 So we are going to start with probability, okay, so probability that is going to come for your competitive exam that will also include the topics that you have done in class 11th. Okay, so in class, in class 11th. What are the two topics in probability that you had done of course we had talked about events, the types of events. And we had majorly talked about the use of permutation combination. So the use of P and C in counting in finding the number of favorable events and the number of events in the sample space, right? Because normally we require that in the classical definition of probability because probability of an event is nothing but number of favorable events associated with that by total number of events. So this was done in class 11th. In class 11th we had also, we had also covered up the concept of addition theorems. Just a second I think, sorry the power was fluctuating. So addition theorems of probability. If you recall we had done something like PA union B, PA intersection B, sorry PA complement intersection B complement, that means de Morgan's theorem etc. So addition theorem of probability was done. In fact today's session also I will begin with few basic questions just to test your understanding about these two concepts. So what are we going to study in 12th? Okay and the 12th part is the major part of your probability topic. In 12th we are going to talk about, we are going to stress upon independent events, we are going to talk about conditional probability. Conditional probability is the main thing of class 11th, sorry class 12th probability. This is the major topic. Okay, everything is revolving around conditional probability for you in class 12th probability chapter. So other concepts that will be taught in class 12th would be the aiding concepts to this concept. So this is the major concept. So even the topics which will be I'll be listing down in some time they will also be associated in conditional probability. Okay, so we'll be talking about conditional probability. We'll be also talking about conditional probability in sync with multiplication theorem. Okay, so this will be in sync with along with multiplication theorem. Along with multiplication theorem. Multiplication theorem. Okay, post that we are going to talk about law of total probability, law of total probability. And finally the application of conditional probability which is in Bayes theorem. So Bayes theorem is nothing but it's basically a conditional probability only. Okay, we also call it as a reverse probability. Reverse probability. After having done these topics, there is one small part left, which is called PDF probability distribution function. Probability distribution function. Okay, for random. Discrete variables. Discrete variables. So under this we are going to study binomial probability distribution binomial PDF. This is the important one by the way. However, I think the last year from CBSC they had excluded this part. And we'll also talk about poison, poison probability distribution briefly, very briefly. Okay. So this will be our agenda for this topic. As you can see, there's nothing much. Okay, I think one, two, three, four, five, five topics. And these are all revision topics of class 11. Okay. Okay. Now just to begin how many of you are comfortable with the use of PNC and how many of you are comfortable with the use of addition theorems. Mostly everybody. So should we start our discussion with the problem which will test you on these concepts. Okay. So one important topic, by the way, for KVP why if you are a KVP why aspirin very, very important topic. Okay, KVP why loves asking questions on PNC probability, etc. Okay, and this time also if you see the J advance paper, I believe should not be looking at it because you should be taking it as a test. Okay, a lot of questions on probability have come, especially in paper, I think paper one, a lot of people questions on probability has been asked. Okay, anyways, let's get started. So I'll begin with a few questions just to check your understanding of probability. Okay, let's begin with this question. So the question says two distinct numbers. A and B are chosen randomly from the set set has all powers of two starting from one to two to the part 25. Find the probability that log B to the base A is an integer. Find the probability that log B to the base A is an integer. I will launch the poll as well. In fact, if you're done you can put your response on the poll as well. Again, I'm not taught I'm not teaching class 1112 topic here I'm just testing you on 11th concepts. Within, I think 45 seconds, somebody has already responded. Very good. Two people responded. Raghav, any updates on how Prakul is feeling? Is he fine? We spoke very briefly over a call because I was in the middle of a session. Yeah, he said that he has got an occult fracture. So again, I'll repeat once again, since everybody is here, please do not indulge in any sports where you can fall down, you can fracture your bone or anything hits you badly. See, it's not like you'll not recover, but you are not a very critical a junction of your life. A lot of exams you're going to write one after the other. So health is wealth. Be healthy first, be in a sound frame of mind first, then anything else will follow. Sincere request. Okay, we'll wait. We will keep the poll on for the next 30 seconds. If anybody wants to respond, please do so. Okay, five, four, three, two, one. Go. If you want to take a guess, if you want to take a call. Okay, just a while, yes, you can take that. Okay, not many people are responding. So, okay, 18 of you have responded so far. And out of that, half the class says it's option B or half the people who voted say option B. Let's check. So first of all, read the question carefully. The question says, the question says you have to select two distinct numbers, distinct. So you have 25 numbers and you have to select two distinct numbers. So let's say A and B. Okay, so A and B have to select. So this can be selected in 25 ways and this can be selected in 24 ways. And this combination gives you 600. That is nothing but the sample space. Okay. Now, if you see, you want to select such A and B where your favorable event gives you a situation where log B to the base A is an integer. So let's say you select B as two to the power of let's say X. Okay, because all the numbers here are of the nature two to the power X, and you select B as some number two to the power Y. Okay. And this is as good as saying X log two to any base you want to put it divided by why log two to any base you want to put it anyways they're going to get cancelled off I'm just put E for the sake of putting a base. So here you want this guy to be an integer. Isn't it? So if log A to the base B is an integer. Okay, so we'll say let A is two to the power X, let B be two to the power Y because A and B are chosen for these powers of two. So this implies that if this is to be an integer that means your X and Y that means your number one two three four 25 should be chosen in such a way that X is divisible by Y completely. Okay. So how many options you have for that. Okay, so let's let's figure it out. If you choose why as why as let's say one, how many possibilities you have for X. How many possibilities you have for X. So let's say you choose to as your B. Okay, that means you have chosen why as one. Okay, then how many possibilities you have for the power. You can choose any number from two to 25. So your X can be 234-25. Isn't it? In short, your B could be if your B is two to the power one, then your A could be any number from two to the power two all the way to two to the power 25. In short, you have 11 such choices, sorry, 24 such choices. Okay, so 24 choices or 24 ways. In a similar way, if I choose why as to how many choices of X you will have. Can I say they will be all multiples of two other than two to the power two, which is other than two itself so this will be going all the way to 24. Right. So how many such choices? I think this will be 11 such choices. Isn't it? Yes or no? Anusha, why 23? 24, it should be enough for the first one. Anurag also. Yeah. Similarly, Y is equal to 3. What are the choices of X that I will have? I will have 6, 9, 12, 15. 24. So how many of them? It will be 18 by 18 by 3 plus 1, 7 choices or 7 ways to do this. Okay. Similarly, we can go see, I can go at the max to 12. Okay, so Y can go at the max to 12 because after two to the power 12, you will not be able to get a such a Y for which any two to the power X will be divisible. Okay. So we'll go till 12. So basically we'll have four, five, six, seven, eight, nine. Don't worry, the count will be not very great towards the end. 11 and 12. Okay. So this is going to be multiples of four. So this is going to give you 8, 12, 16, 20, 24. In fact, literally, we can count it. This is going to be five. Okay. So for five, you can have 10, 15, 20, 25. So that will be four. Okay. For six, you will have 12, 18, 24. That's only three. For seven, you will have 14, 21. That's only two. For eight, you will have 16, 24. That's only one. And I think beyond nine and all you will only get 111 because only 18 will be there, only 20 will be there, only 22 will be there, only 24 will be there. So these all counts will be 111. Let me maintain the color code here. 111. Okay. Okay, so all together, all together, I think, sorry, I think I've written next one. Yeah. This is two. Yeah. So all together, let's have a count of the total number. One, two, three, four, six. And this will be eight. 11, 15, 20, 27, 27 and 35, which is going to be 62. Okay. So 62 ways is your number of favorable events. This is going to be 62 sample space as we already counted was 600. So the probability of this event will be 62 upon 600, which is nothing but 31 by 300. 31 by 300. Yes, faster way to do this. Faster way to do this is you have to keep taking the gif of these functions. I mean, I mean anyways you're doing the same thing actually, but literally have to count it here. If you have a faster way to do it, please recommend it to me. Please recommend it to the class also. But I feel this is the way we can do it. It is not a lengthy way by the way we can show. These all are considered to be, you know, good enough a way to solve the question. It's not a lengthy way. It's not considered to be a lengthy way. Okay, let's take one more question. Yes. You can write it in order to write it as a symbol, you can write it like that. Why it will not be easy if the power went to 100 or 200, but you can't count number of terms in an arithmetic progression fast. Oh, you say the cases formally more right. So that kind of question may not be framed. Yeah. Okay, look at this question this question says that there is a function from x to y which is having the domain zero to nine. Okay. And why is the code domain which is zero to 100. So that f five is five, find the probability that the find the probability that the function f x, in fact it should be capital X, use a black color. Yeah, to be where B is a subset of y is a bijective function is a bijective function. Okay, so you don't have to evaluate it I think they have given in terms of some kind of summation. Options are in terms of summations in the denominator at least. Correct. So when you're using that floor function, what for there is a there is a summation process for the floor function. You have to literally evaluate it now. He's asking if the number of cases we're running up to 100 how would you do it. Of course you're using for function indirectly, but is there any formula by which you can sum up such for functions also. If you literally put your i values as per your question and trying to find it is question. Can you question is if your numbers, let's say if you had powers from two to the power one to two to the power 200. Then how are you going to do it. Is there any shorter way to that that is what was his question. Okay, so let's look into this. What you're not interested in. Oh, class 11th one. Okay, I'll send that to you. Don't worry. Sure, I'll do that kind of understanding. Okay. See, it has nothing to whatever we are doing, it has nothing to do with class 11th primarily is just that your permutation combination has to be good. The previous problem had nothing to do with class 11th. It was just your number of countings you're doing, but still I'll send you don't worry about it. Okay, so I'll be running the poll for this question as well. Some of you're giving me responses on the chat box, you can also click on the poll button. Very good. So seven of you have already responded. Great. Number of bijective functions, how to find it out we had already discussed this year itself in functions chapter. I have got almost for 12 responses but they are very, very scattered. Almost BCD have got equal votes. Now he has also come in. Almost 30 seconds. Since you don't have