 So law of total probability law of total probability as the name itself says it helps you to find out the total number of probability of occurrence of an event. Okay. Now what is this entire theory, let's discuss it through a simple example I'll introduce it you to this concept so simple in a case study very very simple case study. Okay. Let us say there are three bags. Okay, there are three bags back one back to back three. Okay. Now, let us say there are three green balls here to red balls here. Four green balls, six red balls, one green ball, maybe three red balls. Okay. These all balls, even though they are the same color, they're all distinct. So these are three distinct green balls to red, two distinct red balls. Same goes with bag number two and bag number three. Okay. Let's say this bag is kept in a dark room. Okay, this bag is kept in a dark room or a room where you cannot see properly. Okay. So there is a person who enters this room. Fine. And he puts his hand in one of the bags. Okay, he's not sure which bag he's putting in. And he basically gets a ball. What is the probability that the ball will be a green one? Question is clear. So let's say this is kept in a very dark room. There is no light in that room. And there is a person or you can see there is a blindfolded person. Okay. There is a blindfolded person. Okay. Just like we have in Mahabharata. She was blindfolded. Okay. So she goes and she just puts her hand in one of the bags. She doesn't know which bag she's putting her hand and she gets a ball. What is the probability that the ball would be a green ball? What is the probability of getting a green ball from this combination? Okay, Anusha. Okay, Hariharan. Very good. Very good. Okay. So Hariharan, what's your answer coming out to be? Can you just calculate your answer and tell me? Aniruddha, Anusha and Hariharan. So, Aditya, I think you can further simplify it. You mean 5 by 12, right? Okay. Okay. So I'm getting varieties of answers. Okay. I'm getting varieties of answers. I'm getting answers like 8 by 19. I'm getting 4 by 19. I'm getting 5 by 12. I'm getting 8 by 11 also. Okay. Hariharan, what's your answer? I mean, would you like to simplify it and tell us? Okay. Okay. So he has given some raw figures or numbers. So I wanted him to calculate and tell it to us. Well, okay. Now see here, try to understand the question. How is the situation different? Or how is the situation similar? Are you having a single bag? And in that bag, you have around 8 green balls. And let's say 11 red balls. Are both the scenarios same or both the scenarios different? Should be the same? Okay. All these bags are of the same shapes and sizes. Right, right, right. Now see, this is a question which I would like to ask you in a very interesting way. Okay. Let's say South-speaking languages like, you know, Canada, Tamil, Palyalam. Okay. Now, if I say, what is the probability of you picking up a person who speaks a South Indian language, which is one of these languages from the entire country? Okay. And if I asked you that what is the probability of you choosing? Okay. You know, a person who speaks Canada from one of the random states that you have picked up. So now here, the problem is the chances of you picking a state is equal probable, but there are some states where the density of people speaking these languages is considerably high. Right. So when you're picking the entire country as, you know, a bag and you're trying to pick up. Will that scenario be same as you allowed to pick one state, you know, with equal probability and then trying to pick up a person who speaks the one of the South Indian languages. Of course, there, the concept of potential will come into picture. The chances of you picking up the states are equal, but the chances of you getting a person who speaks these South Indian languages will be more in view of the states as compared to the North Indian state. Right. So what you're doing when you are mixing this entire ball in a single bag is like you're giving them all equal potential that everybody has an equal chance to be picked up. But no, in this case, I have created. I have created small, small, small, small packages with higher and lower densities of these balls. Are you getting my point? Are you getting my point? So you getting a red ball will be, sorry, you getting a green ball, the chances will be high in the first bag as compared to the last or as compared to the bag number two. Right. So you cannot treat this question as if every ball has an equal chance to be picked up because now they are concentrated in different, different segments. Right. So you solving this problem by treating it as if every, I mean, there's a single bag and all the balls have been put in that and you're trying to get a green ball is different from you getting a green ball from this combination. Even though the chances of the bags being picked up is equal, but every bag, the concentration of the green balls is different. That makes the problem considerably different from what you would otherwise do considering it as a single bag with all the balls inside it. So here the probability of picking a green ball would be, you'll say, okay, let's say the person, let's say Gandhari puts her hand in bag number one. That chance itself is one third. Okay. And then getting a green ball from that bag will be three by five. As you can see, the chances is very high if she gets puts her hand in the first bag. So it's a factor of both the things. Right. Then one third is a chance that she picks up bag number two or she puts the hand inside bag number two and then gets a green ball that is four by 10. Then again, one third, the chance is that she will pick up bag number three and then picks up a green ball is one by four. Okay. Yes. The concentration, the density, the potential of green balls is differing from every bag, back to back. It is not like, see, what do I say? You're not mixing all the things. Let's say when you're making a lemonade, there is a high concentration of lemonade in one bottle and there is water in the other. Right. When you mix it, of course, the chance that the mixture that you get or the solution that you get, there's an equal chance of, you know, water and lemonade to be present in every, you know, you can say a centimeter cube of the solution, but not when you are, when not even they are separated. So in silo, they are in different concentrations. Okay. What is this one third here? This one third is the chance of picking the bag itself. Okay. Now, why is this one third, one third? As I told you, the backs are of equal sizes. Correct. But let us say if I have a situation where bag number one is almost double the size of bag number two and bag number three. Let us say, okay. Now you choosing or maybe let's say your chance of choosing bag number one is higher because it is a bigger bag, isn't it? Right. So in that case, this one third will get changed depending upon, you know, in what ratio is the, or what is the chances of you picking bag number one vis-a-vis bag number two and bag number three. Okay. So it's obvious that if there's a bigger bag and you're putting your hand inside, it might go into the bigger bag as compared to the smaller bags, isn't it? Let's say you're, you're blindfolded. You can't see anything. Okay. And you have been given a table which has, which is just scaring these three bags and bag number one is bigger. Let's say the probability of choosing bag number one is two-third as compared to the probability of choosing the other one. So let's say one-sixth, one-sixth and two-third. So this one-third will take care of that fact that your hand goes more in the area where there is a combination of three green and two red balls. Okay. So this entire probability is what is an example of a total probability. That means you're trying to figure out in how many ways or what is the total probability or chance of getting a green ball from this. You can say, you know, three different banks which have, you know, the green balls present in it. Okay. So this answer I think comes out to be somebody who was saying five by twelve. I'm not calculating it. I'm just relying on your result. Is it correct? Five by twelve only. Hariharan and, okay. Thanks. Now the same thing I'm going to now formalize as a theory. So I understand my, this thing, purpose is to tell you that this is the hidden principle. This is the thought process that you should be using to solve these problems. Formula or just, just, you know, numbers and figures that is, that looks good on your examination paper. But behind the scene, this thought process should be going on. Okay. Behind the scene, this thought process should be going on. So we'll, we'll talk about a theory with respect to this particular law of total probability. So let's talk about a theory here. See, let's say there is a, there is a set of mutually exclusive and collectively exhaustive events. Let's say E1, E2, E3, etc. Okay. These are set of mutually exclusive and collectively exhaustive events. Collectively exhaustive events. Okay. Now let's say there is an event A. Okay, which occurs, which occurs in combination with or along with, with these events, even E2, E3. Now just to give you an analogy, your even E2, E3 are basically your three bags. The chances of you picking a three bags, the event of picking up a bag one is even, event of picking a bag two is E2, event of picking a bag three is E3 and so on. Let's say there are only three bags. So I'm just comparing this, this particular, you know, statement with the bags and A is like you picking a green ball. So green ball occurs with even also. So green ball is present in bag number one also green number, green ball is present in bag number two also green ball is present in bag number three also. Now, if you try to see the situation through a when diagram. So let's say even E2, E3 are like your, let's say these segments. As you can see, I am making some segments which are mutually exclusive. Okay. And collectively exhausted. Collectively exhausted means they together make the entire sample space. Okay. So let's say this is your even. This is your E2. This is your E3. And this is your EN. Okay. And there is an event A which is occurring along with all these events. So let me show that with the diagram. Yeah. Okay. So let's say this over shaped structure is your event A. Okay. Now, any question anybody has? Yes. I'm coming to the 31 by one. Don't worry about it. Okay. Now let's say the question asks you what is the probability of occurrence of A? Okay. What are the probability of occurrence of A? Okay. So from this diagram. Or forget about probability. Okay. Can I say A is. A is the union of this part. This part. I'm just sharing it in different different, you know. This part, etc. Correct. So what is this part? This part is A intersection even. Please note that the entire A is not present only in even it's a spread up along these even C1, E2, E3. So this part is A intersection even. What is this part? This part is A intersection E2. Okay. What is this part? This part is A intersection E3. And can I say that the last one, the last part will be A intersection EN. Correct. Can I say event A is the union of all of them. Can I say the language of sets? Can I say A is the union of all these. Does everybody agree with me on that? Okay. Now, can I say PA will be nothing but or NA will be nothing but some of the NAs or some of the PAs over here. Why? Because each of these, you know, shaded part that I have shown they are mutually exclusive. Okay. So if they're mutually exclusive, anybody has any doubt so far. Now, if you remember in the last class, I had written something very similar in the denominator. I will come to that again. Don't worry about it. Now, as per multiplication theorem, as per multiplication theorem, can I say this expression, the first expression can be written as PE1 into PA by E1. Last class, we had done multiplication theorem. Some of you were absent. So for them, I'll just write it down. So multiplication theorem says that probability of occurrence of A and B is probability of occurrence of A into probability of occurrence of B given A has occurred. Okay. So this is your basic multiplication theorem applied to two events. So in the same way, all these probabilities I can start breaking up up like this. And this goes on till the last event PEN into PA by E1. Okay. So this is your PA. Right. So this is something that we call as law of total probability, law of total probability. But again, it comes from your common sense. It is not like, it is out of your logic. Okay. So the same thing I had written over here also. If you see, let's say there were only three events associated with this, you know, event A, then your bag number one, picking bag number one was event E1. So the chances of picking bag number one was one third. Given that you are picking bag number one, given that you have picked up bag number one, what are the chance of you getting a green ball? That was three by five. Okay. Yes or no? Same thing. This was again one third. And the given that you have picked up the second bag, what are the chance of getting a green ball? Four by ten. Basically, we saw the previous problem also. Okay. So this formula is called the law of total probability. Probability of two consecutive events, like two dice. Yes, of course. These two events both will be a part of your probability of A, occurrence of A. So you are basically collecting all the parts together and finding the total probability. Hence the name also given to it. Okay. Yes, we did it in the, yes, we did it in the denominator part of our previous question. Very well identified, where we are trying to find out what is the total number of, what is the total probability of you getting N1, N2, N3 in such a way that I to the power N1 plus I to the power N2 plus I to the power N3 will become one. Exactly. You have identified it exactly in that. Okay. We'll talk about it. The story has just begun. We'll be talking more about it. And we'll be seeing how is this used in a conditional probability, which was actually taken in the question last class. The root is why a base theorem, which is also going to be taken through a similar theory. Okay. Very similar theory. Meanwhile, let's take few questions just to get you started on this concept. Okay. I'll just take a very simple school level problem