 Hi, I'm Zor. Welcome to Unizor Education. We continue solving problems on probabilities, easy problems. I do recommend, as usually, go to the website Unizor.com to this lecture. This is lecture number four among easy probability problems. And try to solve these problems yourself. There is a solution provided as a note for this particular lecture. But I do recommend first to try to solve it yourself. Then you can read the notes, you can listen to this lecture, maybe you can come up with your own ways to solve these problems. Alright, so three problems. Number one. Okay, we have three students who are preparing for exam. Exam contains 100 questions. Now on the exam, the student comes and basically pick one of these 100. Now either he is prepared or he is not prepared for this particular question on the exam. So here is my statistics. The first student, column S1, he is a lazy guy and he is prepared only for 50 questions out of 100. Now the second student is kind of busy, he is reading some very interesting novel. So he didn't have much time, he prepared for 75 questions out of 100. And the third student, he is actually a good student, but at the very last moment his girlfriend called him and he did not finish his work, his preparation. He is prepared for 90 questions out of 100. Now I have a couple of problems related to this situation. Problem number one. We need the probability of all students got familiar question. The one which they have been prepared for. What's the probability of this? Okay, let's think about it. Now we have basically three completely independent events. The first student picks the question out of 100, the second one and then the third one. Now if the first student has 50 questions which he is familiar with out of 100, then obviously the probability of the first student to pick the familiar question is 50 over 100. Now for the second student, the probability is corresponding with 75 100s and the third one is 90 100s. Now what we are interested is, we are interested in all of them, all these three students got the familiar question, which means this event and this and this event. Now since events are independent, the probability of the combined event is equal to the product of the corresponding probabilities. You know that. So the probability is equal to the product of these. Whatever it is, this is one half times three quarters times nine tenths, right? So what is it? Twenty seven eighties. That's the probability. Now question number two. What's the probability of all got unfamiliar questions? Well, these are opposite events actually. The first one to get unfamiliar if he is prepared for 50, then unfamiliar is also 50, so it's 50 100s. For the second one, if he is prepared for 75, he is not prepared for the last 25, so the probability to pick one of those would be this and for the third one would be this. And I am multiplying them again because these are independent events and the probability of the combined occurrence of the three different independent events is equal to the product of their probabilities. So that's the answer. Next one. Now what's the probability of only one probability of one out of three students to have an unfamiliar question? So one of them picks unfamiliar question but others pick one of those which they have prepared for. Alright, this is a little bit more complex and that's why I suggest we go to the basics and let's just deal with events and their probabilities and opposite events, etc. So what's the events which we can consider? For student one, there are two different independent events. One is good when he is choosing one of the familiar questions called G1, good for the first student. And the other event is if he is picking the question which is not familiar, that's the bad one. B stands for bad. Now for the second student also there are two events and for the third student, good being picking the familiar question, bad being unfamiliar question. And we know the probabilities of each of these events, right? So this is 0.5, this is 0.5, this is 0.75, this is 0.25, this is 0.90, this is 0.10, right? The opposite. Now what is this particular event? One out of three students have an unfamiliar question and the other two familiar questions. Well, there are three different possibilities and they are mutually exclusive. So the event which we are talking is either the first one would be the unlucky guy who will have the unfamiliar question, which means it's the B1, it's the bad for the first student, but the other two are good and they are independent. So we have to really have, let's put the word end or set theory, intersection sign. So the two is good and the three is good too. So that's one event, then mutually exclusive event when it's the second one which is unlucky, which means the first one gets the question which he is prepared for. The second one is unlucky and the third one is the lucky one. And finally, the third, again, mutually exclusive event is a combination of first one is lucky, second one is lucky and the third one is unlucky. So we need the probability of this combined event. Now let's just