 Hi, I'm Zor. Welcome to Unizor Education. I would like to spend some time solving very, very classical and very easy problems in theory of probabilities. Classical in the terms of they basically correspond to classical definition of the probability as a measure basically. We were talking about certain distribution of probabilities among elementary events and then we were choosing an event which combines certain number of elementary events out of the total number of elementary events and basically the proportion of the measure allocated to these chosen elementary events to measure the entire set of elementary events constitutes the probability. It's like number of good events versus number of total events, something like this. So these are classical concepts of theory of probabilities and I would like to spend some time solving a few very, very easy problems. These problems and this lecture presented on Unizor.com, that's where I suggest you to watch this lecture and the preferable way to approach any kind of educational material dedicated to problems is first try to solve these problems yourself, check the answer against whatever the nodes contain to this lecture. They are right to the right of the video image on this website and then you can actually listen to whatever I'm saying in the lecture because your approach to solving the problems can be different. Okay, so the problem number one is the following. In chess the game is on 8 by 8 board. On every cell you can put some piece and we are talking about rooks. Now if this is your chess board and this is one rook and this is another rook, one is let's say white another is black. Now in this position they cannot capture each other but if they are in the same horizontal line or the same vertical line then depending on which turn it is either this can capture this one or this can capture this one. So being on one horizontal or vertical line for the piece which is called a rook is important for being able to capture something. Okay, so here is the condition of this problem. We have this board and we have two rooks. One is white another is black and the problem is if I randomly put it on the board two rooks what's the probability of one being able to capture another? Or another word what's the probability of two rooks on the 8 by 8 board to stand in the same horizontal or vertical line? Now I started with something which is very important as I said that two rooks are randomly placed on the board. Now what the word randomly means? Well in this problem and in probably many many others if the word randomly is specified without concrete definition of how randomly it's distributed it implies actually the even distribution which means if I have 8 by 8 64 cells 64 squares on the board and I randomly place one particular piece on the board basically the probability of each is 164. So all probabilities are evenly distributed. Now that's easy to say when you are in a discrete distribution case. What if you are facing the distribution which is called continuous? Well in this case we are also implying that random means proportional to some natural measure which is already defined. For instance if I'm saying that a random variable takes random variable on and the segment from A to B it means it's basically proportion is proportional to the any kind of a sub-segment from it. So the probability to be within this segment is proportional to its lengths. The probability to be within the entire segment is one so the probability to be in any other segment is basically proportional to the ratio of this length versus this length. If I'm talking about random distribution of a point inside a square then basically it means that if I will take any kind of a figure inside the square which has boundaries. So the probability for the point to be inside this particular area is basically a ratio between the area of this figure divided by the area of a circle. So random means evenly. Alright so we have defined this thing. Now let's go back to our problem. Now the classical definition of the probability as I was saying before it's basically a ratio in case of discrete even distribution of probabilities. It's basically a ratio of good versus all. So in our case what we are saying what is good? Good is the position when two rooks can capture each other which means they are on the same horizontal or vertical line. Now all events is all the different positions of two rooks right? Now so let's just count one and count another. Let's start from the sample space. What's the whole set of elementary events? The whole set of positions of two rooks on on the board. Well for the first rule let's say the first is white doesn't really matter. We have 64 different positions and for the second we have all the other positions there are 63 left. So this is number of elementary events in our random experiment. Now which one of these are good? Well let's just think about. Whenever my white rook is in any of the points any of the squares on 8 by 8 board you have let me just draw it. So this is the point where my rook is. Now all these all these all these and all these are good places to put another rook so they are in the same horizontal or vertical space right? Now if the whole board is 8 by 8 it means there are 7 here and 7 squares here. Altogether I have 14 different positions of the other rook to be on so they can capture each other. So for each of these and there are 64 of them there are 14 candidates for the position of the second rook. So what I'm saying is that my probability in this particular case should be equal to the number of positions which we consider to be our good positions which we would like to count as candidates for our event. So that number of elementary events constitute the event which we are looking for event when two rooks can capture each other and we have to divide it by the total number of elementary events. And again we are talking about random which means in this case even distribution of probabilities that's why we just count the number of these events because each one of these events has the same probability. So the ratio is obviously 1463 which is 2 9s. So that stands with the probability of 2 9s which is not a small probability if I just drop two rooks on the board the probability 2 9s that they will be able to capture each other. They will be in the same horizontal or vertical line. Now again I would like to emphasize the classic approach here total number of events in the denominator and the number of good events in the numerator of the ratio that's what gives you the probability in case the probability is evenly distributed among all the elementary events which we have actually said from the very beginning. Number two we have 26 letters of English alphabet from A to Z. Now let's consider that I randomly put all these 26 letters into a sequence one letter another the third one etc. Randomly again randomly means that any particular position of these 26 letters in one line has exactly the same probability as any other. So let's just start from the from the very beginning how many different elementary events we have that's how many permutations we have from 26 letters which is 26 factorial right so that's the number of permutations now fine that's okay let's go further now we are talking about number of only those positions of these 26 letters when letter A precedes letter Z not necessarily just somewhere is letter A but Z should be to the right of this letter okay so in this particular case let me just make this test a little bit easier we really don't care about all other letters BCD etc. We care only about two letters A and Z which means we can just forget about 26 different letters we can just concentrate on these two however we would like to say that these two are supposed to be positioned among 26 different places right so let's forget about what we had that we have some other letters we completely disregard them for this particular problem so let's just concentrate on these two so we have 26 positions and two letters A and Z can actually be in in any positions among these 26 now so how many different positions of two letters can be if you have the 26 places and only two