a probability of from 6 to Ace. So, what's the probability to pick one card which has this rank? Well, that's where we actually can use the theorem of addition of probabilities because the probability as we know is an additive measure. If you have certain events which are completely non-intersecting the probability of either or of those is equal to some of their probability. It's like a measure like area of this thing. If you have non-intersecting pieces the area of the whole thing is equal to some of these areas, right? Alright. So, what are non-intersecting events which we are talking about? Well, these are events of picking exactly 6, picking exactly 7, picking exactly etc. up to King and Ace. Now, what's the probability of each one of them? Well, we know that there are 4 different 6 out of the deck of 52 cards, right? We have a 6 of space, 6 of hearts, 6 of diamonds and 6 of clubs. So, there are 4 different 6s out of 52. So, the probability is 1.13. And the same is the probability of 7 etc. up to Ace. And the sum of these probabilities is, let's count how many of them 6, 7, 8, 9, 10 Jack, Queen, King and Ace, 9. So, the sum is 9.13 and this is the probability of picking a card which has a rank greater than 5. All we have to do is add 1.13 which is the probability of each rank 9 times. Alternatively, instead of this, you can choose to calculate different probability. What's the probability of opposite event? Card less than or equal to 5, which is probability of 2, 3, 4 and 5. We are just smaller number of these events. And the probability of each of them is still exactly the same, which is 1.13. So, the probability of the whole thing is equal to 4.13. Now, the probability of this event which is opposite would be 1 minus 4.13 which is 9.13 exactly as we had before. So, I basically divided the entire sample space into two different categories. Rank greater than 5 or rank is less than or equal to 5. And whatever is easier to calculate, this is slightly easier because the number of these is 4 whereas number of those is 9. So, it doesn't really matter. Not much of a difference. Ok, that's the second problem. Now, the third problem is let's say you are in the casino and the casino is offering you a new game. They give you 4 dice and they are saying if you roll and one of them or at least one of them will show 6 then you win 4 dice. Is it a good game for you or not? So, how can we basically decide this particular question? We have to evaluate the probability of having at least 1,6 out of 4 dice. And then we can just think about if probability is greater than half then it's fine. You can play this game as one-to-one which means you are betting one unit of currency and they are paying you one unit if you win or you are losing this one unit if you lose. So, that's basically how you should approach this problem. Now, if probability is different, I mean it's greater than half or less than half, then you should decide maybe you should just change the payoffs or bets etc. But what's important is to evaluate the probability. So, what's the probability of having at least 1,6 out of 4 dice? Well, in this particular case it's easier to calculate what's the probability of not having 6 out of 4 dice which means that the first die will really be either 1 or 2 or 3 or 4 or 5 and the probability of this is 5,6 right? The probability of the second dice to be non-equal to 6 is also 5,6 so it's the third and so is the fourth. They are all independent so the probability is supposed to be multiplied together so the probability of not having 6 on any of the 4 dice is this one and the probability of having 6 on at least one of those, that's the opposite of M, it's this one 1 minus 5 over 6 to the power 4 which is approximately 0.5177 which is greater than half. So, the probability of having at least 1,6 out of 4 dice is greater than 1,5 which means you can actually bet a bitcoin and expect a payoff of the bitcoin and you'll be definitely happy if you will play a significant amount of time. Now, what if it's only 3 dice? Well, with the 3 dice the power will be 3 here, right? So the probability will be 1 minus 5,6 to the third degree that's 0.4213 So this is not a good game for you, this is a good game for the house. If again the payoff is exactly the same as your bet then you would lose more than you win. So that's basically the solution Can it be done differently? Yes we can. You can use certain combinatorics to find instead of opposite event to find direct event. I just don't want to stop on it but if you want you can just send it and send it to me and I'll put it on my website as your solution Ok, the last problem is slightly more difficult. I would still consider it to be an easy because it's just one direct formula and you get the solution. So here is a very long explanation of what it is and then a very short solution There are 3 manufacturers which are making t-shirts. Now t-shirts are of 2 different kinds. Whites and non-whites. So the manufacturer makes 75% of white t-shirts. That's how its production is done. B makes 50% whites and C 25%. Now then there is a department store which buys wholesale from these manufacturers to sell to the public. Now here is how the department store buys from manufacturer A 25%, manufacturer B 35%, and manufacturer C 40%. So 25% of all shirts which are bought by department store are from this guy. And obviously 75% of them are white and the rest are non-white. Now 35% of everything which department store buys are from the manufacturer B out of which 50% are white and other 50% non-white. And then the same for C. Now some of these by the way is equal to 100%, 25%, 35%, and 40% which means there are no other manufacturers. Now you come to a store and you pick actually randomly a t-shirt. So what's the probability of this t-shirt being white? Well this is a typical problem on total probability. Now the total probability formula is basically based on the concept of conditional probability and the idea basically is very simple. So if you are buying a t-shirt, so what's the probability of buying a t-shirt? Well the probability of buying t-shirt is some of probabilities of being white and manufactured by the manufacturer A. Or it can be white manufactured by B or it can be white and manufactured by C. So these are kind of conditions which somehow must be taken into consideration. But we cannot really add these probabilities obviously because each particular manufacturer is not represented in the same way. They are represented in certain percentages right? So basically the formula here is you have to have the probability of this particular short made by manufacturer A and then you can multiply it by the conditional probability which gives you the probability of the shirt which is manufactured by A and white. That's what it is. Similarly here we have to multiply it by the probability of B and then you will get the probability of the short being white and being manufactured by B and being white. Same thing here. If you multiply it by the probability of picking the C you will have the probability of C and white. So these are actually the probabilities which you really should add together to get the probability of picking the white t-shirt. It's again very easily showing on this kind of graphical representation. If these are t-shirts, now this is white and this is white. So this is the graphical representation of white t-shirts out of the whole set of different t-shirts and obviously the area is equal to this area plus this area plus this area. This is A, this is B and this is C and this is white. So this is A and W. This is B and W and this is C and W and that's why we have this particular formula. And this particular formula actually since we don't know any of this but we do know these guys. And since this equals to this this is the classical formula of conditional probability. We can calculate each one of those corresponding to this and this. So we know all this and therefore we can very easily calculate the probability as sum of them. So this is the formula of total probability. This is pW equals to sum of these which is the probability of picking the t-shirt manufactured by A is obviously 25%, right? So it's 0.25 because 25% of all shirts which are bought by department store are from A. That's why the probability of picking the t-shirt from A is 25%, 0.25. And now we know that conditional probability of picking white which is manufactured if we already know it's manufactured by A is 75%. So we have to multiply it by 0.75. Plus this one. Since 50%, right? No, 35%. 35% of all shirts bought by department store are from B. So the probability of picking the t-shirt manufactured by B is 0.35. Now we know that if we know that it's manufactured by B then the probability of having white is equal to for B it's 50%, so it's multiplied by 50%. And finally the probability of picking a t-shirt manufactured by C is 40%, right? And out of those 25% are white. And that's the answer which is equal to 0.4625. So that's the solution. And this is the formula of total probability. P of W is equal to sum of these. Actually again we start with sum of these but each one of those equal to this as a formula of conditional probability. Okay, that's it. That was my last problem for today. I will probably try to put a little bit more problems into this category of easy probability problems. And then I hope I will have some time to put something for advanced problems. Just a little bit more difficult. I don't think we should really put at difficult really difficult problems into this course because the theory of probability is actually in its real capacity is introduced only in the colleges. But anyway for high school it's some kind of introductory and these formulas simple calculations whatever they are quite appropriate. So thanks very much and good luck.