to perform any operation it is just the options if you see they have kept it very raw form. They have just written, you know, the unsolved format of that counting. So you don't have to count it. Okay. All right, so let's begin the countdown five. Four. Three. Two. One. Go. In case you want to take a guess work, please do so. Just to see if you had taken a guess work. Would it come out to be correct or not? Don't shy away from taking a guess work. Don't worry. I don't get to see who is taking part. Okay, I will check it out. I am stopping the poll. C is the most chosen response. Almost 41% of you say C. Of course B and D are also quite close. Let's discuss this out. See, first of all, for a bijective function. See, you already know that five has to participate because they have already mentioned that F5 is five. This is a information which we cannot deny. So let us say five is definitely a part of your choices. Okay. Now, first of all, let us try to find out how many bijections I can create. In order to make a bijection, you should have the same number of element in X and B, right? Do you all agree with me in order to have a bijection? Bijection is what? Bijection is a one-one onto function. It's a one-one and onto function, right? In order to have one-one and onto function, number of element in X and number of element in B should be same. If anyone is less or more, it will not become a bijection. Are you getting my point? If let's say number of elements in B is less, then what will happen in order to use the entire elements of X, you will have to do many to one mapping. So it will not be a one-one function. And if you have more and more number of elements in B, then what will happen? Some elements in B will be left unmapped because you have to do a one-one mapping only. So that will become into function. So into function is not onto, right? In order to meet this criteria of a bijection, you must have number of elements in X and B same, exactly same, right? And now this fellow has already sacrosanct. It is already having 10 elements. In fact, in between there is a five also. So I'll just write it down because this mapping is fixed. So in order to make a bijection function, I need to choose how many more elements for B, nine more elements, from how many, from how many, right? So see, zero to 100, 101 elements are there, right? Five is already used up. So 100 remain. From 100, I need to choose nine more. Okay, so that I can choose in 100 C9 ways. Simple PNC doesn't require your, you know, a class 12th, the topic knowledge also, correct? But when you have chosen that, how many, let's say you have now chosen nine other elements. That means you have altogether 10 elements in B. Okay. So how many bijections can be formed if you have a function mapping to another function, both having 10, 10 elements and one element is fixed. How many bijections will you have? But one is fixed, no, in this case, five is only mapping to five, you can't change this. If there was a free hand, it would be 10 factorial. Right. In this case, it'll be nine factorial because, because, because five is only mapping to five. So you have to map the, you have to do the permutations of the other nine elements of X with other nine elements of B. Okay. So everybody is fine with this. Now the problem is two of the options C and D say the same figure. Right. The answer has to be somewhere between C and D. So A and B is anyways, ruled out mostly. Okay, we'll see. Now this is your favorable events and E. Okay. Now let's find out NS total number of samples base. Okay. So how many total functions can you make? So you can have a function made with as less a number as just a five in B. Isn't it? Okay. So let's say there is only five in B. How many functions can you make? One only. Okay. So let's say your X is zero till nine and you only have a five. Only five. How many functions you can make? Of course. I'm talking about functions. I'm not talking about bijection by favorable cases. I've already counted. So I'm counting how many total functions you can formulate. So how many such functions you can make where B is a subset of white. This is what I'm counting. This is the count which I'm doing total number of such functions. Number of functions. Okay. So this is only one in number. Agreed. Okay. If I have two elements in your set B. Five and one more. That's it. One more. Let me name it as maybe B. Okay. Then how many functions can you form here? Think and answer. Think and answer. Think and answer. Think and answer. Think and answer. Think and answer. Who will choose B? If there are two elements. Who will choose B? No. 100 C1. It is a 100 C1. 100 C0. Okay. One to the power nine. Okay. I mean, just to make it as a one. That's it. Anyways, let's say now, let's say now I want to have three elements in B. Five is already there. Okay. One more thing I forgot. Five is already match matching to five, right? Five will already map to five, correct? Oh, yeah, I've already accounted for it. Yeah. Every other elements have got two options each. So zero has two options. One has two options. Only five has one option. So it'll be two, two, two, one, two, two, two, two, two like this. So only one of the two's, sorry, only one of the two will actually be a one. Okay. Actually, I use the number of elements in co-domain to the power of domain, but I was lucky to get it right because I counted it in my mind as a nine. But here the point is C. Zero has got two options, either to map to P or to map to five. One also has got two options, either to map to P or map to five. Like that, all the elements other than five will have two, two, two, two, two options. So two to the power nine, only five has one option to map to five only because the question has already said F five is five. Okay. Yes. Similarly, can I say if I have, if I have three elements here, first of all I have to choose those two elements. Now, again, zero will have three options. One will have three options. Two will have three options. Only five has got one option because it has to map to five. So this will be three to the power nine. Now, I think the trend is quite clear, isn't it? The trend is quite clear. So even if you go to the last counting, that means if you have, okay, what will be the last, let's say, this thing, let's say if we have around, you know, all the 100 elements used here. Okay. Can I say it will be 100, 100, each one of them will be having 101 options here, other than five. So it will be 101 to the power of nine. Okay. Or you can say 100 C 100, 101 to the power of nine. Correct. So in short, in short, the total number of functions that you can form will be 100 C one or you can start with one also. 100 C, sorry, 100 C one, two to the power nine, 100 C two, three to the power nine, all the way till 100 C 100, 101 to the power of nine. Can I not write this as summation R to the power nine, 100 C R minus one R starting from one to 101. Correct me if I'm wrong. So this is your sample space. Is this sitting in the denominator of any term? Option D. Okay. So option D. What is subjective case? Subjective case means the most broader case you're talking about. So favorable event is 100 C nine into nine factorial sample space will be summation R to the power nine, 100 C R minus one R equal to one to 101 for total events. It will be total events. F five is five is true for sample space also it is true for your favorable events also. This you cannot violate. There is a function which is satisfying F five equal to five and it is operating from X to B. How many of them are bijective functions? So F five equal to five is an universally applicable condition, whether you applied for favorable, favorable anyways have to apply. You also have to apply it for the sample space. I think many of you didn't account for that you thought that F five equal to five is only for favorable case. No favorable case if it was, then in the last part of the function they would have written how many of them are bijective and F five equal to five. No, that is a misinterpretation of from your side. Question is very much clear about that. Okay, we'll take one problem on your addition theorem also because I think most of you would have forgotten that as well so just a simple topic will take up. A and B are two events such that PA intersection B is 0.3 and PA complement intersection B complement is 0.6. Find the value of PA intersection B complement or A complement intersection B. That is the question. Excellent somebody has responded good time 40 seconds is what he took. Good, good, good, good. So after this problem we are going to resume with our class 12th probability. So those who are not there with us last year, don't worry, we'll be starting this thing from the basics of class 12 so please pay attention. But having said that, being in symptom or not in symptom, you should be aware of the concept, right, you can't give a reason to the JVN exam sir I joined symptom, you know, much late so I don't know these concepts you have to remember it. You have to know the concepts. I'll close this in the next 20 seconds. See let's say this is your A and this is your B. I think all of you are already aware that this zone is called A intersection B complement. Okay, and this zone is called B intersection A complement or you can say A complement intersection B. Okay, just use this fact and see whether you are able to solve this. Okay, should I close the poll now? 5, 3, end of poll, most of you have gone with option number D. Let's discuss this out. See as you can see from the diagram, if somebody is asking you what is the union of this? Okay, let's recall our very first formula of addition theorem. We know that probability of the occurrence of a compound event, okay, P or Y is basically Px plus Py minus P intersection Y. Very simple, very similar to the cardinal property of sets that you would have studied in the very first chapter of your class 11. But here if you see, if you treat this as your X and if you treat this as your Y, okay, so it will be P, A intersection B complement, A complement intersection B. Okay, why did I put extra brackets here, not required. Okay, and what about their intersection? As you can see from the Venn diagram, these two sets will never, these two events will never have anything in common. So can I say this last part here would be a zero, okay. Now, in order to find this, can I also use this diagram and say it is actually P, A union B minus P, A intersection B. Because if you need the shaded area, it can also be written as, in fact, the shaded area itself would be combinedly written as A union B minus A intersection B, isn't it? So from the union, you remove the intersection part that will leave you with A intersection B complement, union, A complement intersection B. Now, are these two informations given to us directly or indirectly in the question? Let us see it. Yes, I think P, A intersection B is given to us as 0.3. P, A union B we have to find out that also we can find out by making use of De Morgan's law over here. So we know that P, A complement intersection B complement is P, A union B whole complement. Correct. So this is nothing but one minus P, A union B. And P, A union B, what am I writing? Yeah. So P, A union B, I have to make the subject of the formula. So that will be one minus P, A complement intersection B complement, which is one minus 0.6, which is 0.4. So use this 0.4 here. So answer will be 0.4 minus 0.3, which is option number D clearly. And I think most of you had said D as well, very good. Is this fine? Any questions, any concerns? Now these things we already done in our last year. Okay, so we'll not be spending too much time on this. So now we'll start with our independent events, independent events. The very first topic of our class 12th. Normally in school, they don't start with independence even rather they start with conditional probability directly because in board exam, most of the questions will be around your conditional probability. But for JE and all independent events, questions are asked very, very commonly. So I would like to ask from you, when is a compound event? What is a compound event where there is a multiple events happening? Okay. So when is a compound event A intersection B, an independent event, when will be this event called as a, when will this be an independent, sorry, when will your A and B be independent events? You can apply to multiple events also. So when will you say A and B are independent events? When intersection is a null set, no, that is exclusive events. Anusha, you need to be very careful about the uses of independent and exclusive. Exclusive event means their intersection is a null set. That doesn't make them independent. However. Okay. All right. A and B are independent events. If occurrence or non occurrence of one occurrence slash non occurrence of one of one does not