 Hi, I'm Zor. Welcome to Unisor Education. We continue lectures on theory of probabilities within this advanced math for teenagers course. The course is presented on Unisor.com and that's where I suggest you to watch this lecture because it has notes on the side of the video and the notes basically contain the same material which I am right now presenting. But what's important is this particular lecture and many others actually are about problem solving which means that I would strongly recommend you to start with notes to this lecture and read just the problems, try to solve them themselves. That's the most important part of it and to tell you the truth, that's the purpose of the entire course to teach you how to solve problems and the only way to teach to solve problems is to solve problems. Now I will show my solutions and obviously you can come up with your solutions. You can send it to me in an email, I'll be more than happy to put it on the website as your solution with your authorship, etc. Alright, so these are a few easy problems on theory of probabilities So, let's just start. Alright, the problem number one is about climate change. Well actually it's about climate not changing. Let's assume that we have the probability of snow on a Christmas day, 25th of December, so it's a white Christmas and the probability of this event equals to 0.4. It doesn't matter how we obtain this number based on whatever theories or statistics or whatever, it's given. The probability is 0.4 of snow on Christmas day, on 25th of December. Now, let's assume that climate is not changing and this probability is exactly the same for the next 10 years from now on. Now, I'm interested in the following event. What's the probability to have snow on 3 out of 10 years? On December 25th. I'm not saying which 3 years out of 10, I'm just saying exactly 3. Now, first of all, we have a freedom of choice. Which 3 out of 10 is snowing? Which means that, let's just apply our knowledge of combinatorics, it's number of combinations from 10 to 3. We can take the first 3 years, the last 3 years, the middle 3 years, or the first, the 7s and the 8s, I mean whatever. And the number of these combinations is exactly number of combinations of 3 from 10, which is actually 10 times 9 times 8 divided by 1 times 2 times 3. That's the formula for number of combinations, right? Now, let's assume that our combinations 3 years out of 10 is chosen. Now, if I have chosen these 3 years, and I know that these 3 years are supposed to be snowing, the probability of this is 0.4 in each year. Years are considered to be independent from each other, which means that during these 3 years the probability should multiply. So the probability on all 3 of these years is 0.3 in the power of 3. What about the rest of the years? The other 7 years? Well, the other 7 years it must not snow, because I have chosen exactly the 3 years when it's snowing, and then the other 7 years are not supposed to snow. The probability of not snowing is obviously 0.6, right? If snowing 0.4. So the other 7, each of them has the probability of 0.6 of not snowing, and there are 7 of them. So, I have this number of combinations of 3 years out of 10. Now, this is the probability that the snow will go on these 3 years, and this is the probability of not snowing on these 3 years. And I have to multiply them together to get the total probability of some 3 years be snowing on December 25th and some others not. So the product of these, which is C10, 3 times 0.4, 3 times 0.67, this is the answer to the problem. That's the probability to have snow on December 25th on some 3 years out of 10. Now, what was important in this problem? Well, we actually made 3 decisions here. Number 1 decision was we had to choose which 3 years out of 10. Years are exactly the same. So if I'm saying on any 3 years is to be snowing, it means I have to choose which ones. And that's why I have to have this combination of 3 out of 10. Now, then, when my 3 years are chosen, no matter what they are, for instance, the first 3 years, the probability of snowing on each of them is 0.4 and events are independent year from year, and that's why we multiply the probability. So the snowing 0.4 and not snowing 0.6, and I have to multiply 3 of these and 7 of those. Okay, next. Next is more formal, not really related to any kind of a natural phenomenon. I have a variable, a random variable C, which takes value 1, 2, 1, 4, 1, 8, 1, 2 to the power of n, etc. So it can take an infinite but countable number of values. Is it possible? Of course it is possible. The only thing is I have to take the probability of this to be equal to something. Obviously, not constant because the sum of these is supposed to be equal to 1, right? So I have chosen this probability. So the probability of my random variable C to take the value 1, half is 1, half. The probability to take 1, fourth is 1, fourth. I mean, I'm basically the author of the problem, I can do whatever I want as long as the problem is correct. Now, is it correct? Let's just think about it. Is it possible that random variable takes these values? Yes, in mathematics infinity exists. This is the countable infinity and I can actually demonstrate geometrically how it can be done. Let's say you have a square, which has an area of 1 of something. I divide it by half. Then I divide it the half of this by half and then the half of this by half, half of this by half, etc. And I'm going the smaller and smaller and smaller, right? So this is 1, half. This is 1, quarter. This is 1, 8th. This is 1, 16th. This is 1, 30 seconds. 