letters well obviously this is the number of combinations we have 26 places where letter A can be positioned right and Z should be positioned on any other and the others are 25 so this is number of different elementary events and so we instead of this we have reduced the number of elementary events to this number and obviously the probability of each of them is 1 over 26 25th 26 times 25th all right so these are the number of elementary events which we actually care about now the number of good elementary events are those when A precedes Z now let's count them so we have counted how many elementary events we have all together any position of A and any position of Z by the way if you start from Z it will be exactly the same any position of Z and then the A will be on any of these positions so it doesn't really matter how to how we start now how many positions satisfy our condition that A is preceding B Z sorry well for a we have 25 different places right number one number two number three etc these are places a can be here can be here here here and here a cannot be on position 26 because there is nothing to the right of it and Z has no place right so for a you have only these positions from 1 to 25 now if a is positioned here how many different positions for the Z are this this this this this this so 26 minus 1 which is 25 positions if a is here Z is supposed to be to the right of it and there are only 24 now different variations right now a can be here in which case I have 23 etc and if a is here I have one position for the Z which is to number 26 some of these is actually the number of elementary events which satisfy our condition of a preceding Z okay what's this sum well this sum is equal to you know what it is right the arithmetic progression if you don't remember the formula you have to write it in opposite order and sum together this is 25 sorry so you have 26 26 26 in each pair right so you have 25 times 26 but this is 2s so you have to divide it by 2 so that's the answer right so my answer number of positions which number of elementary events if you wish which satisfy our criteria is this and this is the total number of elementary events of all the different positions of the letters a and z amount 26 places so we have to divide this by this and obviously as a result you have one half now why did I actually go through all these troubles let's just think about it differently a and z can be positions can be positioned either a precedes z or we can always change them and z would precede a so the number of elementary events when a precedes z is exactly the same as number of events when z precedes a right because for each of these there is each of those which means that the total probability one should be evenly divided between a precedes z and z precedes a so that's why we have one half which we can actually come up with just through this logical explanation but again I wanted to solve this problem through some kind of classical approach number of elementary events and number of elementary events which satisfy our condition and take the ratio just as in the illustration alright next problem okay next problem is you take only 16 cards out of a standard deck you take four jacks four queens four kings and four aces okay reduced deck if you wish now we can actually put these cards in any order what I'm interested in I'm interested in the probability of randomly putting them in the order which is four jacks then followed by four queens followed by four kings followed by four aces which means they will be in the rank order and I don't care about suits I mean hearts spades doesn't really matter as long as I have four different jacks followed by four different queens followed by four different kings and for four different aces that's okay with me and question is if I randomly have these 16 cards and put it in in in a line one after another what's the probability of getting this particular order that they will be in a rank order in an increasing rank order okay let's just think about it again we have to consider the sample space which contains all the different elementary events which are obviously 16 factorial number of all the permutations of these 16 cards now let's think about which of these permutations are good for us well let's just think about it I have four jacks in the very beginning but they can be in any order so it's not just one particular order it's actually a four factorial all different permutations of these four different jacks is good for me now with each of these I can have four different permutations of queens and with each of them four different permutations of king and four different permutations of ace so that's my answer because this is the number of good combinations when the order is preserved and this is the total number of all the combinations and if I will divide one by another I will get the probability so it's four factorial to the power of four divided by 16 factorial which actually I have approximately 1 over 63 now it's not approximately it's exactly more than 63 million so 1 over 63 million is the probability of this type of a random position of my 16 cards okay and last one is an example of continuous distribution so before we were dealing with discrete distributions and we were basically counting elementary events good elementary events which are supposed to be combined together into event we are interested in the probability of and the total number of elementary events and since we are talking about even distribution of probabilities the word random was used without any specification of how random which means it's evenly it's assumed it's even now we are talking about continuous distribution and this is a very simple example but I think it's very educational let's consider you have a circle of a diameter R and you inscribed a square with diagonal R R right so my question is if you take a point and randomly throw it inside this particular circle what's the probability of this point to be inside the square so as I was saying before if you have something like a continuous probability you really have to use some kind of a Nate natural measure of the events now in this case measure is actually area so if I'm talking about randomly throwing a point inside the circle it means that the probability for this point to be inside any kind of a figure is proportional to its area so the area of the whole circle if you have a diameter air R is what pi and radius is r divided by 2 so it's pi r divided pi square divided by 4 that's the area of entire circle now what's the area of of this square but if diagonal is R then the side is R divided by square root of 2 and I have to square it to get the area right so this is the area of a square and this is the area of a circle and they have to divide one by another to get the probability so it would be R square divided by 2 square root of 2 square divided by pi R square 4 obviously this is reduced this is 2 is equal to 2 over pi so this is the probability pi is like 3 and 14 so it's about what 0.7 something 0.6 0.6 it's less than less than two-thirds okay so it's about 2.6 0.6 the probability to get inside the square if you randomly put drop the point inside the circle randomly means evenly proportionally to the area okay that was my last problem again all these problems are easy I do recommend you to go to Unizor.com and review them again just solve yourself without looking at the notes and then check against the notes and all of these problems are about the same thing it's a classical definition of the probability as a measure measure of good versus all measure of whatever we are interested versus whatever exists basically and in case of discrete events we usually just count all the elementary events which constitute our event we are interested in and divided by the total number of elementary events because the measure is distributed evenly in case of continuous distribution like this we have to think about natural measure of the subject of our distribution in this case if I'm talking about a point which is dropped inside the circle randomly it means that the distribution is proportional to the area that's it for today thank you very much and good luck