influence or does not affect or does not affect. The occurrence or non occurrence of the other occurrence or both non occurrence of the other. Okay. And you can scale this definition to multiple number of independent events. So let's A, B, C, D, are independent events. If occurrence or non occurrence of one does not influence or affect the occurrence of or non occurrence of any other event in that. Okay. Now, if two events are independent, we normally say that the probability of occurrence of their compound event intersection B will be P A into P B, not zero unless until one of them is zero. Okay. So please note this down. This is very, very important. If A and B are independent events, the occurrence of both A and B is as good as both of them are occurring. You know, independent of each other. A simple example for the same would be, let's say you are, you have a coin and you have a die. You're tossing a coin and you're rolling a die. What is the probability of you getting heads on the coin and the number six on the die? What will you say? Does getting a head on the coin influence getting a six on the die? Will it be like, okay, now I'm head, you can't be six. Is there any kind of a hindrance given by the occurrence of a head? No, right? So they're independent. So it says that it is as good as you getting a head on the coin and as good as you getting a six on the die, which is half into one by six, one by two. Okay. Isn't it? Or if you would have done these, you know, questions in class 11 where you are drawing two walls from a bag with replacement. With replacement questions become independent events. Okay. So another example I can cite here is, let us say there is a bag which contains three distinct red balls. Three distinct red balls mean they're all red. But let's say they are all different from each other in terms of hue. Maybe some is a faded red, some is blood red, something like that. And let's say there are four blue balls. Let's say you're drawing two balls out of the bag. Okay. But with replacement. With replacement means you're drawing a bag and then you're putting it back again. Okay. So my question is, what is the probability of you getting first a red? Okay. And second a blue ball. Okay. So what will you say? So first set is three by seven. Now you're putting it back. Then you're drawing another ball. So what is the chance of it being blue? It's again four by seven. Simple like that. So 12 by 49. So it's as good as you drawing a red ball from the bag and as we're drawing a blue ball or you know, drawing a blue ball probability and multiplying both the results. So occurrence of a red in the first draw will not affect the occurrence of the blue in the second row because the red has been replaced back in the back. So these two events will be independent. Are you getting my point? Now, please keep this as your guiding principle for knowing whether two events are independent or not, because a lot of questions will be asked. And in a very, you can say difficult to understand language. Okay. They'll mix some, you know, language and they will say, okay, are these two events independent? If you're able to figure that the two events are such that the compound occurrence of A and B, that means the probability of occurrence of A and B is the product of the probability of occurrence of A and B. Then only you claim that, then only you claim that A and B are independent. Okay. We'll take some questions. Don't worry based on that. But don't judge the question just by looking at it. No, I feel A and B are independent. No, they don't run by your wins and fancies. If you feel any of your current then nothing like that. I mean, that means you have skills. You have, you're blessed with the art of finding the dependent independent nature without solving the question, but rely on this operation to basically know whether A and B are independent or not. Okay. Is this fine? Are these two examples good enough? So now let us look into some properties of dependent and independent properties of independent event. Okay. Let me, let me ask this as a question rather than giving you a property. Okay. So let's say there is an event A and there is a null event. Are these two independent? Yes. In that case, it will become a conditional probability. We'll talk about that. Is event A and a null set independent events? The answer to this is yes. They will be considered to be independent events. Okay. For the simple reason, they satisfy P, A intersection, a null set, which is actually what? Actually zero only, right? And this also is a zero. So this will be also a zero. This will also be zero because this is going to be a null set only. Okay. And zero will be equal to zero. That condition is satisfied and that's why they are independent. Okay. What about A and the sample space? Is A and the sample space independent or dependent? Let me ask you like this. Are they independent or not? Independent. So what is P A intersection S? It is as good as P A. Okay. And what is P A into P S? You'll say it is P A into one because sample space occurrence is one. Are they both the same? Isn't it? So this implies P A intersection S. Is P A into P S? The moment this criteria is solved, that means P A intersection B is P A into P B, immediately you can say, yes, they are independent. Okay. Not a very useful property, but you should be knowing it because you can be asked in a multiple option correct, which of the following options are correct. Okay. Now, if A and B are independent events, if A and B are independent events, then what can you claim about A and B complement? Will they be independent or will they not? I mean, will they be independent or you can't say? No, I don't think so they will. Can't say means it may be independent may not be. Okay. Depending upon the cases. If A and B are independent, will A and A complement be independent? Okay. Shambhu doesn't necessarily have to be. Okay. Okay. Shatish is saying it has to be. Pranav is also saying it has to be. Many of you are saying it has to be, but some of you are saying need not be. Okay. Let's discuss. So as of now the verdict, I don't know. Okay. What is this? I don't know. I'll just put a question mark, but I'll just try and check my understanding of the fact that does this give you P A into P B complement? That's the tests that you should always perform. If you're not sure whether, you know, two events will be independent or not. Don't take a call based on your, you can say gut feeling that I feel they're going to be independent or I feel that they're going to be dependent. I feel like, you know, there's nothing we can say about. Just take a call basis of what comes out as a result of this operation. Does it give you P A into P B complement? If yes, my answer to this is yes, they are independent. If it gives something else, no, we can't claim that it is always independent. Okay. Let's check it out. So P A into P B complement. Okay. So we all know that from our diagram that this is the area which shows, this is the area which shows P intersection B complement. So can I say it is as good as saying P A minus P A intersection B? Yes or no? Correct. Now, since A and B are independent, since A and B are independent, can we not claim that P A intersection B is P A into P B? We can claim, right? If they are independent, this condition will be true. Okay. So in light of this, I'm just rewriting the expression back. In light of this, can I say this term, this term, I couldn't write it as P A into P B because, you know, A and B are independent. If you take P A common, it will be one minus P B. Correct. And what is one minus P B? What is one minus P B? P B complement. Correct. So this I can write it as P A into P B complement. So what did I figure out, my dear? Those of us saying, not sure. We figured out here that the moral of the story is that this was giving you P A, I'm so sorry. P A into P B complement. That means the verdict here, the question mark here now is resolved. They have to be independent. So even these two events are independent events. Is it clear? Any questions? Any concerns? Good enough. Okay. Let's say I call this as three A. Let's move on to three B. Do I have to explain anything over here or everything is clear from this operation? Again, I'm using my addition theorems to come up to these results. So class 11 becomes a pivot point for me. Or you can say it becomes a prerequisite for me to get to these results. These are all addition theorems, whatever I'm using. Okay. All right. We'll move on to third B property now. Third B. Again, I'll list it down without proving it because it is very obvious. Similarly, A complement and B. They will also be independent. So if A and B are independent, then even A complement and B will be independent events. Same logic. Just that name is changed from A B to B A. Nothing else. Okay. Now this C part I would like you to again through this. Do you think their complements are they independent? What do you think? Yes. No. Can't say. If A B are independent events, if A and B are independent events, is a complement B complement independent or nothing can be said. Not independent. Okay. Sharan, anybody who thinks otherwise? Okay. Vavav has a different opinion. Okay. Good. So there are two levels in the session. Vavav M and Vavav S, right? Okay. Okay. Vavav M has kept M in the name. I think that was the dance also. Okay. Okay. Good. Good. Good. Two gaitries also would be there. No, one of the guy. They are, they are not okay. So both the answers are coming. Okay. Let's start from basics. Okay. Let's not take a judgmental call here. Let's take a logical call. See. All I need to show to the examiner or show to myself more importantly that if this is equal to P a into P b, then only I will claim it to be independent else not simple as that. Now we all know that by D Morgan's law this is nothing but P a union be whole complement. That's nothing but one minus P a union be correct. Now what is P a union be P a union be is P a P b minus P a intersection be correct, which is one minus P a minus P b plus P a intersection be okay. Now you know and we are independent if and we are independent this guy this guy could be written as P a into P b. Okay, now we know that since and we are independent independent independent P a intersection be is P a into P b. Right. So this becomes this becomes one minus P a minus P b plus P a P b. Okay, so take P b minus P b common from the seconds last two terms you'll get one minus P a. Okay, so basically factor is it so you have two factors one minus P a coming and one minus P b coming one minus P a is P a complement one minus P b is P b complement. That's what I wanted no right. In fact, I wanted to verify that whether this is coming or not so what did we conclude that if this is coming out from two such events. Okay, what does it mean, it means that these two are calm are independent events these two complimentary event these compliment and be compliment are sorry independent events. Okay, so what are the model of the story if a and b are independent it makes a be compliment both events independent a compliment and be also independent and a compliment be compliment both also independent. Okay, so these questions have been directly asked in several competitive exams. Yeah, sure, if you want to copy anything down please do so. Do you remember this property I will never say to remember anything. This is actually I mean, sometimes we are hardwired to think also in this nature. Okay, but many times when we don't. I would always suggest you use your basic addition principles use your basic definitions and get it done. If you know it if you remember it you save time. But if not if let's say there was a question where it had figures instead of a and b it had situations instead of a and b and the question you asked me whether these two events are independent. Then I would not you know think of this formula first maybe I will you know go from my basics. So I will try to see whether PA into PV whatever other events whether it is written in terms of compliment or not compliment is it PA into PV. If it is satisfied then yes they are independent. But learning, if you are able to see such, you know exact expressions in the options, then probably remembering the formula would have helped you. Okay, else you have to solve it as if you have been given some neutral question. Okay, now based on this there is one more property and fact. It's actually an extension of property number three last part this is called the complementation law or complementation to complementation by my handwriting is not that good today. So compliment in complementation rule says that if and we are independent events are independent events. Then PA union B will be one minus PA compliment into PV compliment. Okay, now the proof for this is same as what we did just now. The proof is same as what we