think about it. What is this? Why did they put pluses here actually? I suppose, well actually I should put or in this case. That would be better from the probabilistic standpoint. But pluses is good too because in this case we are talking about mutually exclusive events and the probability of or on mutually exclusive events is the sum of their probabilities. It's basically like a measure theory. If you have certain area and we have two different sub areas which are mutually exclusive, which means they are not intersecting, the area of the sum of these or or of these is equal to the sum of the areas. So in this case it's sum of these. Now inside of each of them, so basically I can put return back to the plus, it would be plus, plus probability of this guy, plus probability of this guy. Now inside each probability I have the end condition basically, right? This is intersection which means end. These are independent events and that's exactly how it was in the previous problem. The probability of combined event is equal to the product of their probabilities because they are completely independent. So basically I can, instead of this, I can put the multiplication. It's the probability of D1 times probability of G2 times probability of G3. And same thing here. Times probability of this, times probability of this, probability of this, multiplied by probability of this, multiplied by probability of this. So that's the final formula. Product of these three probabilities, plus product of these, plus product of these. And we know all of them. Like D1 for instance is equal to 0.5. G2 is equal to, it's a success which means 0.75. G3 is equal to 0.90. This G1 is equal to 50.5. B2 is 0.25 because there are 25 unfamiliar and this is 0.9. This is again 0.5. This is 0.75 and this is 0.1. So that's the answer. And finally the last one about the same thing, the same problem, is probability of at least one student will get an unfamiliar question. At least one student gets unfamiliar question. Now what does it mean? Well, let me just formulate it differently. It's the opposite to none of the students get unfamiliar question, right? At least one gets unfamiliar which means once or two or three. It's just an opposite to the event which we can very easily calculate, the event when none of them gets unfamiliar questions, or all of them get familiar questions, right? So basically it's the one-linus probability of all get familiar question, which we have already calculated in the very beginning that's the first task, which is the product of these three. So it's 1 minus 0.5 times 0.75 times 0.9. That's the answer. Okay, that's my first question. Well, first problem I should rather say. It had four different questions. Now the second problem also has two different questions and they are very much look-alike. And what's interesting is even the results are the same. However, the logic which goes is slightly different. Actually it can be different. All right, so you have a standard deck of cards, 52 cards from 2 to 10 in four different suits. Spades, cards, diamonds and cloths. And then you have Jack, Queen, King and Ace also in four suits. Let's call these guys numerics. And these guys I will call pictures. So I have two categories basically. Forget about the cards. I have two categories of objects. I have from 2 to 10 it's what? I have 36 numerics and I have 16 pictures. That's it. That's all you want to know. Basically it can be 52 different balls, 36 white and 16 red or something like this. Doesn't really matter. Now what I'm doing is I'm picking up two cards, but I'm picking up differently. My question number one is, what is the probability of second card which I picked from the deck to be numeric if the first is numeric? And I'm not returning back by the way, cards. That's very important. So I pick first one and that's the condition. It's numeric. I put it aside. Now from the deck I'm randomly picking up the second card. What's the probability of this card to be numeric? Well, this is actually very easy to solve logically. Let's just think about it. If I have already extracted one numeric card, I have out of 36 numeric, I have only 35 left. So I have 35 numeric cards left. Now the whole deck contains obviously 51 card because it used to be 52, but I took one away. So the probability to get numeric and there are 35 numerics out of 51 card is 3551. That's the answer. Such an easy problem, right? I would like, however, to approach this problem in a more theoretical way if you wish because it will be very useful for the next problem. So what's my theoretical approach? My theoretical approach is, okay, let's just think about what's my sample space? What's the probabilities of each elementary event which is constructing, which is making up this sample space? And then we will start from that. Well, basically my sample space is different combinations of numeric and pictures to be among the two picked cards, right? So what kind of combinations I have? I have NN, numeric, numeric. I have numeric picture. I have picture numeric and I have picture picture. That's what I can get if I pick two cards