here. Okay. There is a purse which contains four copper and three silver coins. Another purse contains six copper and two silver points. Okay. One coin is drawn from any one of the two purses. Okay. But here, please note that the chances of you picking the two purses is equally probable or equally likely. Find the probability that it is a copper coin. Okay. Let me see your understanding by your response on the poll. So poll is on. We covered independence. We covered condition. Oh, my God. Very good. Within 45 seconds, 10 of you have responded and that too with one of the options only. Great. So I think 30 more seconds should we wrap this up? Okay. Five two one. Okay. Okay. Just there was a one spoiler. Somebody has pressed option C. But 96% of you have said option D. Okay. Let's check. So what are the chance? So now, again, whenever, whenever, see this topic may come in your semester to exams and that exam may be subjective in nature. So many people ask me, sir, how do you write these, you know, expressions and use the formula, et cetera. So let's define the events. So even is the event of you. Okay. Picking or selecting purse one. Okay. Picking or selecting purse one. Okay. Picking or selecting purse one. Picking or selecting purse one. Picking or selecting purse. Okay. Choosing purse one. E2 is the event of choosing purse two. And a is the event of getting a copper coin. Okay. Drawing a copper coin. So, as per, or you can say by LTP law of total probability. The probability of you. basically these are all cases, you can see that. Okay, so probability of you choosing a purse, one is one half. Now given that you have chosen purse one, the chances of you getting a copper point from there is four by seven. Then probability of you getting purse two is again half and the chance of you getting a copper point from there is six by eight. Six by eight is three by four, correct. So let's simplify this. So this comes out to be two by seven if I'm not mistaken. And this comes out to be three by eight. That comes out to be 16 plus 21, which is 37, upon 56, right? Option number D is correct. Plain and simple. Any questions? Okay, we'll take one more. Okay, so let's move on to the next problem. Let's take this one. Yeah. Three groups A, B, C are contesting for positions. Oh, yeah. Three groups A, B, C are contesting for positions on board of directors of a company. The probability of their winning are this, this, this. If A wins, the probability of introducing a new product is this and the corresponding probability of B and C is this, this respectively. Find the probability that the new product will be introduced. Should I put the poll on? Which new product? It could be a toothpaste. It could be a new MacBook Air. But why do you want to know which is that new product? You know how a company works, right? So in company, there are board of directors, okay? And board of directors decide which product to launch. So for a company, there are three groups which are contesting to become board of directors. If group A comes into power, the chance that the product will be launched is 0.7. Maybe group A is more inclined towards launching it. But if group B comes into power, their probability of launching the product is 0.6 like that. Okay, so the product is ultimately going to be launched. What is the probability of that? That's why board of directors are same, but these three groups are contesting to become board of directors. They're not being chosen yet. So there will be a polling done, election done, and one of the groups will become board of directors. It's like the normal elections that happen in our country. There are BJP, Congress and other political parties who can test for a position. And they'll say, if we come into power, I will remove Berojgari and I will remove hunger and all those, I don't want to use the word also. So, yeah, so what is the chance that Berojgari will be removed? That is what is the question. I'll fix all the potholes. I'll give extra P periods to students, right, Lagaf? Yes, so 18 of you have voted and exactly on one of the options that shows everybody has understood the problem. So can we stop the poll? Yeah, some people are coming as spoilers, marking some other options. Okay, so five, four, three, two, one, go. Okay, so good, good number, 24 of you have voted, which is much more in comparison to our previous class polling participation. Most of you have say option A. See, it's very obvious again, what is your E1, E2, E3? E1, E2, E3, remember, they will be mutually exclusive and collectively exhaustive. So your A coming into the event of A being selected is event E1, B being selected is event E2, C being selected is event E3, right? They are mutually exclusive. Two groups cannot be the board of directors and they're collectively exhaustive also because only these three groups are participating. So whenever you are deciding on your E1, E2, E3, ask these two questions to yourself. Are these events mutually exclusive? Are these events collectively exhaustive? Okay, and what is your A here? A is the fact that or the event that the product is ultimately getting launched. Okay, so ultimately people are getting some PE period. Okay, so the chances of A to happen as per the, you know, I will not use the word A here because A is basically for the group. Okay, let's use a different word. Maybe, maybe, maybe, maybe, let's say R maybe. Oh yeah, Raga and all, yeah. So product R, the chances of it being launched is let's say A wins the board of directors election that is 0.5 probability. And given that A wins the board of directors election, the chances that the product will be launched is 0.6. Okay, no, sorry, 0.7, not 0.6, 0.7, yeah. Then 0.3, okay, one more way to figure out that these events are mutually exclusive is that they will add up to one. Okay, this will add up to one for sure. If they're not adding to one, okay, then whatever choice you have made for mutually exclusive and collectively exhaustive events, that is a faulty choice. Okay, however, those mistakes are very rare. Nobody makes that mistake. Now, chances that B comes into power is 0.6 and the product is launches, sorry, 0.3 and the product is launched is 0.6. The chance that C comes into power is 0.2 and given that C comes into power, the chance that the product will be launched is 0.5. So overall, the total probability will be 0.35, 0.18 and 0.10, okay? That comes out to be 0.63, option number A is correct. Is it fine? That's what I told no origin. One third, because they were equally likely here. ABC are not equally likely, A is more preferred group. Maybe they have more influential people, more hardworking people. So chance of A group winning the election is more. Why I was giving that story, that if let's say a bag is bigger, the chance of you putting the hand inside that bag will be higher as compared to other bags, okay? Great, so I think more or less the agenda is clear. Maybe we'll take one more question. And we'll start with Bayes theorem and then we'll take a collective combination of whatever we have learned so far because Bayes theorem is basically a topic where all the concepts that you have studied so far would be applicable. Let's see this question. Missile is fired from a ship. Okay, in India we say Missile. A missile is fired from a ship. The probability that it is intercepted, or iron dome concept is one third. The probability that the missile hits the target given that it is not intercepted is three fourth. If three missiles are fired independently from the ship, what is the probability that all three hits the target? All three hits the target and put the poll on for this? Poll is on. Good, seven of you have responded. Think carefully and then answer, okay, don't be in a rush, 30 more seconds. Okay, five, four, three, two, one, go. Go, go, go, go, go, go, go, okay. So 51% of people have voted out of which 14 of you. I mean 58% of the people who have voted say option number B, okay. Now this question is actually a mix of two things. One is first of all, a particular missile, a particular missile, whatever you want to name it. What is the chance that it hits the target? And then there are three independent missiles which are fired. So let's say even if the chance that it is intercepted, that E2 is the chance, it will not be intercepted. And as you can see, they are actually complimentary and therefore MEC events, A is the chance that it hits, correct? So what is the chance that it hits? If it is intercepted, it will not hit, unless until you doubt the technology, okay. So assume it's happening in Israel. So if it is not intercepted and the chance that if it is not intercepted, it'll hit as three by four, okay. So overall this comes out to be a half, correct? Now you're firing three independent missile and you want each one of them to be hitting, hitting, hitting. And that option, that will be half, half, half, okay. This hits and this hits and this hits, half, half, half. Answer is one by eight, which is option number B. Well done, I think most of you got it right, great. Any questions, any concerns, any doubts, any clarifications, do let me know. Could you please repeat? Okay, so I'll repeat once again, Sidhanth. So here are, so hitting is associated with two events, whether getting intercepted or getting not intercepted, right? So even in E2, play the role of those MEC events, which I was discussing with you in the theory, okay. A is the event of the target getting hit by the missile. So under two cases, we are going to talk about it. One is one third the target is intercept, as the missile is intercepted, but if it is intercepted, it'll not hit. So the probability of hitting given it is intercepted is zero, if it is intercepted, how will it hit? Why this three fourth part was not needed? Can you tell me Arjun, why it is not needed? If it is not intercepted, it is sure that it'll hit. No, it will fall on some other place. If it is intercepted, means missile is destroyed in the air. It cannot hit anybody now, correct, isn't it? That's how a Jew country protects itself from all the surrounded Muslim countries, isn't it? And that technology has to be super. But let's say if the intercept doesn't happen, it may not fall on the target. It may fall on some places, which is not expected by the target. Yeah, see, many people will not prefer to write this Shatish, okay? They'll say it is not intercepted and hits. That's the only way it is going to hit. But this is just to give it a shape of the law of total probability. No difference, no difference actually, whatever you are saying, okay? Is this fine? Now, Siddhant coming back to you. So the missile gets intercepted, it will not hit the target. And if it is not intercepted, then it will be hitting the target with the probability of three by four. So total way of hitting the target is going to be half. Yeah, if it is not intercepted, it hits the target with three by four probability. So not intercepted, chances two by three, right? One minus this. That's why you got PA as the same thing, no? So you are saying PA is three by four, that is what you are saying? No, that is not correct. PA is the chance that the target is, so PA is the probability of the target being hit. But the target being hit is also a factor that the missile was intercepted or not intercepted. So you have to take this two by three into account. Are you getting my point? If you're not taking this two by three into account, it is as good as all the balls were in the same bag and you're trying to pick up a green ball. Don't make it equi-probable, okay? Let's take another one. Achai throws two fair dice. If the numbers showing are equal, he adds them together and gets his final score, okay? Good. On the other hand, if the numbers showing are equal, if the numbers showing are equal, he throws two more. I think the numbers showing are not equal, it should be, right? Something is missing, no? It should be not equal. He throws two more dice and add all four numbers showing to get, okay, his final score. What is the probability that his final score will be six? So let's do this question. Read the question carefully. He wants to get a final score of six, but this final score of six is associated with two MEC events. One being he gets equal numbers. Second, he doesn't get equal numbers. Let's put a poll on for this. I think the question is other way around. Just give me a second. This is actually not, the first one is not equal. Yeah, this one is not equal. Yeah, correct earlier. The first one is the one where the dice are showing two numbers which are not equal. Sorry about that. I'll give you extra time for that. Don't worry. What, sir? We started working with this original question. Okay, so please consider that if the numbers are not equal, he adds them together. But if the numbers are equal, he has to throw two more dice. Good, good, good. Four of you have responded so far. Not with the same answer. No, I'm getting mostly option number dash. Dash I will not reveal because people will put that only as their answer. But now you are correct actually. Okay, should we discuss now? Okay, let me give you 10 more seconds. Or you want me to wait for one minute. Fine, fine. I'm fine, it's okay. Meanwhile, let me just write few things just to get started with the question. Meanwhile, you can keep solving. Even is the event of getting equal numbers. Equal numbers on two dice. One different numbers. A is the event at the sum of the final score is a six. Okay, should we discuss now? Ready? Everybody? Fine, so I'll stop the poll now. Most of you have gone with option number D. Okay, option number D. Now see, in order to find out the total probability of occurrence of A, we need to see in how many ways or what are the probability of you getting equal number on the dice? So equal number of the dice could be one, one, two, three, four, four, five, five, six, six. That is one sixth of the case, correct? Out of 36 case, the chance of you getting equal number is one sixth. Now given that you get equal number, what is the chance that, okay, let's say not equal numbers. So not equal numbers will be five by six. Let's take that. Yeah, because the first part is not equal, sorry. So I will say this is PE2, okay? So this is PE2, okay? Now what is the chance that, given you have not got equal numbers, your sum is showing to be six. Now not equal numbers, the sum will be six, will be only happening in one, five, five, one, two, four, four, two, yes or no? So there will be four cases out of 30 cases of them being not equal, where you will get the sum of six. So it's four by 30. Arjun, my friend, listen to the question. E2 is the event that they're getting different numbers. How will they get three, three? Three, three are different numbers for you. Listen, listen. Okay, so the probability that your sum will come out to be six, given that you have got different numbers is four by 30, because out of 30 different combinations, six only are equal, one, one, two, two, three, three, four, four, five, five. So total 36 are there. Out of that six only are, you know, you can say, both the dice showing same numbers. 