did just now we basically did the fact that if a and b are independent events so would be their compliments right. That means we learned this from our previous property. Okay, so from this is from our previous property, previous property. Correct. And what is this, this is PA union B whole compliment. So this is just a new version you can say another way of looking at it. So one minus PA union B is PA compliment into PV compliment. So basically, you get the proof as PA union B as as one minus this. Okay. Now, do you recall those questions where you still have this in class 11 that let's say and be both are trying to solve a problem. The chances of a solving a problem is that say, you know something chances of be solving a problem is something what is the chance that the problem will be solved. So what do you used to do is to say, okay, let the problem be not solved by both of them. One minus that will be the problem getting solved by at least one of them. So this is the same thing actually written in a, you can say mathematical, you know, jargons. So I say some people are hardwired to solve this question even without knowing these properties. So indirectly they will be using this property without knowing that they were using this property just because they could frame the question to logic. So logic is above everything. All these formulas properties are, you know, evolving from that logic. Never say, you know, keep a side your logic and work with formula formula is going to ditch you long time, like a big time. Are we assuming nb are exhaustive? No, that was never said here. And we are independent events, exhaustive events work when they are, they are events coming from the same, you can say bigger event. Right. So it may be like tossing a coin and rolling a die. So how do you talk about exhaustive? Are you? No, it's always true. Are you? Whether a is exhaustive or not exhaustive this, this thing will always be true. No. Oh, sorry. I wrote a compliment on top. Yeah. This, this is always true. No. See, independent and exhaustive. See, I mean, you have to define an event like this exhaustive means they together will make the sample space. The sample space of that entire experiment that you are performing. Let's say you are tossing a coin. Head, tail, they're exhaustive events. Okay. Are you getting my point? And they are exclusive events. You can't use the word mutually. You can use the word independent there because they can't occur together when the experiment is performed. Are you getting my point? Your question is, can there be events which are independent and exhaustive? So independent events. I mean, ask yourself independent events. Do they come from the same experiment? This is tossing a coin. Right. If I toss a coin. Or if I roll a die. Can you tell me two events here which are independent of each other? But can you tell me two events which are, which are exclusive and exhaustive? Exclusive. I think getting a one, two, three, four, five, six, they're all exhaustive events because if you together add them, they will give you a one. Try to understand the definition of all these things. I think I need to go back to class 11. When you say exhaustive, even E2, E3, let's say these are the events associated with an experiment. They are exhaustive and they add up to give you a one. Okay. Now many people think PE1 plus PE2 et cetera, they will give you one. That is only when they are exhaustive and mutually exclusive. That happens in case of complimentary events. Oh, don't get confused here. So I'll just give a simple example. Let's, let's forget about even E2 till Ian. Even an E2 are exhaustive. They will satisfy this condition. Correct. If even an E2 are exclusive, then they will satisfy this condition. Right. So basically this is equivalent to saying even union E2 is your sample space. Right. But that doesn't mean they are exclusive of each other. Correct. Here, even an E2 are exhaustive if they are disjoint. Correct. Even an E2 are complimentary. Let me write it down here. Even an E2, sorry. I'll erase it. Even an E2 are complimentary, complimentary if they are both actually. That means even union E2 is your sample space. And even intersection E2 is a null space. If both these criteria are satisfied, then they are complimentary. Now, if they are complimentary and they're satisfying this. So first of all, you can say that this is equal to one and you can say that this is equal to zero. Now, using this fact, you can say P E1 plus P E2 minus P E intersection. Even intersection E2 is equal to one. And because of this, this fellow becomes a zero. And because of this, you end up getting something like this. And because of this, there is a property that if even an E2 are complimentary, one will be the probability of one will be one minus the probability of the other. That means if let's say if I write X here, this will be one minus P X compliment. So if an event X you're talking about with its complimentary, no matter whatever is the event, this condition will always be satisfied. Understood. I think there is a lot of doubt related to exhaustive, exclusive, complimentary. And now you're learning independent also. So independent means that PA intersection B is PA into PB. And A and B need not come from the same experiment. Getting a point. So they will not come from the same experiment. Like tossing a coin rolling a die independent. You being tall and you being good in studies independent events. It height doesn't reduce your brain power. Many people say shorter people are intelligent. So all the girls must be intelligent. There's no correlation like that. Okay. Is this fine? Any questions? Now this property can be extended. That means you can generalize this property. And you can say that if you have even E2, E3, etc. Some set of events which are all independent of each other are independent events. Then their union, their union would be one minus the product of their compliment probability. Okay. So you just have to remember this property and you just have scaled it up to this. Okay. So in general, this property is going to hold. Okay. This is called the complimentation law. And again, I'm telling you guys and girls. This is only when the question gives you, you know, sets kind of a thing in the question. If they give you some kind of a situation, this can automatically be obtained from logic. You don't have to remember it. So I'm giving you all these property because sometimes they will say, okay, A, B are independent events, which of the following options are correct. A and B compliment are independent. A compliment, B are independent. A compliment, B compliment are independent. PA union, B is one minus PA compliment. So in that situation, if you remember this property, you are through. You can save a lot of time. But normally they give questions where there is a wording, there is a situation involved. And in that case, these formulas will automatically be obtained from the logical part. You don't have to remember it. If you remember it, good. Not also, it will not hit you. Let's take questions. I think we have discussed our independent events to a considerable depth. So let's discuss them out. Again, I would like to ask anybody, is there any confusion related to exhaustive events? If you want to give me examples, I can give you. For example, exhaustive events are let's say, you getting, let's say I take a situation. Let's say you're rolling a die. So even is you getting one, two, three, four. Okay. E2 you are rolling a die and you're getting let's say three, four, five, six. Are these two events exhaustive when you're rolling a die? You'll say yes sir. Because if you take the union of both of them, you get a sample space. So they're exhaustive. So in this case, even union E2 will be a one. But are they mutually exclusive? No, because they have three and four common. So let's say I give you a mutually exclusive events. So rolling a die mutually exclusive events like you getting an odd number and you getting an even number. But right now I have taken a case where they are mutually exclusive, but yet exhaustive. But you could have a case like this also. Let's say I don't put a six. So this is mutually exclusive events. They do not have anything in common. So this is mutually exclusive, but not exhaustive. Mind you, because six is not there. If I put a six there, then exhaustive and mutually exclusive. So here their intersection is a null set. Simple as that. What is complementary event? Now again, these two are complementary also because let's say if I put a six. Let's say one, three, five and two, four, six. Then these are complementary events because together they make the sample space and mutually exclusive. They are mutually exclusive, nothing in common. Correct. So that is why you say the probability of you getting an odd number is one minus you probability of getting an even number. Get this distinction clear. Right. But if I say, are they independent events? So you say, let's check. Are these two independent events? Let's check. So what is the probability of even intersection E2? Even intersection E2 have only two events. Correct. So their probability will be one by, sorry, two by six, which is one by three. Is it equal to PE1 into PE2? Let's figure it out. PE1 is what probability of you getting one, two, three, four is four by six. Probability of getting three, four, five, six is again four by six. Multiplication will give you something like four by nine, not equal. So they cannot be called as an independent events. What's the point? Is the idea clear now everybody? Okay. Please do not be in any confusion. Okay. Because later on you'll be going to hit very hard when typical problem comes on this. Oh, sorry. You have not done conditional probabilities. So mad question to pick up. Sorry about that. I think I've taken some. Oh, let's take this one. Now I'm not putting any poll for this because I'm not very sure whether multiple options may be correct or not. If you think multiple options are correct, please give me all the choices that you think are correct. Is there any person sitting here who can actually read this and say, sir, this is often is correct without picking up the pen. It's difficult actually, right? I mean, knowing to know into events are independent by just reading the question. See, many times it is obvious and I don't deny that. But most of the time I feel reading the question will not say, ah, this seems to be independent. Anybody who basically has that power of guessing like that? Prandah wants to try. Okay, Prandah, go ahead. Very good, Kinshok. Excellent. Okay, Anil. Okay, Raghav. That depends on the person also, Kinshok. Okay. So Anusha, you think they are pairwise independent also and all of them together also are independent. Okay. Okay, Ruchita, Ritu, Gayatri, Rohan, Prandah. Okay, should we discuss it, guys? Anybody else? Anybody? Gaurav. Gurman. Gurman has been quite for long time. Where is this guy, Oshik? Oshik Dhar. Is he all right? Why doesn't he come for class? Oshik, Oshik from Kormangala. Yeah, yeah, ask him to keep sleeping. He'll wake up in a very, with a very bad dream then. He has not been coming to class and same goes with Arabi. Arabi and yeah, I've been observing few of you have, are not coming for the regular classes. There is a guy called Raghuram. Raghuram in YPR, he also same thing. Okay, let's discuss it. See, your kismat is in your hands, right? No teacher, no coaching institute can change it. Not even your parents can change it. You can only change it. See, first of all, E1. E1 is what? E1 is the event where the ticket that you have drawn has the first digit as two. Okay. So can I say that favorable event will be two if you pick 211 or 222. So favorable will be two, total number of tickets is four. So probability of E1 will be half. Probability of E2. Neat. He's considering what he will write. No. So am I not teaching board stuff? This is all board stuff. Yeah, so E2 will be what? Probability of you getting the second digit on the ticket that you have drawn as a two. So there are two such cases again, 121 and 222. So that will again be one by two. Now probability of E3 is the third ticket being two. That is this and this again. So again, one by two. No doubt. Okay. Now, what is the probability of even and E2? That means the first and the second digits both are 22. That can only happen in the last case. First and second both are 22. So one by four. Is it equal to PE1 into PE2? Yes, it is because both are half, half each. So even and E2 are independent is absolutely correct. So option number A will be right. Now E2 and E3. So the second digit and third digit both should be two. The probability is again one by four because only in the last case it is happening and this is also equal to half into half, which is this. So E2 and E3 are also independent. That means option number B will