1, 64th. 1 over 128th, etc., etc. Now, the area of these rectangles, the areas of these rectangles are exactly these numbers. If I am continuing this to infinity. And if I drop a point randomly inside of this square, the probability to get into this piece, which has an area of 1, half, the probability is 1, half, right? If the whole thing is 1, this is half of this, so the probability is 1, half. So that's exactly the model of this particular problem. Now I have a few questions about this. Well, question number one. Is it a correct problem from the theory of probability standpoint? Which means if I will add up all the different probabilities for all the different elementary events, will I get 1? So if I will add all the probabilities, will I get 1? Let's put it as equals to infinity. This is the geometric progression with the first member, 1, half. The factor, when it's diminishing actually, the factor is also 1, half. And I know how to calculate the sum of this geometric progression. Now, obviously, I don't like the formulas, which somebody is supposed to remember. But I do remember the geometric progression when you would like to summarize. The very easy way is to multiply it by the factor. So s times 1, half equals to 1, 4s, 1, 8s, etc. And if I will subtract them, obviously all these will cancel each other. And I will have s minus s divided by 2 equals to 1, half, right? Which means 2s minus s equals to 1, which means s is equal to 1. So everything is fine. My sum of all the probabilities is equal to 1, which means the problem is basically correctly stated. So I have a certain number of variables. I have a certain number of probabilities of these variables. Sum of all probabilities adds up to 1, which is good. Next problem. Next problem is I would like to know the probability of my random variable to take values between 2 to the power of m and 2 to the power of n, assuming that m is, well, let's say m is less than n, but then I supposed to basically put it in reverse, right? Because it's diminishing, right? So if m is less than n, then 1 over 2 to the power of m is greater than my variable and this is greater than 2 to the power of m. All right. So how to calculate the probability of my random variable to be between these two values? Well, basically considering these values are completely unrelated to each other, these are elementary events, right? So this is just an event which can be represented as sum of events which are in between these two, which means I just have to add the probabilities for each elementary event which constitute this particular inequality. So I have to add the probability of c is equal to 1 to the power of m plus probability of my variable taking 2 to the power m plus 1, etc. Plus, and the last one would be probability of my variable equals 2 to the power of m, right? So I have to just summarize these probabilities of elementary events which constitute this big event. I know the probability of each of them, right? It's basically it's 1 over 2 to the power of m plus 1 to the power of 2 m plus 1, etc. to the power of m. The probability of each value is equal to the value itself as we were staging in the condition of the problem, right? So how can I calculate this? Let's do exactly the same thing. I don't remember the formula obviously as I said many times. I will multiply it by 1 half and I will get this. 1 over 2 to the power of m multiplied by 1 half would be 1 over 2 to the power of m plus 1, etc. Now I will have this and I will have this, right? So this would be from the previous prep member and to the power of m will give me to the power of m plus 1. Now if I will subtract, all these would go out. So I will have s minus s over 2 which is s over 2 equals this remains and this remains. 1 over 2 to the power of m minus 1 to the power of 2 m plus 1. Now I have to multiply by 2 to get s. It would be this. That stands, right? If I will multiply by 2 I will reduce by 1 this and reduce by 1 this. Now is it the correct formula? Well, let's just think about it. If my m is equal to 1 which means I'm starting from the very beginning and my n is equal to infinity which means I don't have anything at the end. So I'm basically summarizing the whole thing. I should have 1, right? Do I have 1? If m is equal to 1, m minus 1 is 0. 