30 cases will be where the dice will be showing, the dice will be showing different numbers, isn't it? Out of those 30, only these fours are there where the sum will add up to be six. Yes or no? Next, one sixth is the chance that your numbers will be equal. Correct? Yes or no? Right? Now one sixth is the chance that the number is equal. Now when this is happening, this person is throwing two more dice. Okay, now let's say, let's say the chance that he got one comma one. Okay, one comma one would be a case where he could have got equal numbers. If one comma one case is a chance where he has got equal numbers, then in order to get a six as a sum, he needs four more. That means after this, he should have got either two two or one three or three one. Am I right? If he would have got two two, then after this, he should have got one one. I think that's the only way, okay? No other way. If he gets three three, will that be counted another case? No, okay. So now what should I write next? Given that he has got, given that he has got two equal numbers, what is the probability, okay, that he will end up getting a sum of six. So what should I write? Six into six into six, is it six into six into six four times or three times? Three times, okay. So see, the question is he has already got two dice. He has already got two same numbers, right? So if he has got one one, then he must get two two and one three on the other. Yes or no? So out of this one six, if he has got, again, please understand here, if he has got one comma one, then he must get two comma two or one comma three or three comma one, correct? So how do you take this into account? So three out of 36, three out of 36, correct? So one six is the chance, let's say he gets a number which is, one comma one. So three out of 36 is what he will need. Correct? Yes, again, one six is a chance he gets two comma two and then again, getting a one comma one is a matter of one by 36, agreed? See, if he is getting, let me explain this earlier, if he is getting equal numbers, it could be one comma one, two comma two, three comma three, four comma four, five comma five. So I'm just breaking all the events like that. So if he is getting one comma one, whose chance itself is one sixth, oh, sorry, one by 36, one by 36, sorry. Yeah, yeah. So the chance of him getting two comma two on the, or you can say the chance of him getting these three on the other two throws will be three by 36, correct? Same goes with him also, sorry. And if he gets, if he gets two comma two, okay? Then he's getting one comma one is one by 36. Now the other chances like, you will say, sir, five by six, there should be one by 36 six times. Two you have only covered, why not you have covered three, three, four, four, five, five, six, six, because for them, the chances will be zero, zero, zero, zero, zero. So I need not write it, okay? Makes sense, does it make sense? That means I do not want to write this. That's why I did not write any further. I think one more will come, one, two, three, four, five, and one more will come. Is it fine? So overall, what should be the answer that should be coming, let's try to find that out. And this is going to be a six once again. So it's four by 36. This is again four by 36 square, and this is one by 36 square, okay? So this is going to be 36 into 36, which is 144. So 144 plus four plus one by 36 square. 36 square is 1296. So it's 148 by 1296. Option number D, option number D. One by one, I think a lot of you are asking a lot of questions. One by one, how 148? 149, oh yeah, 149. No, but I think this is three, sorry, this is three. My counting mistake, it is 148 only. This is three. So I'll repeat the question once, I'll repeat the solution once again. So what happened in this question was if you got two different numbers, okay? The chances of that happening is five, six, and given that you have got two different numbers, the chance of there some being six is four by 30. That was a simplest case. But let's say given you have got equal numbers. So equal numbers could be one, one, two, three, four, four, five, five, six, six. So equal numbers, I'm basically breaking up as one by 36 chance of getting two, two. One by 36 chance of getting three, three. Sorry, one by 36 chance of getting one, one. One by 36 chance of getting two, two. One by 36 chance of getting three, three. One by 36 of four, four. One by 36 of five, five. One by 36 of six, six, okay. So given that you have got 1-1, the chances that you will end up getting a sum of 6 by the further throws of 2 dice will be 3 by 36 because you have to get 2-3 and 3-1. So when you throw 2 dice out of 36 cases, there will be 3 cases only where you will end up getting either 2-2, 1-3 and 3-1. So 1 by 36, given you have got 1-1, so this is the probability that you have got 1-1. And this is the probability that you have got 2-2 or 1-3 or 3-1 given you have got 1-1. Same with this fellow also, given that you have got 2-2, the chance of you getting a sum of 6 will be if you get a 1-1 on the 2 dice which you are throwing separately. So 1-36 of getting 2-2 and 1-36 of getting 1-1 from the remaining 2 dice given that you have got 2-2. And same goes with the others but others there will be 0-0-0 because once you have got 3-3, you cannot achieve a, you know, you will always get a number more than 6. So it will become an improbable event for us. Is it fine? So with this we are now ready to start with our understanding of Bayes theorem. Bayes theorem, many books call it as reverse probability or inverse probability also. Bayes theorem takes a lot of its ideas from the law of total probability. So what is Bayes theorem? Let's say I give you a simple example, very simple example. Let's say a person, he thinks he has got some disease. Let's say he thinks he has got TB. Now he goes to a doctor. So the doctor says the chance that a person of your age gets TB is this much. He gives some chance percentage. And the chance that my report says or my report is correct is this much. So see basically what is happening, I'll just give you the scenario in writing. Let me take a scenario in a slightly interesting way. Let's say a man is known to speak truth. He says truth 3 out of 4 times. Now what this man does, he rolls a dice and he reports it's a 6. Reports to be a 6. What is the probability that it was actually a 6? Are you getting my question? The question is this man is known to speak truth only 75% of the times. One fourth he speaks lies. So what does this man do? He just throws a die and he says, hey, I've got 6. What is the chance? Now you're going back. Reverse. What is the chance that he actually got a 6? So what is the probability that he actually got a 6? Because just that this man says I've got a 6 doesn't mean he's got a 6. He may be lying also or may be correct also. Who knows? So how do you solve such kind of a question? That is what we are going to talk about base theorem. Let's take one more example. I definitely saw this question. Don't worry. I'm just giving you examples where base theorem is basically perceived. Many people will say it sounds very similar to your conditional probability. Given that he reports a 6, what is the probability that he actually got a 6? So the word given that he reports a 6, what is the probability that he got a 6 is basically indicative that it's a conditional probability. So base theorem is nothing but conditional probability only. I always say base theorem is a very special name given to some certain kinds of conditional probability questions. Another example I can give you, let's say I take the same example of that bag which I started with in total probability. So let's say 3 green. I don't know. I don't exactly remember the combination. I'll just try to write some numbers. 1 green, 4 red, anything. I mean, doesn't matter. 6 green, 3 red. Now let's say I give a question in this way. There's a man who goes blindfolded. He puts his hand in one of these bags whose chances of being picked up is equally likely because all these bags are of same dimensions. And he gets a red ball or a green ball. What is the probability that this green ball came from bag number 3? There's a reverse path you're taking. You know the final outcome. Final outcome is known that he has got a green ball. But what is the chance that this green ball actually came from bag number 3? If you're trying to solve such questions, then you're basically dealing with base theorem question. Are you getting my point? So once the event has happened, you're trying to figure out that event A happened with which of E1, E2, E3, etc. in your mutually exclusive and collective events. That answer is basically given by base theorem. Is it fine? Okay. So I think Kinshukh you were talking about that cheating question. See a person can get marks by several ways. He can either know the question. He can know the answer to the question. Or he may guess it in an objective based exam or he may cheat the answer from his neighbors. Now let's say I was correcting the paper of this guy and I found that one particular question he got it right. Okay. Now I want to know what is the probability that he actually cheated it. That's base theorem. Are you getting my point? So you're going back in the event and trying to figure out out of which E1, E2, E3, etc. That event A actually happened to occur. Okay. So I will solve these questions but before that I will give you a theory and then I'll come back to this question. Okay. So let's go back to the same diagram which I had made in the previous page. In fact, I can make it once again. So theory behind it is let's say there is a set of events E1, E2, E3 till EN. Okay. And these events are mutually exclusive and collectively exhaustive mutually exclusive and collectively exhaustive. Normally I use a single symbol for it, MEC. I call it as M-E-C-E. Books do not use this word MEC. Normally I have made a abbreviation out of it. So let's say E1, E2, E3 till EN are MEC events and there is an event A which occurs along with these EIs event. Okay. Now I want to find out given this event A has occurred, what is the probability that it occurred along with EK? Okay. So let me make a diagram once again. Okay. So let's say these are my sets of E1, E2, E3, etc. Like that. Okay. Now, of course I have to make some bifurcation here also. Yeah. This is your EN. This is your E1. This is your E2, etc. Okay. So given that this A has occurred, what is the probability that it occurred along with EK? EK being let's say some event over here. So let's say this is your EK part. Okay. So now as per conditional probability, as per conditional probability, can I say that this is as good as saying, so last class we had done conditional probability formula which comes from your multiplication theorem. So this is EK intersection A by PA. Correct. Now your numerator is what? PEK into PA by EK. Can I not write it like that? Again, my multiplication theorem. And what is your denominator? If you remember, denominator comes from your law of total probability, which was PE1 into PE, sorry, into PA by E1 plus PE2 into PA by E2 and so on till PEK into PA by EK. Right? So you can see this particular formula reminds you of all the concepts that you have studied. It involves conditional probability. It involves multiplication theorem. It involves law of total probability. That is why I keep telling students that if at all a question comes in your board exam, there is a high probability that it comes from Bayes theorem. Because in Bayes theorem, the examiner gets an opportunity to test you on all the fronts of this chapter. Okay. Is it understood? So again, this is something which also comes from common sense. The moment you do this, the moment you do this, you don't have to do anything further. I mean, this formula is just an eye wash. If you write in the exam early, you will not lose that independent marks which is given in the board exam. So your formula basically ends at a conditional probability formula. Nothing else. So you just need to figure out what is the probability that A and EK occur together divided by what is the total probability of occurrence of it. That's it. All these things is an eye wash. Okay. So let's use this formula and just check and just find out what is the answer to those two examples which I cited. Ayo, sorry. Thanks. It was supposed to go to EN early. My mistake. I wrote EK. Okay. In between, you'll get EK also. Okay. So let me just rewrite it once again. In between, you'll get some EK also. Let me write it like this. I mean the same term that you have written on the numerator that will be in one of the terms that you'll be getting. E N. E A by E N. Thanks. Sorry. I missed that term. Is it fine? Okay. Now coming to the two examples that I took. So there is a man who is supposed to, who speaks two, three out of four times. He reports a six. What is the probability that he actually got a six? Okay. So what is your E1 and E2 here? What is your E1 and E2 here? What is he speaking truth? Okay. And E2 is he speaking a lie. What is A? A is his reporting a six. Reporting a six. So now your question is, or you can many people say he actually got a six is even and doesn't get a six is E2. So actually got is this. That's it. Actually got six. And E2 is didn't get a six. Did not get six. Okay. As you can see even E2 or MEC events. Now what is the question asking you? Question is asking you. What is the probability that given he reports a six. He actually got a six. Okay. Now here, what do you do? You'll first write down what is the number of, what is the probability of. What is the probability of him getting or him reporting a six under which situation