be correct. And I'm sure E3 and even will also be independent because third digit and first digit being two is also happening in only one case out of four, which is actually half into half again. Okay. So this is also independent. ABC is PAKKA correct. But are all of them correct? That means all the three are independent. So let's check. So even intersection E2 intersection E3, that means all of them, all of them being 222 is only one by four. That's the last number. But this doesn't match with half into half into half. That means it doesn't match with this. So claiming the last option to be correct will lead to a negative mark. So only option ABC are right options. These are the only right options. Any questions here? Any questions, any concerns? Next question. Okay. I'll give you a question from my side. They are mutually independent, but not taken together. How come two events be mutually exclusive, but three may not be? Are you getting my point? See, sorry, let's take a question here. Again, this is a question which is based on your word problem. So there is a family which has got three children. A family has three children. Event A is that a family has at most one boy. Event A is that the family has at most one boy. Has at most one boy. B is the event that family has at least one boy and girl. At least one boy and one girl. And even C is family has at most one girl. Which of the following options are correct? A and B are independent. A and B are not independent. A, B, C are independent. A, B, C are not independent. Simple options. Okay. Pranav, Anural, Archit. Good. Quick to answer. Okay, Raghav. Raghav, you are consistently giving incorrect answers. Last also you got it wrong. You said D only. Where D was the only option which was not correct. Of course, here also you are wrong. IMO and all the probability is the main key ingredient of these IMO and whatever camp you are trying to be. Okay, Shatish. See, first of all, I would like to, you know, irrespective whether you have solved it or not. Okay. Let's try to answer the following questions. Okay. What is the probability of A to happen? Just tell me what is the probability of A? The family has at most one boy. Right. So how do you find it C? Now the family has got three children. Right. Child one, child two, child three. Okay. You want at most one boy means either the family doesn't have a boy or the family has exactly one boy. Am I right? Okay. So can I say this is as good as finding no boy or exactly one boy? What do you are finding? So what are the probability of no boy? See, by the way, these two events are mutually exclusive. Okay. So you can directly write it like this. So you can break this event as probability of no one, probability of exactly one boy. Okay. What are the probability of no boy? One by eight. So out of total cases, see each of them can have two genders, right? Male, female, male, female, male, female. All of them have to be female, female, female. So one such case out of eight, or you can say half to the power of three. That is also a good way to talk about it. Okay. Right. Actually you're using the fact that child one, child two, child three, their sexes will be independent of each other. Isn't it? So half chance of being a boy, half chance of being a girl. So half into half into half. The probability of exactly one one of you, you have to choose which one you want to keep as a boy out of the three children, which one you want to be a boy that itself can be chosen in three C one way. And then again, half to the power of three. Correct. So yes, I'm so sorry, this is power of three, not power of two. Distinct. They're distinct. You having one boy and that boy is the first child is different from you having one boy and the boy is the second child and you having one boy and the boy is the third child. That's distinct Aditya. Why not? You're dealing with distinct objects, human beings, children. Okay. So the answer is one by eight plus three by eight. Answer is four by eight, which is half. Let's see. I mean, does it matter in the question that is independent? Right. Now what is PC? Same will be PC also because you getting one at most one girl is like you're into saying you're getting one one boy because the situation is same that to one girl now. Okay. Now. What about the B case? At least one boy and one girl. So one boy and one girl is saying you have one boy plus two girls scenario. Or you have one girl plus two boys scenario. Okay. So again, we can say they are mutually exclusive. So you can just say this is as good as this plus this. Okay. So the probability that in the family, there is one boy and two girls. Now, here also, very important. You have to first decide which two children will be girls. Of course it's all in the hand of God's. So three C two way of deciding which children will be girls. Okay. And then we'll be half into half into half. So you can say half cube again. Oh, so sorry. Half cube. Same goes with this also three C two into half into half into half. So here in this case, it will be three by eight into two. That's three by four. Is it fine? Okay. Now, having known P a, P b, P c, let's try to address the question. The question says A and B are independent. Now, what is A and B? A and B means you want, you want at most one boy. But at the same time, you want at least one boy. That means it has to be a one boy two girl case. Am I right? Am I right? This has to be one boy. Why am I writing one in? Okay, one boy and two girls case. Correct. So what is that to happen? I think we have already figured it out over there. That is going to be three by four. So is it equal to, is it P a into P b equal to three by four? Let's try to figure it out. Does it equal to three by four? Let's try to figure it out. P a was half. P b was three by four. Sorry, this was three by eight. I believe, I believe, sorry, sorry. Yeah, I forgot. I'm just dealing with half the expression. Wonderful. So yeah, it is equal. It is equal. Okay. So what does it mean? It means P a intersection B is P a into P b. So they are independent. Option number A is right. Option number A is right. Got it. So B cannot be right. Let's check C now. Are ABC independent? So ABC means what you want? You want at most one boy. You want at least one boy, one girl, and you want at most one one girl. So which is the situation? Can it, can it happen actually? Can all the three be simultaneously? Can all the three be satisfied that you are having? At most one boy. Okay. At most one girl. Correct. And at least one by one girl. So let's say one boy and one girl A and C are already satisfied. And you have to have three children, right? So the third one will be a boy or will be a girl. If it is a boy, it will violate A. If it is a girl, it will violate C. So together they cannot happen. It's my coin word together. Together or together. Together they cannot have, right? And this doesn't match with PA into PV into PC, which is, which is equal to half into three by four into half. So they are not independent. They are, they are, they are not independent. So option number A and D are correct in this. There are some people who solve this question pretty, you know, pretty fast. How did you solve it so fast? And you take, you have, you can share. I mean, even if I don't write all the steps or you use one of the previous methods. Okay. Is it fine? I think we have done enough with our independent. Now let's move on to condition probability because that's where your, you know, major crux of the question or major part of the question are going to come. Okay. So I'm getting a lot of chats here. I'm just asking so many questions. Can you just unmute and talk because you have written so many things and I have to scroll up and read all the stuff. If you have any questions still, you can unmute and talk. Oh, you did a mistake. You realize it. No worries. It's fine. Okay. So before I move on any questions related to independent events, so you'll be able to figure out whether two or more events are independent or not. So just follow whatever properties and things we're doing. And as you can see in these questions, reading, reading them that people read. So A is this B and this. Oh, yeah. They're independent. That may not be possible every time unless until you are trained to think like that, which I don't know how will it happen. Maybe Kinshuk has some idea. But if I were to do this one, I will literally find PA, PB, PA intersection B and check whether PA intersection B is PA into PB or not. Okay. So moving on to conditional probability. Conditional probability. Okay. Okay. See conditional probability. The name itself says there is some condition which has occurred. Okay. So whatever probability you have, you're trying to find out. You have been told some condition. Okay. Under which you have to evaluate it. See, let me give you a funny story. Let's say there are, there are two contestants for 100 meters. So once they let, let's say one of the contestant is, okay, Charan. Another contestant is Mr. X. Okay. So Mr. X and Charan are, both of them are running for 100 meters together. What is the probability of Charan winning the 100 meters? What do you say? What is the chance of Charan winning that 100 meters? On a serious note guys, please. Half. Correct. But what if I say Charan is running with Mr. X, whose name is Usain Bolt. So he's running with Usain Bolt. What is the chance of Charan winning it? It's a zero set. And this guy will. Okay. So here is the answer changed because you knew a condition. Right. You knew something. I mean, not exactly the same as what I've given an example. This is just a funny example for you to get connected with the topic. So if you know something about the event or something that occurred before the event, then the probability of that will depend on what occurred. It'll depend upon that condition that occurred. For example, let us say, again, now I'll give you a realistic example. Let's say there are, there are three red balls of different types in a bag and let's say four blue balls of different types in the same bag. Okay. You are drawing a, you're drawing two balls, one after the other, without replacement, without replacement. Okay. Correct. My question to you is, what is the probability that the second ball is a blue ball? What are the probability that the second ball is a blue ball? Now here if you see, which he normally does, I think in one of the races he pulls a muscle and he abruptly stops. Yes. So now because the question says there is a without replacement scenario, second ball being a blue depends on what has been drawn as the first ball because if you had drawn a red ball before, then your answer will be different. If you had drawn a blue ball before that second ball, then answer will be different. Isn't it? It is not a with replacement that your answer doesn't depend upon what was drawn in the beginning. Had you put the ball one back again in the bag, then the probability of you drawing the second ball blue would have been as good as a four by seven. But now actually it depends on what you had drawn in the first go. Correct. So the probability of you drawing a blue ball basically will depend whether the previous ball was a blue or the previous ball was a red. Okay. Now I have written some symbols on your screen right now. PB by B, PB by R, et cetera, et cetera. Okay. So what is this? Let me explain you all those things. So probability of A by B, now it is written as A by B, but it is to be read as like this. So when you're reading it, you will read it like this. Probability of occurrence of probability of occurrence of A, given B has occurred. Okay. And this result will be different. If your information about, you know, if they're independent, I mean, if they're dependent, answer will be different. If they're independent answer will be different. If A and B are independent, then this value of P, A given B has occurred is as good as A. Like the one that will happen when this, the balls were replaced back. Okay. The first ball was replaced back. Right. But if the first ball is not replaced back, let's say the first ball was a blue, then the answer will not be the same as what you would have got when you had taken these two events as independently. Right. Right. So we'll talk about all those things. Okay. So please note that when you say P, A divided by B, you read it as probability of occurrence of A, given B has occurred. And this is not equal to P, A, unless until A and B are independent events. Okay. And of course, in this case, your P, B should not be zero. Okay. Because if B hasn't occurred at all, then the statement doesn't make any sense that probability of occurrence of B given B has occurred. Okay. So B is an impossible event that is you're trying to say. So B has to be some possible event. Then only you can talk about probability of occurrence of A given B has