one after another from the deck, right? So let's call these A, event A, event B, event C, and event D. Now, what's the probabilities of this? Well, let's just think about it. First of all, the total number of different combinations of two cards which I pick out of the deck of 52 is 52 times 51 if we are actually considering them to be an ordered pair, right? We are talking about ordered pair. So the number of the different cards which I can pick on the first pick is 52 and on the second is 51. So their product is the total number of ordered combinations of pairs. So it's 52 times 51. Now, how many combinations are of these? Well, it's 36 for the first one and 35 for the second, right? So it's 36 times 35. This is 36 for the first and picture is out of 16, so it's 16 for the second. This is 16 for the first and 36 for the second and this is 16 times 16. Now, if you will add these together in theory, you have to have the same number as this one, right? Because that's all the different combinations. Well, you know what, I actually would like to do it. I think it's very educational and just to check myself. So if I add them up together, I will have... Let's add these two. I will have 36 times 35 plus 16, which is 51. Let's add these two. That's 16 times 52, right? Which is... Well, let's call it 51 plus 1. Okay. So if I add them up together, I will get... So this is 61... 16 times 51 plus 16. If I add these two, I will have 52 plus times 51. Now, something is wrong here, right? I think I did not really add it up correctly. 16... Oh no, of course not, because it's 16 times 15. The first one is picture and then we have only 15 pictures left, right? So this is 36 times 51 and this is 16 times 51. And if I will add them up together, I will get 52 times 51, which is this. Okay, everything is fine. Now, let's continue. So we know the probability of this is 36 times 35 divided by 52 times 51, etc. So we know the probabilities of all these events, A, B, C, and G. Now, what are we interested in? We are interested in the second to be N numeric if the first one is N. So the first one is N is a condition and we are looking for a conditional probability of the second to be N, which is this and this, which is A or C. Under condition that the first one is N, which is these two, A or B. Now, let's think about it. What is the conditional probability? Let's just remember our theory. The conditional probability, let me just remind you. Let's say this is our sample space. This is my A and this is my B. What is conditional probability of B under condition of A? Well, basically you take only, you consider whatever is inside A to be basically a new sample space and whatever falls inside A, which is part of B, is basically the events which we are interested. But we have to know which part of the whole area, which is area of A, is taken by this particular common part. So it's P of A and B divided by B of A. This is conditional probability of B under condition of A. That's what it is. That's basically the definition. So now let's think about this. So our condition is A or B that goes to the bottom. I shouldn't use A and B. Let me use X and Y that would be more appropriate because A and B I'm using here. So this is X, this is Y, so this is X and Y divided by X and this is the probability of Y under condition of X. Now, the X is our condition. X is this and Y is this. So what's the numerator of this? A or B and this is A or C. Now A, B and C are usually exclusive, right? A, B, C and Z. So if I'm or A or C and I intersect A or B. So obviously I have only A, right? So this is A. And on the bottom I have A or B, the probability of X which is A or B. And A or B again, because they are mutually exclusive, their probabilities are added together. So A or B we should add together. Equals. Now what is it equal to? Well, both of them are this divided by this. So this denominator is mutually cancelled out. So I can basically compare these numbers only. So P of A which is 36 times 35 divided by P of A plus P of B which is some of these which is 36 factor out times 51. So it's 36 times 51. Now 36 is cancelled and I have again 35 over 51. Exactly the same as I have just received using my logic. Like considering I have already picked one numeric, I have only 35 left out of 51 cards. So we have exactly the same answer but more I would say theoretical, classical approach, etc. And why do we need it? For the next problem. So it's more significant to have this particular approach because there is no proper logic like in the first problem. So let me just again write down my events. This is numeric and numeric. B is numeric and picture. C is picture and numeric. And D is picture and picture. And the numbers are 36 times 35. 36 times 16. 16 times 36. 