will he report a six? He'll report a six when he actually is truth and he got a six. Or he will report a six when he's lying and he did not get a six. Correct. These are the only two occasions when this person will report a six. Yes or no? Correct. No. Okay. Now here's the question that is the total probability of you, of him reporting a six will be. He telling a truth. When he actually got a six. And he telling a lie when he did not get a six. Of course, if he's selling a truth. And he did not get a six. He will not report a six. Are you getting my point? And if he's selling a lie and he got a six. Then also he will not report a six. Okay. So now as per the previous result that we saw. It's P a into P e one into P a given even has occurred. And please note that the numerator part is your first expression that you have written over here. So your answer will be three a by four into one by six divided by total thing that you have written, which is three by four. Into one by six divided by total thing that you have written, which is three by four into one by six, one by four into five by six. Okay. Maybe let me let me solve this question separately on the next page. Here it looks quite modular. So I'll solve it separately in the next page. Don't worry. So give me one second. I'll just solve it. Or maybe I'll solve it over it. So the solution for the first one is. Let's say even is he actually got a six. Actually got a six. E two is when he did not get a six. Did not get a six. Okay. Is the event that he reports a six. Okay. So probability of him. Actually getting a six when he reports a six. As per the conditional probability. Okay. See, I'm writing it step by step. Okay. So I'm writing, I'll write the formula for condition probability first of all. The one which we did yesterday and the numerator part again could be broken like this. Okay. My PA. Now, every time when you're solving this question first write down your PA. This is the first thing you should write. Because while you're writing PA. The numerator term will appear within that PA. So while you're writing that expression of PA. As I told you here in the formula also this numerator term will be appearing here while you're writing your PA. So first understand what is the total number of ways he will report a six. So he's selling a truth and he got a six. Then only he'll report a six. He's selling a lie and he did not get a six. Then also he will report a six out of that. This is your numerator part and this whole thing is your denominator part. So I'll write it down here. This is your numerator part. This is your denominator part. So the answer will be three by four into one by six upon three by four into one by six plus one by four into five by six. I think this answer is going to come out to be if I'm not mistaken, three by eight. Rohan has a question, sir, but when he lies isn't there he reports a number other than. Then six to what is that my concern. Read the question question is. What is the probability he got a six given he reports a six. So given is something you can't go beyond it. Correct. Given he has reported six is what you have to take as your basis and then move forward. Whether he reports a mother number or not, that's a subject matter of a different question altogether. In this question, the question says he reports a six. What is the probability that he got a six? So when he when he's lying. There is a five by six chance that he will he'll get some other number right other than six. So that is accounted for. No. Why does not accounted for it is accounted for. Okay, Rohan. Now here coming to this question. Given that he has got a green ball, what is the probability that it came from bag number three. So given that he has got a green ball. What is the probability that it came from bag number three. So first write down what is the total probability of him getting a green ball. One third three by five. One third one by five. One third six by nine. Out of that, which is the case where you had chosen bag number three and got a green ball that was the last case. So that will come on the numerator. So this is how we normally solve based theorem question plain and simple. Very easy. I'll give you more questions. Aditya has a question, but sir, if he lies in a show he so the chance of him getting some other number is five by six. Correct. Other than other than six, the chance that he getting some other number that probability itself is five by six. Right. So he lies one fourth times and one four times on that in a five by six chances that he's not getting a number six that has been factored in our total probability. But aren't we assuming that he reports a six every time he lies. Sir G Rohan G that is the question my dear. What is the question says heroes that die and reports a six. So where is the assumption here. It is given in the question it is stored at this fellow is reporting a six. Where is the scope of any assumption in the question. See you have to only figure out those cases where he will report a six, not any other case. Let's start with this question. And this question is definitely going to come in your school exams. For sure. I mean there's a question one is like they will say London and Clifton in the case of Krishna Giri and Dharma Dharma Puri. And in one case they put the name as a Tata Nagar and Calcutta something like that. Don't worry the process is the same. See read the question carefully the question says a letter is known to come either from Krishna Giri or Dharma Puri. On the post mark only the two consecutive letters are I are visible. Okay. Now what is the chance it came from Krishna Giri. See the look and feel of the question the question is basically giving you an end result that the person is only saying R I consecutive letters everything has got wiped out. Maybe these two are very rainy places and by the time the post arrived all the ink got wiped off. Okay. And this is a person who has got relatives only in Krishna Giri and Dharma Puri. Okay. Now he's wondering what is the probability that you know my relative from Krishna Giri has sent me this letter. Okay. So how will you solve this question again name the events. First name the events. It's very important especially from your school point of view. So we are also getting trained for our school exams also right. In case there is a subjective paper. That's definitely there he should open the letter and see who's it but let's say everything is wiped off. Let me know my name. Only the envelope contained the name of the person got erased. Should I put the poll on. Okay. Let's see whether you're able to solve this. I'll put the poll on. I think people would definitely like to give it a try. Now again don't don't think like you just have to count. There is a concentration of the R.I. word different in different different you know Krishna Giri and Dharma Puri are different concentrations of the word of the consecutive letters R.I. It's like the green ball in different bags. I think I've given enough hints you should be able to solve it with these insistence only R.I. is visible only and that to consecutive not that this R and this I is visible. So looking at the letter the person knows consecutive letters are visible not like this and this. So that's why they have written to consecutive letters R.I. Of course looking at it you can know these are consecutive letters or no there was some letter in between like that. So he showed that they were consecutive letters R.I. See one R.I. Yes only one R.I. Is see that is not possible. Either we raise it or leave it but do not displace the letters other than solving the problem you people think of everything else know. So he should read the name later the rain pushed it to some other corner. What is what are you Ram? I am here from Dharmapuri. Because the last name is almost slimy. Okay jokes apart. Anybody? Okay a poll is already on I think 12 people have responded. What about others? Okay we will stop the poll in the next 30 seconds. So if you have solid piece based on the response on the poll and if you haven't please solve it by the next 30 seconds. Okay 5, 4, 3, 2, 1. End of poll. But I would like to know from one of you what is his or her response. Let's hear it out from Atharva. Atharva what do you think was your response? Which response did you choose? Atharva are you there? Still solving. Okay so let me ask Ritu. Ritu which option did you go for Ritu? Okay so Ritu has gone for option C. Okay Ritu let's see. So C is the most chosen response by the way but let's see. Now see let's first name our even and E2. Even is the event that the letter came from. The letter came from Krishnagiri. Krishnagiri. Okay E2 is the event that the letter came from Dharmapuri. Okay and A is the event that only two consecutive letters R i is C. Now let us try to write down in how many ways or what is the probability that if you see two consecutive letters R i what is the probability that that two consecutive letters R i will be you know will be left off. So it can happen under two situations. Number one the letter came from Krishnagiri. Now given the letter came from Krishnagiri what is the probability that if you pick up two alphabets those will be R i two consecutive alphabets mind you two consecutive alphabets that will be R i. So Krishnagiri how many two consecutive letters are there? Let's count. One. Two. So I am just taking Kr then R i then I s. Okay. I s. S h. H n. N a. A g. G i. I r. R i. Okay. So two consecutive letters what is the total sample space you have? One. Two. Three. Four. Five. Six. Seven. Eight. Nine. Ten. Yes or no? Yes or no? Eleven sorry eleven yes correct. One. Two. Three. Four. Five. Six. Seven. Eight. Nine. Ten. Kr. R i. I s. S h. H n. N a. A g. G i. I r. Nothing. So one. Two. Three. Four. Five. Six. Seven. Eight. Nine. So one of these is two by ten because there are two R i s here. Out of ten combinations that you can have of two consecutive alphabets there are two such combinations where R i will be there. Correct. Okay. Then another chance will be when you when you receive the letter from Dharmapuri and in Dharmapuri how many consecutive two letters can be picked up? Again let's count it like this d h h a d h h a a r r m m a a p p u u r r i. So again one two three four five six seven eight nine only. Out of that nine there is only one such case where you will end up getting R i. Correct. Now the moment you write this your job is ninety nine point nine nine percent done is just that you have to write your final expression for the base theorem. So the chance that the letter came from which place they want us to come from Krishnagiri. Krishnagiri is I think even even. So the chance that the letter came from Krishnagiri given that the two consecutive R i s are visible is the one where you took Krishnagiri. I think Krishnagiri was in the first this thing half into two by ten divided by the total probability. Total probability is pa itself. Okay. Let's simplify this. So this is one by ten one by ten plus one by eighteen. So I think if I'm not mistaken this will give you eighteen by twenty eight eighteen by twenty eight is nine by fourteen nine by fourteen is nine by fourteen in the options. Yes. Option number C is this fine. So as I told you the approach is pretty straightforward well defined. Sorry. Write down your total probability in that total probability there would be one such you know expression which will be coming on the numerator and the entire thing is in the denominator. That is basically based on that's it. Let me tell you it is one of the easiest topics. People don't normally lose marks in this unless until the question is very complicated especially in school exams. Of course J. J. Advanced questions can be much more trickier. Any questions any concerns. Okay. Now many people will think there are two. R. Is in Krishna. And there's only one R. So the chance that the letter came from Krishna. Will be two by three. Which is not the case. And I think two by three will also be there in the options. Option B is there. Right. So people who are not aware of this seed it's not an distribution. Right R. R. Is not equally distributed. Are you getting the point? The same way as the balls were not equally distributed or the density of the balls were not the same in all same in all the banks even though they were in silos which were equally distributed. It's like let's say all the states have equal area also but the concentration of let's say south speaking languages people are different in different states. So you can consider that the south speaking language, south language speaking people are uniformly spread throughout the India. No, let's take more and more questions because this is the important part. Let's take another one. Okay, let's take a passage-based question. This came in J advance exam 2006. That time they didn't call it as advance, they call it just ITG exam. So the question says there are n earns containing n plus 1 balls such that ith earn contains i white and n plus 1 minus 1 red balls. Okay, assuming these balls are all distinct, they have to be because the word identical is not being used. Let ui be the event of selecting ith earn and w denote the event of getting a white ball. Okay, let's talk about question number 12 first. Question number 12 says probability of ui is directly proportional to i. Okay, that means probability of picking a u ith or you can say ith earn is directly proportional to the the number of that earn. Okay, what is the probability of getting a white ball where n tends to infinity. I'll be giving a poll for this. I'm not trying it. I'm not solving it, Raghav. Okay, I'm putting a poll for this just for the 12th question, please. Very good. I have already got one response, sorry, two responses. There is somebody by the name of Guhaan. Is it Parvati? Somebody by the name of Guhaan. Okay, sorry. So I mean, I was just seeing the attendance who all are present. Gaurav is absent today. Yes, almost going to be two minutes, five people have responded. I can give another half a minute or maybe one minute. Okay, five, four, three, two, one, go. Okay, not many people are interested. So equidistribution of votes between A and B almost and some few votes have gone to C and D. Okay, let's check. See, first of all, probability of you getting a white ball will depend upon which earn you are picking up. So we have to be very careful about what is the probability of picking the ith earn. In the question they have said that it is proportional to the position of that earn. Okay, of course, you cannot equate it. You have to put some proportionality constant, let's say Ki. Okay. And one more thing is that they are mutually exclusive events. So if you sum up these i's from 1 to n, that should give you a 1. That would give us the proportionality constant, by the way. So k1, k2, etc. till kn is going to be 1. In other words, in other words, k value is going to be 2 by nn plus 1. So putting back over here, the probability of picking up any earn is going to be 2i by nn plus 1. Am I right? If I missed out anything with respect to finding pui, please highlight everybody. Okay. Now comes to loft total probability because here they're asking you in how many ways can you get the white ball. So white ball, how in how many cases we will get if you pick up earn 1 and then pick up white given that you picked up on 1, that will give you a white or pick earn 2 and get a white ball given that you have picked up on 2 and so on. And this will go up all the way till nth earn. Am I right? Am I right? So what are the chance that you picked up earn number 1 that itself is 2 into 1 by nn plus 1. Now the chance that you are getting a white ball given that you are picked up earn number 1 will be 1 upon n plus 1. Yes or no? Same with this. This will be 2 upon n plus 1. Then 2 into 3 by nn plus 1 into 3 by nn plus 1 and so on. If you see the look in the field of this expression is something like this 2 by nn plus 1 whole square and you're doing 1 square, 2 square, 3 square till n square. Am I right? Correct me if I'm wrong. Okay. That's nothing but n plus 1 the whole square n into n plus 1 into 2n plus 1 by 6. This is your probability of getting w. Okay. So let's cancel out nn goes for a toss. This also goes for a toss. So that becomes 2 times 2n plus 1 by 6. Now the question setter has asked, what is the limit of this? The question setter has asked, what is the limit of this as n tends to infinity? You tell me what's the limit of this as n tends to infinity? 