1 over 2 to the power of 0 that's 1. m is infinity which means it goes to 0. So that's true. Now what if m is equal to 1 and n is equal to let's say 2? Which means I'm adding up the first and the second members only. So I'm supposed to get 1 half plus 1 fourth which is what? 3 quarters, right? Well, let's just see. If m is equal to 1 that's 1 because this is 2 to the power of 0. Minus 1 over 2 to the power of 2 which is 1 quarter. 1 minus 1 quarter is 3 quarters. Exactly what we're supposed to be. So this is just a checking. I do recommend you whenever you get some formula or some result try to check it in elementary cases like this. All this helps. Now the third problem related to this particular random variable is what is the expectation? What is expected value of my random variable? Now expected value of the random variable is basically a sum of value which it takes times the probability plus another value times another probability, etc. So if I will summarize all values weighted with their probabilities I will get the expected value or mathematical expectation or average if you wish. Alright, so that's the definition of the expectation. Now in this particular case what does it mean? Well, we know that our random variable takes value 1 half with a probability 1 half. 1 quarter with probability 1 quarter, etc. Which means in this particular case it's equal to 1 half times 1 half plus 1 quarter times 1 quarter plus 1 eighths times 1 eighths plus, etc. 1 to the power of 2 n times 1 to the power of 2 n plus, etc. Now what is this? 1 quarter, 1 sixteenths, 1 sixtieths, etc. 1 to the power of 2 n, etc. It's also a geometric progression. The first member is 1 quarter and the factor is 1 quarter, right? So what I will do I will multiply it by 1 quarter and I will get what? 1 quarter times 1 quarter is 1 sixteenths, 1 sixteenth times 1 quarter will be 1 sixteenths fourths, etc. To infinity. So if I will subtract all these guys, oops, will cancel out and I have s minus s over 4 is equal to 1 fourth. So this is 3 quarters of s is equal to 1 quarter from which s is equal to multiply by 4 it's 3s is 1 so s is 1 third. So the average value of this random variable the expectation is equal to 1 third. That was the last part of this problem. That's it. And I have the very last problem for this lecture which is this following. Okay, let's assume you're trying to dial a phone number. It's a number of your friend but you forgot the last digit so you remember everything else but not the last digit. Well, you don't remember it which means you can try. There are only 10 different digits which you can try which means you can definitely succeed in no more than 10 attempts, right? So now my question is what is the probability to succeed in no more than let's say n attempts? Where n is some number from 1 to 10, right? Now what's the probability to hit the right number on the first attempt? Well, obviously 1 tenths, right? Because you have 10 different digits you just randomly choose any one of them and the probability obviously is 1 tenths that you get the right number. Alright, so the probability in case your number of attempts let's put it k is equal to 1 is 1 tenths. Okay, now what's the probability of hitting your right number on the second try? Well, it means that the first time you try you are wrong and the probability of this is 9 tenths, right? Because there are 9 wrong numbers and you choose one of them so the probability is 9 tenths. Then the next try you obviously don't try the same which you already tried so you have only 9 different choices from the remaining numbers, remaining digits which you didn't really yet dial and let's assume that you have succeeded. This is number of attempts is equal to 2 which means out of the remaining 9 numbers you hit the right one and the probability is 1 ninths. And the result is 9 is canceling, it's 1 tenths again. What's the probability to be successful on the third trial? Well, it means the first trial will be unsuccessful. Now remains 9 wrong, 9 numbers out of them is only one right and 8 wrong. And you made a second attempt and you made a wrong choice which means out of 9, 8 are wrong so the probability is 8 ninths. And then on the third attempt when you have only 8 remaining numbers you hit the right number and the probability is 1 eighths, right? 1 tenths again. As you understand, obviously the probability will always be equal to 1 tenths of succeeding in the fourth, the fifth, etc., etc., including the tenths. So the probability to be successful on N's try is 1 tenths and the probability to be successful on any number of tries not exceeding N, so from 1 to N would be obviously N divided by 10 because it contains 1, 2, 3, etc., up to N. Each one of them has 1 tenths the probability and we just add them up together. Well, that's the last problem. I do recommend you to go through these problems yourself right now. Use the notes on Unisor.com and that's it for today. Thank you very much and good luck.