16 times 15. Out of the total 52 times 51. Now here is the second problem. Now remember the first problem was what's the probability of the second card to be numeric if the first is known to be numeric. Now I'm reversing the situation. What's the probability of the first card numeric if... Let's consider I picked the first card and put it aside and I don't know where it is. But now I picked the second one and I open it up and it's numeric. So now my question is what's the probability of the first card which we don't see to be numeric? Well, it's not exactly the same as the previous problem as you see. And I don't see the same kind of very easy logic which would allow us to come up with an easy solution. I mean maybe there is one. I just didn't come up with it. So I decided to go using the heavy artillery. Exactly the same approach as I did before. Now in our case what am I supposed to find out? I'm going to find out the conditional probability of the first one to be numeric which is a plus b, a or b, under condition that the second is numeric which is a and c. So you see formula is very close to the one which I was using for the first problem. In the first one it was a or c under condition a or b. Here it's basically condition is the conditional probability and whatever was in conditional probability becomes a condition. Alright, now there is no much difference. I'll do exactly the same thing as I did before. Which is the probability of their intersection and their intersection is probability of a divided by in this case that's probability of the condition which is probability of a plus and they are mutually exclusive the probability of c. So the formula is always the same except it's p of c probability of event c rather than probability of event b. That's the only difference. But now let's just think about it. b and c have exactly the same chances to occur. So it's either the first one being numeric and the second picture or the first picture and the second numeric. So the probability of both of them is exactly the same and that's why the answer is exactly the same. Which is 36 times 35 divided by a and c. 36 times 35 plus 16 which is 35 first. As you see the probability is the same as the first question was. But it's a completely different problem, completely different approach. Whatever was the event which we are interested in becomes a condition and whatever was a condition becomes an event which we are interested in. But it's the same answer quantitatively. So that's basically the only interesting thing. Well actually I do have another problem in the same general setting. So again I have the deco 52 cards some of them are numeric some of them are pictures. Now question is if I'm picking the cards one after another one after another. What's the probability of the last one to be numeric? Well in this case let me just approach this using again very simple logic. It's basically equivalent to picking up one particular card and basically call it the last one. Which means everything else I'm just choosing before it and then leaving this one as the last one. And the probability obviously is equal to how many different numeric cards. There are 36 numeric cards out of 52. So that's the probability which is what 9.13. That's easy. Now I have the problem which also involves a deck of cards 52 cards. But now this is more closer to those who play bridge. You know the bridge is played by 4 people. 52 cards are divided among 4 people 13 each. And now I am interested in the following distribution of cards. So I would like to have the probability to have one particular hand to have all 4 aces and all 4 kings and all 4 queens and 12 in one and one jack. So if you're playing bridge and this is the game without the trumps without the trumpsuit this is the highest combination possible right? All 13 cards are the top cards in each of the suit and there is one extra jack. That's as much as it can be. So what's the probability to have to pick 13 cards? Let's say I'm playing one of the players so I'm asking what's the probability of me to have these 13 cards so I can win the game without the trumpsuit. Alright that's actually quite easy. The total number of all the different combinations of 13 cards out of 52 is obviously which is yeah it's 52 factorial divided by 13 factorial and 52 minus 13 39 factorial. So that's the total number of all the different combinations of 13 cards which I can get in theory right? Now which of them are good? Well the combination which contains all these cards and these are fixed basically and this is the only card which I have a choice. Either it should be jack of spades or jack of hearts or diamonds or clubs so there are 4 different variations. If these are the total number of different combinations of 13 cards and 4 of these combinations are those which I'm interested in well that actually makes the probability equal to 4 divided by number of combinations from 52 by 13. So that's the answer. And that completes my lecture. I would certainly suggest you to go to the website Unisor.com again and look at the notes to this lecture. Especially notes where I am talking about all these different events related to numeric and picture, ABCG, different conditional probability, the theoretical approach which works in those two cases. It's very educational and it actually brings some clarity into your understanding of what actually the conditional probability is. So that's it for today. Thanks very much and good luck.