4 by 6. Isn't it? So it's actually 4n plus 2 by 6n plus 1 or 6n plus 6. You divide throughout with n. That will give you 4 plus 2 by n, 6 plus 6 by n and as n becomes very, very large, this term will go for a toss. This term will also go for a toss and the answer is going to be 2 by 3, which is option number, option number b. Let's check the response once again. How many of you? Oh, five of you gave b. Great. So I think most of the work has already been done in the first 12th question. So let's go ahead with the 13th question and we'll do one thing. Since I've already utilized maximum part of the screen, I'll take the question once again in the next slide. 13th question. Let the poll begin. If pu i is c, that means fixed, okay, c is a constant, then what is the probability that given you have got a white ball, it came from the last earn? Oh, people have started giving the answer also. Good, good, good, good. Okay. Last 30 seconds, 5, 4, 3, 2, 1. All right. Mixed response. Some of you have said a, b, c, d, all of them have been picked up. Okay. See here, this itself is a base theorem question because you have already picked up a white ball. Okay. Not enough. You already picked up a white ball. What is the probability it came from bag number n or earn number n? Now see, probability of picking up any earn is constant. That is c. So first of all, if you want to get a white ball, it can come from picking earn number one, whose probability is c itself. And the chance of you getting a white ball from there is 1 by n plus 1, then again, 2 by n plus 1 and so on till c n by n plus 1. Okay. That means, I mean, if you simplify it, this is going to be c by n plus 1. This is n into n plus 1 by 2. So that gives us c n by 2. So c n by 2 will come in the denominator. And that case where you're, you know, you have picked up the ball from the nth earn, that probability is c n by n plus 1. Guys, come on. This is so simple. This is 2 by n plus 1. Option number a is correct. Any doubt? Any doubt? Any concerns? I have a question in denominator. Yes, tell me. Can it be made more than one? See such a value that it is not more than one. That is what, I mean, let the tension of taking c be on them because they are not mentioned. What is the c value? Denominators, the denominators that you have, you're saying can't they add up to give you more than one? That's what you're trying to say. I don't think so. They will add up to more than one because then you're trying to say probability of getting a white ball is more than one, which violates the definition of the restriction that we have on probability itself because denominator is what? Pw, no? See, what is this? This term is what? This term is PUN intersection w by Pw. So you're getting earn number n and picking a white ball from there divided by 10 to the power 16. How can probability go to that extent? You're confusing with the count. Count and the probability are two different things. Oh, n can be made 10 to the power 16 you're asking. Okay. How do you know c value then? c value will be reciprocal of that. No. See, Aditya, even if you find that, does it matter? That anyway is going to get cancelled. No, c is not appearing in our answer only. So whether you write one by n, whether you write it c, it doesn't appear in the answer itself. So why are we worried about c being one by n? Let it be one by n, let it be not one by n. I mean, let it be just a c, right? Of course, ideally, if you have a cc value, should be one by n because it's saying it's a constant. Okay. So that constant should all add up to give you a one. So nc is one, so c is one by n. But let's not worry about it too much because c is not going to appear. All right. Last part of the question, should we go to the next page once again? Because this is all cluttered. I'll put the question once again. 14th part of the question says, if n is even and e denotes the event of choosing even numbered earn and probability of choosing an earn UI is one by n, the probability of choosing even numbered earn is one by n. Okay. What are the probability that you will choose a white ball given or even number earned is chosen? I'll put the poll. No worries, Anusha. I'll put the poll. Shouldn't this be one by i actually? What do you think? One by n or one by i? If n is even, oh, total number of earns they are given. Okay. And the number of earn is even. Okay. And the probability of choosing each earn is one by n. Okay. Let it be as what they are given. Okay. Yeah. Now I got the part of the question. They're saying that there are even number of earns and probability of choosing any i turn is one by the total number of earns equally likely. Good. I have got four responses so far. Okay. Can I stop the poll in the next 30 seconds? One. Okay. Most of you have said option number b. Six out of ten people have said option number b. Let's discuss. So what are we trying to find out? We're trying to find out probability of getting a white ball given an even numbered earn is picked up. So basically as per your conditional probability formula. Okay. We'll have to use this. Okay. So probability of choosing an even earn. Okay. What is the chance of choosing an even earn? Okay. So let's say I want to choose the second one. What is the chance of choosing a second one man? Okay. And what is the chance of you getting a white ball out of it? Two by n plus one. Similarly, the probability of choosing the fourth one is also one man and you getting a white ball out of it is that. Okay. So the numerator part will become something like this. Okay. And the last earn, I think it will be n by n plus one. n is an even number. Mind you. If you want, I can write a 2k for n also if you want. Should I write 2k for n? You want me to do that? Okay. I'll go. No, I'll write it as the, you know, as you suggest. Okay. It's fine. And what are the probability of you choosing an even earn? What are the chance of you choosing an even? What is the chance that you choose earn number two? Okay. So this will be your chance of choosing these ones. So as per this, as per this, can I say the numerator would be one by nn plus one into two, four till n. Now remember, n is an even number. So how many such terms will be there? n by two. Isn't it? So can I say the sum will be n by two by two, two plus n, isn't it? So in an arithmetic progression, the sum of n terms is what? The total number of terms by two into first and the last term. So total number of terms is n by two. So n by two by two, first and the last term. Okay. And down in the denominator, remember, there are n by two, one by n. Hope I have not missed out anything. Do let me know. So a lot of things will get cancelled. By the way, from here, I think one by n will get, oh, sorry, one by n will get cancelled with one by n, n by two will get cancelled with n by two. So the answer that you will get is n plus two by two n plus one from this entire exercise, which is clearly option number B, B for Bangalore. Again, let me check the poll result. Oh, most of you said B, well done. Great. So I think this passage was a good eye opener for us in terms of refining our concepts with respect to total probability, base theorem, etc. Okay. We'll take one more question and then I'll give you some five minutes break and pause the break. I will start with probability density function that will not take much of our time. Maybe 45 minutes or so. That would be good enough for us. Let's take this question. This question came in ITJ 2005. Question says, a person goes to office either by car, scooter, bus or train. The probability of which being, okay, so the probability of him using these modes of transport is this. Probability that he reaches office late if he takes car, scooter, bus or train is... Given that he reached office in time, what is the probability that he traveled by a car? Poll is on. Good. I can see six responses so far. Can we have the poll closed in the next 30 seconds? So please vote by then. One, good number of responses for options. C, as you can see, 85% of you have said C will be the right option. Okay, great. So first of all, what are our mutually exclusive and collectively exhaustive events? So him taking a car, him taking a scooter, him taking a bus or him taking a train. So these are your mutually exclusive events and he's the event that he's on time. So these are your MECE events and this is your final event that he's on time. So my question setter is asking us what is the probability that he took a car given that he was on time? So this is what they're asking us. So for this, what did I tell you? First, write down in how many ways can he be on time? The denominator you write down first. So he will be on time if he had taken a car and given that he has taken a car, he's on time, whose probability is 7 by 9 or he had taken a scooter and given that he had taken a scooter, he is on time or he would have taken a bus and given that he has taken a bus, he's on time or he would have taken a train and given that he has taken a train, he's on time is again, sorry, is again 8 by 9. Okay, please note that the probabilities of E1, E2, E3, E4 are adding up to 1 or not. Are they adding up? Yes, as you can see, if you add these four, they will add up to give you a 1. Okay, out of this, the numerator part is where he had taken a car. So this guy will be on the numerator. That is what we want to find out. So you can safely remove the denominators, which is 63 in all of them and just deal with their numerators. So this will be something like this. Okay, so answer is going to be 7 by 31, 41, 49, which is 1 by 10. Any questions? So option C is correct. One more question I will take up. Kinshuk will not leave me unless and will I take that cheating question? What Kinshuk, you're waiting for this question only. From that time he's saying, when will you take that cheating question? Okay. Alright, so let's take this question. In a test and examinee, either guesses or copies or knows the answer to a MCQ question with four choices. Only one option is correct in there. That also should be mentioned because MCQ can be multiple options correct also, but this is only one option correct. The probability that he makes a guess is one-third. The probability that he copies the answer is one-sixth and the probability that his answer is correct given that he copied is one by eight. Find the probability that he knew the answer to the question given that he correctly answered it. After this we'll take some big, small big. Yes, Raghav, you were saying something. Okay, Raghav is getting some answer. Okay, good. I'm not given any options here, so please give me a response on the chat box. So end result is what? He has got it correct. Are you going back? Reverse. That's based theorem to check whether he actually knew the answer. Okay, Aditya has a different opinion. Raghav, answer and Aditya answer are different. So G is the event that he guesses. G is the event that he guesses. C is the event that he copies. K is the event that he knows and R is the event that he got it right. Okay, the answer was right. Okay, so what are we asked? We are asked that given that he has got it right. Given that he has got it right, what is the probability that he knew the answer? So what do we do in such cases? We first figure out how many ways can he get the answer right? The denominator part. Always write the denominator part first. So he will get it right had he guessed the answer. Now, the chance that he guesses the answer is one third and guessing gives the correct answer is one fourth, isn't it? Because there are four options. Only one option is correct there. So if you're guessing it, the chance that you get it correct is one fourth only. So given that he guesses, the answer is correct is one fourth and the fact that he guesses itself is one third. Okay, or he knows the answer to the question. Now, that answer is not given to us rather the question setter has given the probability that he copies the answer. So now copies, guesses and knows should add up to one because they are MEC events. So guessing is one third, copying is one sixth. This is I don't know. So can I find PK from here? Yes or no? Yes or no? So what is PK? Half. Is it fine? Any doubt related to how did I find PK? Clear? Because they are MEC