occurred. Okay. So now my next question is, okay, said this symbol is fine. We have understood what this symbol say. How do you write down the formula for this? So in order to write down the formula for this, we read this conditional probability in light of multiplication theorem, which is very easy theorem again. See all these names have been put in order that you refer to it while you're writing your answer. Most of these things are intuitive. But in the exam, you cannot say no intuitively thinking this is the answer. Right. You have to support mathematics is a very, very logical subject. It's not like humanity. You have to give reason, answer for everything that you're writing. So in order to say something, write something in the examination hall, of course I'm talking about your board point of view because your second semester may be subjective one. You have to use these mathematical jargons because that, that increases your chances of getting higher percentage. Okay. So multiplication theorem says that if A and B events occur. Okay. It can occur like this. A occurs and then B occurs given A has occurred. Okay. Or you could also write it like this. P B occurs into P A occurs given B has occurred. By the way, here, please note that P is not equal to zero here. And in this case, P B is not equal to zero. Then only. Okay. This will make sense. Right. So this is conditional probability. So I was asking you the question that what is the probability of getting? So let's say again, I'll, I'll quote the same question. Three red distinct balls for no distinct blue balls. You're drawing two balls without replacement. Okay. What are the probability of you getting one red ball and one blue ball? Okay. So when red ball, if you want, you can, you know, take this event in two ways. Either the first ball is a red. Whose probability is three by seven and given the first ball is a red, the probability of you getting a blue ball will be four by six. Are you getting my point? Clear. So this is what I'm calling as P A and this is what I'm calling as P given, you know, B has occurred. Okay. Or you can also say, if you want a red and a blue, I mean up at the end of the, you know, this thing. When I say red and a blue, this could be red here and this is blue. Second ball is a blue. First ball is a red and second ball is a blue. Is it fine? Any questions? So this is saying a red ball. First ball is a red. Second ball is a blue. Now, can this multiplication theorem be extended to multiple events? Yes, we can generalize this. We can generalize it to saying that if you have A1, A2, A3, etc. Okay. You can write it as P A1 into P A2 given A1 has occurred into P A3 given A1 and A2 has occurred into P A4 given A1, A2 and A3 has occurred and so on and so forth provided. Remember provided these guys are not null sets. Okay. They are not null sets. Okay. And so on. So using this multiplication theorem, you can always find out the expression for this. So I'm using this multiplication theorem part. And from here I can say that your P A by B is given by P A intersection B by P B. Okay. Please note this formula down. This is the formula which you are going to use, you know, N number of times in this chapter. Okay. So probability of occurrence of A given B has occurred is the probability of occurrence of both A and B divided by probability of occurrence of B. So in the previous part, we did the same thing, but we did not change. You're talking about when the replacement is done. No, in the previous part, P A by B became P A only because A and B were independent. It had nothing to do with whether the first ball was a red or not. So the second ball is a blue. It had nothing to do with the first ball was a red or a blue. Okay. But here what is happening you have a red ball as the first ball and blue ball as the second ball. So the probability of you getting blue ball as the second ball, given the first ball was a red will now become four by six instead of four by five. So that's that figure actually became higher because you will know, you know that there is something occurred before this event which increases the chances of you getting more blue balls or the chances of you getting a blue ball. Are you getting my point? It's like I gave that question. If one of the person is Shanmold, you know the chances of meaning is almost zero. So if you know there is a red ball drawn before it, the chances of you getting a blue in the second draw becomes four by six instead of four by seven. Higher chances. Okay. This is what is called the conditional probability. Any questions, any concerns here? I'll be taking word problem based on this, not to say, but before that I will take some properties of this formula or properties of conditional probability before I move on. Now here, many people ask me, sir, can we ever write this as PA by PB? So note, can you ever write this as PA by PB? If yes, under which situation will this happen? Tell me, tell me, tell me. Under which situation can you write this? Right. When A is a subset of B. So note, this will happen, happen when A is a subset of B. Obviously, why? Because if A is a subset of B, A intersection B will be A. If A is a subset of B. So the numerator part, which is PA intersection B, that will become PA only. So many times, you know, this question is also framed like this, that can this be a possible answer for PA by B? The answer is yes, under the situation where A is a subset of B. Okay. Now let us go to some properties of, let us go to some properties of conditional probability, properties of conditional probability. First property here is that probability of occurrence of this is actually PA itself. Okay. It is obvious. The proof is very simple. PA given as has occurred is PA intersection S by PS. PA intersection S is PA because A is a subset of S. PS is one. That proves the, okay. That is the real property, but you should be knowing it. But on the other hand, probability of occurrence of S given A has occurred is actually a one. Again, the proof is very simple. Same way. PS given A has occurred is PS intersection A by PA. S intersection A is A. So PA by PA, that will actually be a one. Next, probability of A complement given B has occurred is one minus probability of A given B has occurred. Now, please don't get confused here. Many people use this property incorrectly. They say probability of A complement by B complement is one minus. No, it's only A complement on top. No B complement. Okay. Can you prove this in the interest of time? I'll be only doing it. So PA complement given B has occurred is PA complement intersection B by PB. Okay. And what is PA complement given B? Now remember again the Venn diagram. Venn diagram are very useful stuffs. Okay, let's say this is A and this is B. This zone is your B intersection A complement, isn't it? This zone is your B intersection A complement or A complement intersection B. Doesn't matter. It's a commutative operation. So can I say the numerator is PB minus PA intersection B? Okay. So if you individually divide, you'll be getting one minus PA intersection B by PB and PA intersection B by PB is PA by B or P of occurrence or probability of occurrence of A given B has occurred. Is it? Okay. Now, obviously a question will arise in your mind. So sir, how can we write PA complement intersection B complement? See, for this, you have to again use your basic funda. Don't start guessing anything. Okay. Please apply your basics. So basic says that this is going to be this. Okay. So basic says that this is going to be, okay. I have not done that part, but still I will just write it down. Okay. This is again one minus PA union B by one minus PB. Okay. Again, this depends upon, this depends upon your expressions. Okay. So please do not get confused. This and this are not the same things. So they're different things. Okay. Next property that I would like to discuss here is what is PA union B by C? First copy this down. If you have any questions, anything that you would like to ask. All right. I think I can take it. So this is a simple. PA union B given C has occurred is given by PA given C has occurred, PB given C has occurred minus PA intersection B given C has occurred. Okay. As you can see here, my event on the numerator part over here is a union B. Okay. So how do we prove this simple? In this case, we will again rely on our basic definition of PA given B has occurred. So I can say this term is going to be P C intersection A union B upon upon PC as per our definition of conditional probability PA union B given C has occurred is this. Okay. Now you can write this by using your distributive law. So you know how to distribute intersection over union. So this will become C intersection A. Okay. Union C intersection B. Now, let's use our addition principle on this, treating this as X in our mind, treating this as Y in our mind and then I write it as P X, P Y minus P. Now see this intersection this. Okay. If you look at the last term, I'll just rewrite it again once. Okay. If you look at the last term, this is as good as probability of C intersection A intersection B. Am I right? Yes or no? Okay. Are you happy with the writing of this last term like this? Any doubt related to it? Again, from your sets, you know, topic. Now individually divide them by the denominator. This is as good as saying PA given C has occurred. This is as good as saying P B given C has occurred. I'm so sorry B instead of P I wrote up B. Yeah. And this is as good as A intersection B given C has occurred. Is it fine? Any questions? Any concerns? Anywhere you want me to scroll, please do let me know. You want enough? Please scroll up. Oh, sure. Sit down. Scroll up means you want to see this part or you want to see the down part. See, it says that what is the probability of A or B to occur given C has occurred. So if you see the left side, the way you pronounce it in the language of probabilities, what are the probability that A or B has occurred? So it is written as probability of occurrence of A given C has occurred or probability of occurrence of B given C has occurred and from there you exclude the case or exclude those scenarios where both A and B has occurred because they will be a double counting. So this exclusion is because in the counting of those cases where A and B both have occurred when C has occurred, they will already be taken care once in probability of where A has occurred or the number of cases where A has occurred, given C has occurred. So take your thinking process to your sets level. In set what was there? Let's say this is the number of events where A occurred, given C occurred. That means A and C both together occurred. This is the number of events where B occurred, given C has occurred. And this is the number of events where both A and B occurred, given C has occurred. So this is a double count in both of them. Let's move it once. Okay, now what type of questions will I get? So let's take the different varieties of questions that you can be asked. So first of all, I'll be taking up some questions where there is a... Okay, so we'll take... Yeah, let's take this question. I think this is a single option correct question. No, multiple option? Okay. Question is saying which is not true by the way. Not very sure. Okay, Rohan. Let's do one thing. Let's have a poll if this is a single option, correct? Let's have a poll. So again, a reminder that tomorrow will be again continuing with the session. Okay, so Monday, 8 a.m. to 11 a.m. you will have a maths class announcement. Math session, maths class. No idea till when it is postponed. The matter is subjudice. It all depends upon how much time the court is taking but given that it's a career related issue, the court will take it up on priority. The honorable high court is going to take it on priority. And we'll see. When do you think? I don't know. First week of... I think you are not attentive in the last class. I said I predicted to happen in the first week of January. I think with you people only I was guessing no. No. Okay. Fine. We'll stop the poll. I think 13 of you have responded. I'll give you five seconds to please give forth your response. Take a guess if you haven't solved it. Five, four, three, two, one, go. When is the registration deadline for what? Siddhant. Has the J notification come? Has the J main registration come on their website? Nata website? NTA website? It hasn't come yet. Keep looking at the J main. So there is a window one month to register. Okay. So once the notification comes that the exams are going to head on so and so date, they will give you everything. All the red lines will be given their Siddhant. Last day to fill, register yourself. Last day to fill the registration fees. Last day to do the correction in the form. So all those dates will be disclosed to you on the the J main website. J main.ac.in, I think. Okay. We are also