events, their respective probability should add up to one. Okay. Now, given that he knows the answer, what is the chance that he will get it right? It's a sure event. One. Now, don't ask me like, sir, what he thinks he knows the answer, but he doesn't know the answer. Don't go to that complexity. Okay. If he knows the answer means he knows the answer. Okay, he had read it and he's going to get it correct for sure. Next is he copies the answer whose probability is one sixth and the question setter has also provided that the fact that he copies, he gets a correct answer is one by eight. So given that he copied, he gets it correct is one by eight. So this part is also provided to you. Now out of this, your half into one will be on the top. That's it. That's the answer. Can we simplify this? I think we can take 48 as our common this thing. So multiply it throughout with 48. This also multiplied with 48. This also multiplied with 48. So that leaves us with 24 by two plus 24 plus one, I believe. Sorry, four. This is four. So answer will be 24 by 29. This is going to be your answer. Absolutely right. Aditya and Radhov. Good. So a small break of five, six minutes. Right now it's 10, 10. Let's meet at 10, 21. Okay. Six minutes break. Siddhich also very good Siddhich. Sorry, I missed you. Okay. So next 40 minutes, we'll talk about PDF probability distribution function. Oh, sorry, sorry, sorry, sorry. It's 10, 10, not 10, 15 extra time you would have got. See you in six minutes time. So this would be our last subtopic. Probability, probability density function or distribution function. Many people, many books will refer to it as probability distribution table also. Probability distribution table. Now what are we finding here in this topic or who's this probability distribution we are trying to find out? So we are basically trying to find out the probability distribution table for a random discrete variable associated with an experiment. Now use the word which is slightly new to you to a random discrete variable associated with an experiment. Right. Now let me give an example to explain what is the meaning of random discrete variable. What does it mean? See, let's take a simple experiment. Experiment as good as tossing three coins. Okay. So let's say there are three coins, you are tossing it. Okay. Three coins are tossed. Okay. So when you're tossing a coin, let's say I call the number of heads that you obtain. Okay. So simultaneously are solving three coins. Okay. All the three coins have been tossed simultaneously. And the number of heads that you obtain, you're calling it as a discrete variable. Why we call it as a random discrete variable? So let's say I'm tossing a coin, the number of heads that I obtain, this could be an example of a random discrete variable. Why random? Because it is coming from a random experiment, a probabilistic experiment because all the experiment that we perform, they are all random. They are not definite experiment. Why do we call it as discrete? Because number of heads can either be zero, one, two, or three. Their fixed value is assigned to it. Right. So they're discrete in number. Why they're variable? Because I've never mentioned how many heads. It could be zero head, one head, two head, three head. Okay. So this is a random discrete variable. Number of heads will be a random discrete variable. Okay. So if I make a table, if I make a small table where I mentioned the probability against several values of X that could be possible over here. So here I could have a zero head. And what is the probability of getting a zero head? Or I get a one head. What is the probability of getting one head? Or I get two heads. What is the probability of getting a two head? Or I get three head. What is the probability of getting a three head? In fact, I don't need this extra. Okay. So if I write such a table, right, and basically start making a function out of it. So what is your P of X function? Okay. That function, that function will be called as the probability distribution. Let me just write it here. So treat this like a function. So this function would be your probability density function or probability distribution function. So this is what we call as probability distribution table. And this function will be called as a probability distribution function. We'll come to all those things via this example itself. And then probably I will generalize it. Okay. So, yes. So let us try to fill in. Oh, very good. Raghav has already given the answer. He has made my life easy. So what are the probabilities that you end up getting no heads when you toss three coins? One bite. Okay. What is the probability that you get one head when you toss three coins? You'll say three bite. What is the probability that you get two heads? Again, three bite. And probability you get one head is again one bite. Yes, I know. So now you have figured out what is P zero, P one, P two, P three, etc. through this particular experiment. Right. But let us say I want to give it a shape of a function. I want to call it as a probability distribution or probability density function. Can somebody guess? Okay. Of course, not a random guess. What is happening here? And can I make a function out of it? That is fine. That is definitely a bell curve. By the way, in your higher grades, you will study something called random continuous variable. So discrete will change to continuous. Okay. For your J perspective, we are only going to restrict ourselves to random discrete variable. Yes. Can somebody guess what is happening here? No. Okay. If I have to write the probability of getting X heads, just a second. Yeah. Sorry. If I just want to write the probability of getting a X number of heads. Okay. When you roll a dice, what would you actually do? You will first of all think that out of three, I have to choose any X positions where I have to place my head. So you'll say three CX. Correct. And the probability of whether head or not a head for each one of them is half, half, half. Correct. So you end up getting a probability distribution function by putting your X value as a zero, one, two, three, etc. You can keep filling this table. Are you getting my point? So this is what we call as the probability distribution table and probability distribution function. Any doubt anybody has? Okay. Now we are going to talk about two types of probability distributions. One of them being more important than the other, and that is binomial probability distribution. So let's talk about binomial. In fact, the example that you're seeing right now on your screen is an example of binomial probability distribution function. Okay. Binomial PDF. Now what's the binomial probability distribution and when do we apply it? So is this a phenomena seen in several types of cases? Yes. It is seen in those kinds of cases where basically an experiment has or experiment gives rise to Bernoulli trials okay. There's something called Bernoulli trial. Okay. Bernoulli's trial. Bernoulli's trial are basically those trials which always lead to complementary outcomes like head or tail. Like getting an even number, getting an odd number when you're volatized. So these are complementary trials. The target being hit, the target not being hit. Okay. So whenever there is an experiment where the outcomes are complementary to each other. Right. So there will only be two outcomes. Okay. Of course the question will be framed in such a way that you will be having two outcomes out of it and they will be complementary of each other. So an outcome where there is a complementary outcome that is p and 1 minus p kind of a scenario. Normally in the language of probability we call this as a success case and we call this as a failure case. Okay. Please don't take it literally. Don't take it by the actual meaning of the word. See, for example, let's say I say the ship, five ships have left for their destination. Okay. What are the probability that at least two of them reach the destination? Right. So reaching a destination now is a success. Not reaching is a failure. Every ship is associated with two fate, success, when it reaches, failure, when it drowns. But let's say if you're a person who, let's say you're on the enemy side and let's say these are all warships, for you the success will be that they drown. Right. Failure will be when they reach the, when they reach their country and of course the war starts. Okay. So what's the Bernoulli trial? Bernoulli trial in plain and simple words are basically those trials where the outcomes are categorized as success, failure means complementary to each other. Okay. So in such cases, in such cases, let us say you are performing N trials. Okay. N trials are being performed. Then the probability distribution function in such cases is given by, is given by, can somebody complete taking a clue from this coin case? Take a clue from this coin case. Okay. Do you want me to write this in a more proper way? Okay. So three CX, half to the power of three minus X, half to the power of X. Correct. No, that is for the coin, no, a shortage. What about in general, when P is the probability of a success and one minus P is the probability of a failure. Okay. So P is the probability of a success, probability of success and Q is the probability, oh, sorry, one minus P, which is also called Q by the way. I didn't mention that, but now I'm mentioning it. This is called the probability of failure. So take a clue from the previous result and tell me what do we, will you get? N, C, X, P to the power of X, or you can write it like this, Q to the power of N minus X into P to the power of X. Correct. So this becomes your probability distribution function for Bernoulli trial probability distribution. So whenever there's a Bernoulli trial, which is conducted for N times where P is the success. Okay. V is the probability of success means whatever you want comes out, that is your success. For example, I wanted head to come out because I was finding the probability distribution table for the number of heads. So getting a head is a success, not getting is a failure. Are you getting my point? So the probability of getting the success X times or let's say there are X successes and N minus X failures will be given by this expression. Now why it is called Bernoulli, why it is called binomial probability distribution? Because it resembles the way you expand a binomial expression, isn't it? Okay. Since it resembles the way you expand a binomial expression, this is named as a binomial probability distribution. So many people ask me, sir, are there further more probability distribution? Yes, there are many more probability distribution, which you will learn in your higher grades. In fact, today I will also talk about Poisson's distribution. Later on, you'll learn Langrange's distribution. There is something called Z distribution for continuous variable. There is something called rectangular distribution. So I categorically remember my fourth semester in IIT when I was taught probability and statistics. So we were taught almost six, seven types of probability distribution depending upon cases to cases. One of the most difficult course, by the way, in mathematics, probability and statistics. Anyways, so coming to this, let's use this concept to solve few questions. And then I'll also talk about mean and variance. Maybe one question I would like to take on this. Let's say E20 World Cup is going on. So an apt question in this present scenario. India and Pakistan play a five-match test series of hockey. Then the probability that India wins at least three matches. India wins at least three matches. Now, how would like to solve this question? I would like you to make a probability distribution table first of all. So let's make probability distribution table of the number of wins of India, not Pakistan. I hope success is India winning, not Pakistan. Okay, so there are five matches. So five matches means India can win zero. One, two, three, four, five. Yeah, this time Pakistan won no. India World Cup if it doesn't win also no problem, but it should not lose against Pakistan. Oh, by the way, these videos will be seen by people from Pakistan also. Normally I put it on YouTube. They will definitely dislike this video. Okay, let's let's write on the probability distribution table for x being the number of wins of India. Okay, please note they're playing five match test series. So it can't be more than five. So it starts from zero goes all the way to five and they take discrete values, nothing like 0.5 or 1.5 like this. Zero match win, one match win, two match, three match, four match, five match. Okay, now we already know the formula. So we can quickly fill these boxes up, isn't it? Shall we? So this will be five C zero. So now there's 50-50 chance. So P will be half, Q will be half. Okay, chances that India wins is half, chances that India loses is also half. So it's going to be five C zero, Q to the power of five, that is half to the power of five. Okay, then five C one, half to the power of, half to the power of four, four match it loses and half it wins. Okay, so please note that the expression which I'm using here is five C x, five C x, Q to the power of five minus x, P to the power x. Okay, so yeah, five C two, half to the power of three, half to the power of two, then five C three, half to the power of two, half to the power of three, five C four, half into half to the power of four. And then finally, five C five, half to the power of zero, half to the power of five. Okay, now what are they asking us? They're asking us what are the probabilities that India wins at least three matches, at least three matches means they're asking you what is the probability of P three plus P four plus P five. This is probability of winning at least three matches. Okay, so just add this five C three and now this is as good as a 32, five C four, 32, five C five, 32. Okay, by the way, this sum is going to come out to be 10 plus five plus one by 32. That's actually a half. Am I right? Anything that I've missed out, please highlight. Okay, is it fine? Now, one, you know, small observation that you can have over here is that if you add all the, you can say PX, PX is. Okay, so that is something very important observation. Why only three, four, five, why not one, two, three? What is the question, Shradha? Probability that India wins at least three matches. So why should I take one, two, three? X is the number of wins. You wanted to win at least three matches. No, so P three, P four, P five. In this case, it works fine, but my job is to tell you all the things, right? Later on, you start making your own shortcuts. Shradha, does it answer your question? Something else. P is the X, I've already written it. X is the number of wins India has. You're writing the probability distribution for this. Okay. Now, one important thing to be noted down, number one, the summation of all the probabilities, that is, let's say PXIs, okay, that will always be one, always, undoubtedly. Any probability distribution table you're writing, whether binomial or otherwise, the sum of all the probabilities should be zero because, sorry, some of all the probabilities should be one because they're basically all mutually exclusive and exhaustive cases, right? So the chances of India winning zero, one, two, three, four, five, these are the only possible events. There's no other event other than this. And they're all mutually exclusive. It cannot happen that India wins zero at the same time India wins one, right? So they are MECE events again. So there, some of the probability should always come out to be one. Okay. Second thing, even though it is not observed in this question, is something related to mean value. Now, try to give this entire situation a shape of a statistics question. Okay, so think as if this was a statistics table given to you where your zero, one, two, three, four, et cetera, they were all data and the PX were the frequencies, even though literally it is not a frequency, right? So your PXS, PXS are your frequent, you can say the frequency in which those data are is occurring. So as you can see, are thus the two and the three, they are the highest frequency data, isn't it? Because it is 10 by 32. This will also be 10 by 32. So the chances of India winning either two or three is the highest. Okay, so think like that. So instead of the frequencies, you think that the probability of those random discrete variables, they are playing the role of the frequencies. So when you start looking at this whole data as a statistical figure, there is something which will come out from there is a mean value, right? So what are the mean number of times India is going to win? Okay, so that is given by the expected value. So if somebody gives me this probability distribution table and he asked me this question, hey, what do you think? What is the mean number of times India is going to win this match? Okay, so let's say they are going five matches. What is the mean number of matches? What is the average wins that India will get in those five matches? So that is called the expected value. It plays the same role as the mean. Many books will write it as this also. So mean value is given by see, just like our mean was given by summation fi xi by total number of frequencies in our statistics chapter in the same way here it is given by summation xi into Pxi. So if you sum this up from one to the total number of trials or the total number of matches that they are having, that will give you the mean number of matches India is going to win. Is it fine? It's actually 2.5 in this case. We'll figure it out. We'll figure it out. In fact, our next exercise is where we'll be figuring it out. And there's something called the variance. Variance as you know, it's basically a parameter statistical parameter which tells you how scattered is the data. So variance is given by summation of Px into mean value, you can say ex minus x the whole square. Very much similar to like you had used in your statistics chapter last year. Okay, so you can write it other way around also doesn't make any difference xi minus xi minus estimated value of exit or estimated value ex. Okay, this is called the variance. And of course, there's called standard deviation. Standard deviation is under root of variance that you are already aware of since your class 11th statistic days. Okay, now let's talk about a bit of variance in this. I'll be simplifying that expression slightly more. So let's simplify this slightly more simplification. So variance is given by summation of P xi into xi minus estimated value. Let me call that as x. Should I call it as e by the way just to you know, write less actually, I don't want to write e and then bracket x just to write it in a simple sober fashion. Okay, so if you expand it, you'll end up getting P xi. And this is going to be xi square minus two ex I plus e square. Okay, let's open the brackets, you'll end up getting P xi into xi square minus two e is a fixed value. So that will not participate in the summation. And you'll have something like this. And you will have e square summation P xi. Now, as per our understanding of whatever we have discussed so far, this term itself is, is your e, isn't it? So it becomes two e and this term is a one, isn't it? So it becomes something like this, which ultimately boils down to a simpler expression summation P xi into xi square minus e square or minus the expected value square. Do you recall this formula similar formula came in our statistics chapter also? Okay, so please make a note of this as well. See, both are the same formula, but I've just, you know, simplified it to a more simpler extent. Now, let's do a work here. Use the same problem to figure out what is the mean number of wins or what is the expected number of wins for India in this hockey test match. And we'll also find out what is the variance and the standard deviation for the number of wins for India in this hockey test match. Let's figure that out. Please note this down, then I'll take you to the question once again. Note it down. Okay, let's go to this question. So let's find out the expected number of wins for India in this hockey test match. So it's summation P xi into xi. I mean, in whatever order you want to write it, it doesn't matter. So it's zero into whatever number comes out from it. Five C zero. In fact, I'll write it in an expanded version, then one into five by 32, then two into, I think 10 by 32, three into again, 10 by 32, four into five by 32, and five into one by 32. Okay, in short, I can do one thing. I can take 32 common. So this term is anyways a zero. So I'll have five plus 20 plus 30 plus 20 plus five. How much does it come out to be? It comes out to be 50, 70, 80. 80 by 32. Okay, 80 by 32 is exactly 2.5. Yes or no? Yes or no? Cancel by a factor of 16, five by two, 2.5. So expected number of wins for India in this case is 2.5 wins. So you can expect this is the average number of wins it can get. So this is just like a person who is trying to get into a, you can say, an uncertain event like gambling and all. So he can use this particular concept to basically estimate how much will be the income that he will gain by indulging into a, let's say, gambling event. So don't try to use this. Gambling is not a good thing. So people definitely use this expression to figure out how much they're going to win in an event. So depending upon what other stakes for different differences in gambling industry, there are a lot of different types of small, small events and there is a lot of money associated with it. So if you're indulging in a set of events, you know that how much money you're going to win from there by calculating the estimated income from that event. Okay, let's calculate the variance also because variance is also something which we have learned just now. So variance of the data will be summation of xi square into Pxi minus the estimated value square, always a 0.49 chance of winning. Yes, in gambling industry, it is correct. You, the chances of you winning is actually slightly lesser than chances of losing. Yeah, they adjusted very carefully. That's why there are rates on these gambling industry. Ideally, they should be neutral. They should only charge for maybe, you know, the party or the whatever drinks or food they are serving, not for the game. Right on the last scale, they made a lot of profits. Okay, yes, let's complete this. So this is 0 square into 1 by 32. Of course, that is not going to be used up one square into one by 32. Oh, not one by 32, five by 32. Sorry. Then two, I'm just calculating this part by the way. Okay, so two square into 10 by 32, three square into 10 by 32, four square into five by 32, five square into one by 32 minus this part that is your five by two the whole square. Okay, please calculate this. And without calculating, I can tell you what is the answer going to be. The answer is going to be five by four. Convince yourself. Is it coming out to be five by four? I'll tell you, I'll tell you. But is it coming out to be five by four? Check that first. Somebody please confirm. Fast, fast, fast. We don't have much time. Five by four means 1.25. So do this calculation. I mean, if you want, I can do it on my phone as well just to save time. I'm not promoting use of any kind of a phone here. So it's five on the numerator, then you have a 20, then you have a 90, then you have 16 into five, which is 80. And then you have a 25. This divided by 32 minus 2.5 square minus 2.5 into 2.5. Yes, are you getting it? Five plus 40 plus 90 plus 50 plus five. So it's 7.5, this whole thing, 7.5 minus 6. Yeah, 7.5 minus 6.25. That's 1.25. 1.25 is five by four. Okay, now some of you are asking how did I do this calculation so fast? Okay, now I'll tell you some hidden secrets. Hidden secret here is for a binomial probability distribution, the expected value is always, of course, this formula is always there, but there is a shorter way to get it. It's always NP. That is total number of trials into the probability of a success. So the fact that we got 2.5 was because there were five matches, chances of India winning is half. Okay, always. Okay, so note this down because it really saves a lot of your time. Okay, and your variance of the data is always, I'm writing the exact formula which you have learned just now, but you need not use this for a binomial probability distribution because it kills a lot of your time. The formula directly is NPQ. So remember the reason I said five by four is because five matches, India losing and winning is half of each. So answer became five by four. Okay, now what is the derivation for this? We will see that. I will not derive it completely. In fact, I will leave up to you to figure that out. I'll just partially give you a hint. See, how is a binomial probability distribution written? Okay, I just could have used, I'm just making some few. So this is x, this is px, 0, 1 all the way till you go till n. Okay, so n trials are happening. So you can have minimum of x value as 0 and maximum of x value as n. Okay, let me make one more. So we already know that this is nc0, q to the power n, p to the power 0. This is already nc1, q to the power n minus 1, p to the power 1, nc2, q to the power n minus 2, p to the power 2 and so on. So these terms, as you already know, they already come from, so these terms that you have written, they come from the binomial expansion of q plus p. Failure probability plus success probability to the power of n. Okay, so you just expand this. This is your first term of this expansion. This will be your first term of this expansion. Okay, this is the second term of this expansion. This is the third term of this expansion and so on. And remember, I was telling you that the summation of the probabilities come out to be 1. It is because your q is 1 minus p and p is p. So if you add it, it's actually 1 to the power n. That's why that's the reason why summation of pxi comes out to be 1. Okay, now my purpose here is to tell you how does the expected value comes out to be np. Okay, so here I'll just give you a hint. So use the expected value formula that is summation xi pxi. So you have something like 0 into nc0, q to the power n, p to the power 0, 1 into nc1, q to the power n minus 1, p to the power 1, 2 into nc2, q to the power n minus 2 p square and so on till your last term, which is n into ncn, q to the power 0, p to the power n. Okay, I'm not solving it completely. I'm just giving you the hint to solve it. Okay, now just take q to the power n common from this entire game. So when you do that, anyway, this term is a zero, so I will not include it. So the next term will be nc1, p by q. This will be 2 into nc2, p by q the whole square. Third term will be, I mean, the next term would have been 3 into nc3, p by q to the power of 3 and so on till last one is n into ncn, p by q to the power of n. Okay. Now in your mind, just for this part, just for this part, treat your p by q like an x. Okay. And this part will actually remind you of an expansion like this, cx, 2c2x square, 3c3x cube and so on till ncn, x to the power n. Can anybody tell me how do you get this expression? Let's say I call this expression as e for the time being. How do you get this e expression? Can somebody tell me? Right. Now recall class 11th binomial theorem chapter. In class 11th, you have learned that when you write the binomial expansion for this, you should write c0, c1x, c2x square, c3x cube and so on, where c1, c2, c3, c0, etc are short forms of nc0, nc1, etc. Okay. So I'm not writing completely. Then what do you do in order to in order to bring these coefficients in front? You used to differentiate, right? So differentiate with respect to x. So those who were there with us in class 11th, you would know that this process was very much taken up. So this will be c1, this will be 2c2x, this will be 3c3x, c3, c3x square and so on till ncn, x to the power n minus 1, correct? Multiply throughout with x. Okay. So if you multiply throughout with x, you'll end up getting x here, x square here, x cube here and so on till x to the power n. Okay. So this term that you have, this term that you have, can I not write it as q to the power n? Okay. x into 1 plus x, sorry, n will also come, 1 plus x to the power of n minus 1. Yes or no? So just, I'm just using the same reason xn1 plus x to the power n minus 1. I'm just putting my x back as p by q. And remember, there was already q to the power n sitting over it. Okay. So when I do that, let's simplify this. So this will be q to the power n. This will be p to the power n, not p to the power n. This is p to the power p by q and into n. So n I will pull out. And this is p plus q to the power n minus 1 by q to the power n minus 1. Note this, this and this will get cancelled