looking at it, but I think it will not come because they have recently concluded everything. So keep an eye on it. Keep an eye on it. Maybe every day once they say you open your Instagram page or Facebook page, you can open your J main page also and have a look at it. Okay. Nine people as an option B. Okay. Let's check. First of all, we have been provided that PA is one by four. PA by B is also one by four now PA by B is what PA intersection B by PA. Sorry, PB. Okay. This is also given to us one by four and you have been given that PB by is one by four. That means PA intersection B by PA that is one by two. So these are the three set of informations provided to us. Okay. Now, A and B are independent to know this. I need to know what is PA intersection B. So PA intersection B, I can get from the last information. So from here, I can say half into PA. PA is one fourth. So it is going to be one eight. Okay. One eighth is your PA intersection B. This is one eighth PB can be known. So PB is half. Okay. So is PA intersection B. PA into PB. Yes, definitely it is true. That means they are independent. Okay. So option number one is wrong because this is a correct statement. This is a true statement. Hence it cannot be my option because they're asking which is not true. Next B. This is from the property itself. One minus PA by B. Okay. So one minus PA by B is three by four. This is also a true statement. So cannot be a right option. Okay. Because this is also true. C says PB compliment by compliment. So as we had already derived this, I think this was one minus PA union B by one minus PA. Correct me if I'm wrong. Correct me if I'm missing out anything. So what is PA union B? PA union B is PA plus PB minus PA intersection B. PA I think is already given one fourth. PB we just now figured was a half and PA intersection B. Sorry, this word slightly. Yeah. PA intersection B. That also was figured out to be one by eight. So this will give you two plus four minus one. How much is it? Five by eight and down in the denominator, you have one minus PA one minus PA is one minus one by four, which is three by four. So your answer is going to be one minus five by eight by one by three by four. So this is I think three by eight divided by one by four. That's half. So this is also correct. So this cannot be my answer because this is a true statement. So the answer to this question is none of these. Let me check the poll. Only four people said none of these. Why? Any doubt? Any questions here? Anybody? No problem. Should we take the next problem now? Okay. So I'll still take it on the properties that you have done. Let's take, I don't plan to start with the word problems now. I can take this fellow. Should I put the poll for this? Okay. Very good. Two people have responded. Different answers coming up. Okay. So let's hope this is correct. Last one minute. Those who want to poll, please give a response on the poll. In the last 40 seconds now. Okay. Five, four, three, two, one. Go. Okay. End of poll. I could see only 15 of you having responded on this and most people say option C. Very few people say option A. Okay. So Raghav who is saying A very few people are concurring with you on that. Let's see whether they all are wrong and you are correct or vice versa. So first of all, PA is 0.7. Undoubtedly. PB is only given 0.5. PA intersection B is given to you as 0.3. Okay. Now. It is, I think the probability is high. Let's see. Let's see. So P be given a union B complement has occurred is nothing but probability of B intersection a union B complement by PA union B complement. As per the multiplication theorem, this is the definition of probability of a conditional event conditional probability. So this is going to be, if I'm not mistaken, PB intersection a union PB intersection B complement, but this event is a null null event. Okay. So that is as good as taking union with some nothing that is as good as PA intersection be only. So this is going to be your final expression isn't a intersection B is 0.3. No doubt. What about the denominator denominator I can say PA PB complement minus minus PA intersection B complement. So PA is 0.7 PB complement is 0.5. Now what is PA intersection B complement all of you please pay attention. This area is PA intersection B complement. So isn't this isn't this PA minus PA intersection B isn't it. Right. So I think when you open the brackets PA PA will get cancelled off and you'll have this plus point three. So answer is three by eight. Okay. Which option says three by let's check. Oh my goodness. You are right. For the first time today you're right. Finally. Janta how could you make this mistake I mean two people getting this site out of 15. No it's not about I took we took I thought this thought wrong is wrong full stop. How can everybody think like that what do you think. I mean this makes me doubt whether you have understood this whole game of addition problem and condition probability. Where is that out in this entire solution where is the disconnect where did you mess this up people who gave it wrong all the 13 of you gave it wrong. Can I know the reason why you got it wrong. It was very important for me to know the reason it cannot be like he's don't do that you're one of your seniors. Very good 99.56 percentile in jmin very disappointed he got two and a half thousand rank he was aiming for under thousand. You know what was his major regret. Five questions despite solving wrong he wrote some different answer all together and that would be his regret for the whole of his life. You cannot go back and correct it. Unless until he writes J advance once more J advance you can write twice one in the appearing here another in the drop here. Okay. So the silly mistake this rushing through the question. My going to the washroom or mom calling. Okay I okay I hope that is true but if you're rushing through it and you're getting it wrong then again it's going to be you'll be regretting it. The least regret will be when you didn't know it and you know. Got it wrong. And the most happiness is when you didn't know it and you got it right. Okay so they're different phases of. I see it's all about your. You know, see they're incorporated we're given a window. This is what you think. You are and what does the world think about you. Okay. You think you're good. Maybe the world may be thinking you're bad. Right. Here you think what do you think you think you're good. What also thinks you're good. Here is where you are bad but what things you are good. Here you are bad words also knows you're bad. So there are four windows of a person's character. For example, you are not good in mathematics and the world thinks you are good. So that means you fall under one of the windows where you are not good. So he he himself was thinking where we were thinking is ranked is good or not. Or he will be doing well or not he himself was thinking I'll be doing good. So it depends upon the person's character. Okay. Anyways, we'll take some word problems now. Yeah. Everybody falls in some or the other. Okay. Let's take this question. I'll start with easier ones. I think these are all slightly heavier. It's a subjective question. I know option given to you. So please solve this. Give me a response in the chat box. Very good. Very good. Very good. Take the cheating. Cheating is out of question in these events. I mean, are you talking about the happiness part? Oh, there was a question based on cheating. Okay. Okay. Let's discuss it now. Two dice are thrown. Find the probability that some of the numbers coming up on the dice is nine. If it is known that number five always occurs in the first dice. See, this is how you are supposed to read the question. So A is the event that the sum is nine. Okay. And B is the event that the first dice shows five. Okay. You want the sum to be coming out to be nine. Okay. Given that the first I show the five. So basically this is what you are looking for. As for the condition probability, this as for the multiplication theorem, this is the formula. Okay. Now, all of you please pay attention. Can I write this formula as this? Can I write it like this? See, ultimately, what is this? This is an intersection B by NS. This is an B by NS NS NS gets cancelled. So instead of calculating numerator probability and denominator probability, why don't you just count the events on the numerator and denominator and divided. So this is another version of this particular formula noted down. So let us count those cases where you get some of nine also and first throw as a five. So there's only one such case where there is a five and there's a four coming up. So one out of depends upon total number of cases, which I'm not worried about, but ideally it's, it's probability is one by 36. Okay. Anyways, in how many cases will you get a sum of nine? So nine will come. The first thing is three, six, six, three. Correct. Yes or no? Four, five, five, four. Right. Anything else? Anything else that I'm missing up on? Five, one, one, five. Oh, sorry. I want a sum of nine. I'm so sorry. Oh, sorry. I want the first number to be five. P is the total number of cases where first number is five. Okay. So then if this first number, I was counting something. Thank you. So if this is five, this can be chosen in six weeks. So this is six answers on my six. Okay. Is it fine? Okay. Could I reverse the case and ask you, what is the probability that the first number shows a five given that the sum is nine? Yeah. So any intersection we will not change. Now how many ways can you get a sum of nine? I don't know what I was calculating. So I think there was only four such ways. Three, six, six, three, five, four, four. Right. Any questions? Good. Next. And on contains three white, six red, four black balls. Now again, I'd like to repeat here. Unless until stated that the three whites are identical, you will not assume it to be identical. This mistake people do year after year. If they don't use the word identical, then these three whites, they are not identical. Just because they're of the same color, they doesn't make the balls identical. Okay. Those six reds, not identical. Those four blacks, not identical. Please don't make these mistakes. I know make these over assumptions. They will cost you the problems. If they are supposed to be identical, the question center will write it. There are three identical white balls. There are six identical red balls like that. He will mention. So from your end, please don't make them identical for God's sake. Should I put the poll for this as well? I'm launching it. Very good. Three of you have responded. Okay. Last 30 seconds I can give you. So those who want to vote, please do so. Five. Four. Three. Two. One. Two. Okay. So I think. Good people are still polling. Okay. I'll stop the poll now. Everybody's done. Okay. 17 of you have voted out of a 65% say option B is correct. Okay. Let's discuss this question. See, first of all, what is the sample space? Sorry, what are your events? So can I say there are two events here? One is you want. One ball to be red. And other to be white. Okay. And there's another event which says they are of different colors. Okay. Balls chosen are of different colors. Chosen balls. Are of different colors. What are we finding? Read the question. Question says. What is the probability of one ball being red? Another is white. That is probability of a given. They are of different color. That means you're finding this. So the word given. The word if they are all basically indicative that you're using conditional probability. So when extra information is only provided that. The two walls that you have drawn, they are of different color. Then what is the probability that one is red and another is white? Think like that. This is how it is. No to be perceived. So as per our multiplication theorem, this is PA intersection of B. Okay. No. You getting two balls of different color and you getting one red and one white. One red and one white is a subset. Of B. Isn't it? So the intersection will be the subset only. So what is the challenge that you get? What is the number? So let me write it in terms of numbers rather than probability. By the way, it's time calculating the sample space when it is not required. Right. Why will you waste time finding sample space when you can get canceled from the numerator and denominator. So out of six red, you want to pick up one white. So that is 61. And again, the fact that they were distinct, I could have used 61. I could not use 61 if they were identical. I hope you know your NCR definition very well. NCR means number of ways of choosing our objects from distinct objects. Okay. Now B is the other wall being, sorry, one is red and other is white. Other is white. There are three white balls. You are choosing one out of it. Okay. Now B is the event where you have two walls of different colors. So let's