off. And p plus q is already a 1 because p plus q is already a 1. You can write this as np1 to the power n minus 1. That's how this result comes. Okay. That's true. Is this fine? Any questions? Now the variance part I'm not going to do. Variance part proof, please you do it. Same approach. You have to use your differentiation concept only. So I have now proven why the binomial theorem, sorry, binomial probability distribution expected value can directly be written as np. You don't have to use this formula over and over again. This is a universal formula, however. You're going to use it to solve multiple problems. But for case of binomial, you can cut short and use np in place of that. Any questions anywhere you have, please note down if you want to. And for your homework, I'm going to ask you to prove the variance of a binomial probability distribution. By the way, binomial probability distribution sometimes will also be written like this. Okay. And your standard deviation is definitely under root npq. So please note this is a symbol that normally books will use to tell that a particular distribution is a binomial probability distribution. So they basically provide you with number of trials and probability of a success. Okay. So they will say, what is the, for example, they'll say, what is the expected value in B five, two by three, something like this. That means this means number of trials for n, which is five, and this probability of success is two by three. So that is how they are going to put the question. So the answer will be what? Answer will be what? 10 by three. So they will ask you what is the variance for this binomial probability distribution? What is the variance for B five, two by three? So what will you say? npq. Right? So P is known, then Q is automatically known. So npq, answer is 10 by nine. Clear? So in your DPPs and all, if you see such kind of this thing, so they will also, they sometimes use this X follows a binomial probability distribution like this. They will also use terms like this. So read this as if X, random discrete variable, you are finding for a binomial probability distribution. So X, you're finding the probability distribution of X where the event is basically occurring like a binomial probability distribution or Bernoulli trial where n is the number of trials and P is the probability of success. That means for X to happen. Is this fine? Any question, any concerns here? Okay. Last but not the least, maybe I'll take five, 10 minutes of your time. Please bear with me. We'll talk about the Poisson's probability distribution function. Okay. Now Poisson's probability distribution function is actually a, you can say special case of a binomial probability distribution where number of trials is infinitely high and the chance or the probability of success is a ending to zero quantity. Okay. So for a Poisson's probability distribution, your number of trials are infinitely high and the probability of success is very, very small. Okay. Now just to give you an idea about this scenario, I normally quote this funny story which happened with me. When I began my teaching career, I used to go to one of the schools in the morning center that was supposed to be my morning center for the classes. So it was some government school. Okay. And of course, there were some faculty staffs also and it used to be very early in the morning. So I used to use one of the toilets, one of the washrooms and later I came to know that I was using female toilet. I was not aware that it was nothing was written. It was a government kind of a school. Okay. So now let's say I make a story out of this. I make a situation out of it. Let's say normally the washroom is used by let's say three staffs or three female staffs every hour. Okay. Per hour this is your uses of the staff room on an average. Okay. Average uses of that washroom by staff is three staffs per hour. Okay. Now I was in let's say my average time of using the washroom was let's say 10 minutes. Okay. So I use the washroom for 10 minutes. That's it. Now what is the chance that in this 10 minutes, I will be spotted by at least two staffs or two female staffs. So this is a question which is based on Poisson's distribution because time is an infinitely divisible figure. Right. So 10 minutes has infinite amount of you can say time divisions into it. So what is the chance that I will be seen or spotted by two female staffs which are supposedly coming in the same washroom. Okay. So this entire situation would become a case of a Poisson's distribution. I know this is sounding very funny, but just to keep the you know concept fresh in your mind. Another instance could be let's say you all have you know receptionist in your school. Right. They keep getting calls. Let's say you know the receptionist gets around 50 calls in one hour. I think it's too much 20 calls in one hour. Let's say okay. Now let us say that is the average number of calls she is getting. Now let's say she's away from her seat for two minutes. Maybe she went to see you know Madam Principal or maybe to see any you know emergency situation. What is the probability that she would have got three calls in that two minutes duration. That is a case of Poisson's distribution. Okay. Now Poisson's distribution, the PDF or the probability distribution function is given by this expression. It's a slightly you know weird expression. So it is given by e to the power minus lambda, lambda to the power x by x factorial, where lambda is the average number of occurrence, average number of occurrence of the success in the specified time duration. Let me give a simple example for it. Very simple example. Let's say I was talking about, I was talking about the receptionist question. This receptionist receives around around 20 calls in one hour. Okay. And let's say she's away from her seat for let's say six minutes. I'll just take a number which is going to help us divide. So she's away from the seat for six minutes. Okay. What is the probability away for six minutes? What is the probability that she would have got three calls in that six minutes of duration? Okay. Then here, please understand lambda would be the average number of calls that she would have gotten six minutes, not in one hour. Okay. One hour is just an example to confuse you. In this time duration specified in six minutes, what is the average number of calls she would have got? Two. Okay. So two will become your lambda because she's getting 20 calls in one hour. That's an average of one hour. So the same average applied to six minutes, she would have got two calls. Okay. So the lambda value is two. Your x value is three. So the probability of she getting three calls in that six minutes of away time will be given by this answer e to the power of minus two, e to the power minus two, two to the power of three, whole divided by three factorial, whole divided by three factorial. Of course it'll come out in terms of e only. So it's eight by six, eight by six e square. Or in fact, you can say four by three e square. Okay. This will be the probability that she would have got three missed three calls, which she would have definitely missed because she's away from the seat. Okay. Is it fine? Now, many people ask me, sir, how do you derive this? Okay. So the derivation part is basically coming from our binomial theorem only by taking a special case, by taking a limiting case of n tending to infinity and p tending to zero. I will derive it in some time. Don't worry. But do you want me to give you one more example? Should I take one more example? Okay. Let's take one more example. Okay. Let us say on West Quad Road, on West Quad Road, five accidents happen, five accidents happen every half an hour on an average, every 30 minutes on an average. Okay. So there's a road, West Quad Road. Okay. Five accidents happen only every 30 minutes on an average on let's say one of the roads. Okay. I was on this road for let's say 10 minutes. Okay. So I was on the road for 10 minutes. Okay. So let's say some person is on the road for 10 minutes. So in 10 minutes, what is the chance that in 10 minutes, what is the chance that what is the probability that the person would see six accidents? What is the chance that in 10 minutes, let me write it like this, in 10 minutes, six accidents will take place. Okay. So on an average, on an average five accidents happen every 30 minutes. Okay. I was on that road for 10 minutes. Let's say I was taking a walk 10 minutes. What is the chance? What is the probability that I would see six accidents happening in front of my eyes? Not a good sight though. But what is the probability? Just say it done. No need to give me a figure. Just say done if you're done. Okay. Very good. Done. Very good. Okay. So what is lambda here? Who will tell me? What is lambda here? Write down the lambda at least. Right. Lambda will be five by three, not five. Don't don't get this question is trying to confuse you this part. So in 10 minutes, what would have been the average number of accidents? That is a lamp. Okay. So as per our probability distribution, as per our presumed probability distribution, Px will be e to the power minus lambda, lambda to the power x by x factorial. So now this is going to be e to the power minus five by three. Lambda is five by three to the power of x means six accidents. Okay. So in this case, your x value is six. I'm putting six value divided by six factorial. Now, please don't let this entire number would be very, very small. So it's very less chance that six accidents I will see just in 10 minutes. Okay. Right. So this number will be quite small. Okay. Now a quick derivation of how the formula came in. Sorry, I'll take just two minutes of your time. So your Poisson's probability distribution, your Poisson's probability distribution comes from a limiting case of the fact that n is tending to infinity, or you can say p tending to zero in this expression, n c x q to the power of x, or you can write it like this, one minus p to the power of x into p to the power x, because everything should be in terms of x. Okay. So this becomes a limit question, by the way. And let me tell you in many cases, this itself is a question asked. People are not aware that there's a limit question, which is actually asking you to evaluate this. Okay. So this itself could be a possible question. Okay. So now this term, 11 minus x, n minus x. Oh, yeah. Sorry for that. Yeah. So this will be nothing but n factorial by x factorial n minus x factorial. This is one minus p to the power of n into one minus p to the power of x into p to the power of x. Correct. So one minus p to the power x goes in the denominator. Okay. Now all of you, please pay attention. Please pay attention. p is nothing but n by, sorry, p is nothing but lambda by n. Lambda by n, because n p is lambda, because for binomial distribution, n p is to be lambda. See, I'm taking binomial only. Just I'm taking a limiting case when I'm just increasing the number of events to infinity and probability to be very small. But still this relation will hold true now. This relation will hold true. Average value, n p, it will be holding true. Okay. So now what I'm going to do is wherever I see an n of p, I'm going to put lambda by n. So now here, first of all, let's write it down. So let's expand this first. Just leave x factorial as it is. So n factorial divided by n minus x factorial. What all will you see? Can I say I will see n, n minus one all the way till n minus x plus one. Am I right? Yes or no? Okay. And this will be one minus p to the power n. Okay. And this also p to the power x. I will write it as lambda to the power x by n to the power x, n to the power x. I will keep it like this. Okay. And I have one more term. I think which I, yeah, there was one minus p to the power x also. You can write it as one minus p to the power minus x if you want. Okay. P also here, I will change lambda by n, lambda by n. Okay. Now all of you please pay attention. How many terms are there here? x terms. And you have also this, remember x will always be a natural number because you're finding the number of events for that many events to happen. Remember I said six accidents, I see six is a x. She would have received five, three calls, three was x. Okay. So you'll have as many number of factors over here as the number of x over here. Right. So you can do one thing. You can distribute one, one n's to each of these terms like this. This also n, this also n, this also n. This x factorial I'll keep separately. Let's say I like it like this. And this lambda to the power x also I'll keep separately. Okay. Now see what are the terms which are left here? This term and this term. Now, when n becomes very, very large, would you all agree with me that this term will be tending to one, this term will be tending to one, this term will be tending to one, x is a finite number, x is a finite natural number. Correct. So can I say each one of them will be tending to one, one, one, one like that. And this lambda to the power x by x factorial I'm keeping it as it is. Now you tell me as n tends to infinity, what is this limit? This is one to the power infinity form. Remember, what are the limit of one minus lambda by n to the power n and tending to infinity. Let's do it separately. What is this limit? E to the power minus lambda. Right. So basically it is, it is nothing but it is one to the power infinity form. Isn't it? One to the power infinity form. I've already done the limit concert with you. Right. E to the power f of x by g of x. Okay. f of x here is minus lambda by n. G of x is one by n. So it will give you e to the power minus lambda. So this term will become e to the power, this term will become e to the power minus lambda. Now what about this term? Now see, this is a finite quantity and this quantity is tending to zero almost because this is your p. So can I say this quantity is as good as a one again? Agreed? So in light of all this simplification, can I say the answer that I would get to see finally, finally is this and that's the expression for the pdf function for Poisson's distribution. Okay. So with this, we are done with the chapter. With this, we are done with a chapter. I think we will take two more classes and we should be done with our AUC as well as not two more classes, I think. Yeah. Friday, are you ready to have two sessions with me? Morning and evening both.