say one is red, one is white, which itself is written on the numerator or one is red. Other is black. One is white. Other is black. Sorry about that. One is red. Other is one is white. Other is black. That simply boils down to 18 by 18 plus 24 plus 12, which is 18 by, I think, 40, 64. How much does it come out to be? Oh, sorry, 30 and 54. How much does it come out to be? One by three. Which is option number B. Is it fine? Any questions, any concerns related to this question? Okay. Now, if you want to use PA intersection B, then all you need to do is divide the numerator and denominator by 13C2 because out of 13 balls, you're choosing two balls. But 13C2, 13C2 eventually will get cancelled off. If there are identical, do we just take one off each? Yes. See identical balls. If you pick any one of the balls it is going to be a taken care of the situation, right? So only one way to do it. Would you scroll high? I think this is the entire question, Shradha. I mean, the entire solution and question is on the same page. Got it? Okay. See when the balls will become identical, when the balls will become identical, how would this question change? So basically in how many ways can you pick up? I mean identical in every aspect, right? Identical in color and shape and sizes. That's it. I mean, exactly the same. So when you talk about picking up one red and one white. Okay. So what is the chance that you end up picking one red and one white from this bag? Let's say I make it identical from every perspective. Everything will become one, one, one, right? Yes or no? But here the problem will arise is there are, the same ball of same color, identical. I mean, so I, yeah, ball of the same color are identical. Okay. So then white in it isn't a thing, right? Aditya has a question. Shouldn't we also take how light we are to get a white though? See in, in this case, what will happen is you getting a red ball will now be three divided by 13, actually. Because now they're, sorry, six divided by 13, because now there will be six, six red balls, that their numbers will be more, correct? So PA intersection B will be what? In that PA part will be what? See you are picking two balls simultaneously, correct? So what are the chance that one comes out to be red and other comes out to be blue? So picking up a ball simultaneously is, is like picking up two objects without replacement. So for PA intersection B, you'll have to take cases in that case. And in that case, your answer will be six by 13. And then you pick up a white wall which will be three by 12. Or that is plus three by 13 into six by 12. No aditya, see, again, that's the point. If you're picking two balls together, it is like you pick to one and within a fraction of a second, you have picked the next one also. So replacement cannot happen there. Your denominator is 1313 shows that there is a replacement back of the volume of kept. That is not happening. No, I'm getting my point. So you are picking two balls. And this thing can be perceived as if the, the first ball is a white and the second one immediately is a red one. Or the first ball is a red and the second ball immediately is a white one. So it would be, yes, can't you cry? So it would be six by 13 into three by 12 plus the other way round also six by 12, so into two, it will happen. Same will go with the denominator also when you're trying to pick up two balls of different colors. So that, that problem will be much more complicated than this expression. Are you getting my point? It won't be like one, one and one, one, one. It won't be like one by three. All right. All right. So last seven minutes of our class, I would like to take one more question at least, at least let's take this question. Yes. Something by 156 and all will come 18, 18 into by 156 will be the top case and down also you have to take because it's a case of conditional probability. You're just talking about the numerator. I think it should. Okay, let's read this question. The question says a coin, a coin is tossed. If head appears, if head appears, a fair dice is thrown three times. Otherwise a bias die with probability of obtaining an even number twice as that of an odd number is thrown three times. Okay. So there is a coin which is tossed. If head appears, a fair die is thrown three times. Okay. If a tail comes, then a bias die. And in this bias die, the chances of getting an even number is twice as high as getting an odd number is thrown three times. And one into entry is the outcome. Okay. Of course, and one and two and three, they will all be between one to six only. And it is found to satisfy. I to the power and one I to the power and size the imaginary part. Okay. This should be one. Why would denominator add up to one are the exhaustive events. That means you getting ball off to different colors. Is that an exhaustive event? Why I can't get ball of same color. So how they will not add to one ball of different color. No. Always different color you're taking. No. No. Or it is still coming out to be that. No. I'm talking to Aditya, not kinship kinship. Aditya is saying that the answer is still coming out to be one third. As if he had used the same formula. Then I think logical wise. Yeah. I think logical wise it is correct. I think it is simplifying to that extent. And it's. So yeah. Shit. Shit is also getting one third only. Right. Yes. Let's focus on this question. So you have to find out what is the probability that the dice thrown was a fair guy. Okay. So this question is actually a base theorem question. But again, base theorem is a type of conditional probability. So I will try to discuss this with you in very simple way so that you can do the topic. You will feel connected to it. See, let us say I have the event E as the event. Where. And one and two and three are such numbers that they add up to give you a one. Okay. And let's say even be the event that a fair guy was thrown. Even that a fair guy was thrown to achieve it. Was thrown. So you is the event that a bias die is thrown. Bias dies. Bias thrown. First of all, related to bias die, they have given some information that the chances of you getting an even number is twice that of you getting an odd number. So let's say X is the chance of you getting an even number. And sorry to X is the chance of getting an even number and X is the chance of getting an odd number. So either you get an even number or you get an odd number. So X is equal to one by three. Oh, sorry. I think there are, there are three such even numbers and three such odd numbers. Sorry. X is going to be one by nine. I was thinking in terms of points. It's a die actually, right? That means chance of you getting an even number. So a chance of you getting an even number is two by nine and a chance of you getting an odd number is one by nine. Okay. Now, let's come back to the question. What is the question demanding from us? The question is demanding from us. What is the probability that the fair die was thrown? Okay. Given that this condition got satisfied. See, read the question. I found that this expression, this result is going is one. Then what is the probability that a fair die was thrown? So you're finding P e two by e one. Now as per multiplication theorem, this is going to be this. Yes or no. Now, let us focus on the denominator part first. What is the probability that if you're rolling three die, you are going to get N one N two N three in such a way. That I to the power N one, I to the power N two and I to the power N three will add up to give you a one. Okay. So first of all, can we have the cases for N one N two N three for this to happen? What all possible cases are going to occur? Can I say if this is two? Okay. And if this is four, four, it will happen because minus one four four will take care of it. So one combination is two, four, four. Can I say other combination is six, four, four. Even for I to the power six, it will be minus one. Other will be plus one, plus one. Correct. Can I say it can be one, three, four. Yes. One and three. I and minus I will cancel each other out. And can I say it will be three, five, four. Now I can always do a shuffling of it because all of them are, you know, symmetric expressions. So that I'm not going to, I'm not worried about that as of now. Okay. Now you tell me, let us say, let us say I'm throwing a, I'm throwing a fair day. Okay. Let's say I'm throwing a fair day, a fair day was thrown. First of all, what is the chance of you throwing a fair day? Half itself because a fair day is thrown when you get ahead. Okay. So first of all, the chance of you throwing a fair day itself is going to be half. Okay. Now what are the chance that he will occur given a fair day has been thrown. Now see all of you please pay attention to this. When a fair day is thrown, you can have how many combinations of this it can be two, four, four, four, four, two, four, four, four, two like that. Basically can I say it will be three C two combinations and fair day will give one, six, one, six, one, six for all these values that you will be getting. Right. See basically what I did getting a two is one, six, getting a four is one, six, getting a four again is one, six, but they could be a shuffling of these positions. Right. It could be four, two, four, four, four, two, and two, four, four, so three C two such possibilities are there to get it. Similarly, this also can I say three C two into one by six, one by six, one by six. Why am I writing unnecessary brackets just to, yeah. One by six, one by six, one by six. Here also all combinations, but now here please note that this will be three factorial one by six, one by six, one by six. Same goes with the last three factorial one by six, one by six, one by six. Correct. Now let us talk about how do you get when a biased die is thrown. So first of all, the chance that a bias die will be thrown itself is a half because bias die is thrown when you get a tail, when you roll up, when you toss a coin, isn't it? And how do you get E given a bias die has been thrown. Okay. So let's figure it out. The three C two part will be the same, I believe, but remember three C three C two, you have all even numbers over here. So even numbers will be two by nine, two by nine, two by nine, because in a biased die, the chances of you getting an even number is two by nine. Correct. Then again, same three C two. Again, all the, all of these are evens. So again, two by nine, two by nine, two by nine. Okay. Then the third case, three factorial. Now remember two are odd and one is even. So this is one by nine, one by nine, two by nine. Same goes with three factorial, two are odd, one by nine, one by nine, and four is even. Okay. Now, all of you, please pay attention. Coming back to this formula, this formula, I can write the numerator as P e one into P e by even, or you can say this term, P e by even, am I right by multiplication theorem? So this expression by multiplication theorem, can I write it like this? Yes or no? Okay. And P e is the total number of ways in which you are actually getting n one n two n three in such a way that I to the power n one, I to the power n two and I to the power n three is one. So that is nothing but listen to this, that is nothing but it's going to be P e one intersection e plus P e two intersection e. Because you will end up getting that e scenario either with even in combination or with e two in combination. That means either even occurs, that means you are throwing a fair die and that even takes place. Or you are throwing the unfair die, which is the bias die and that even taking place. In short, what you have written over here is I'm just rewriting it once again, even though I'll be taking this subject matter in our base theorem concept. So you are actually writing this. And all these values are very much known to us. P even is half P even even has occurred is this calculation. Can somebody do this calculation because I think it's not that difficult three C two is three, three plus three plus six plus six divided by one by x cube. I think this is going to give you 18 18 is three into six by six cube. That's one by 12. I'm not mistaken. So this part is one by 12. The numerator part. The other term is one by 12. Okay. So down here, this is also half into one by 12. This is half into now, let me calculate this term. This sum is three into eight, three into eight. So two into three into eight. And this is six into two, which is 12. Again into two divided by nine into nine into nine. How much does it come out to be? How does it come out to be? I think it comes out to be 24. 24 plus 48. 24 plus 48 is 72. 72 by nine into nine into nine. That's actually eight by 81. Okay. So when you solve this half half anyways gets cancelled off. So you'll end up getting one by 12 upon one by 12 plus one. Let's simplify this. This will come out to be a very pathetic number by the way. Let's do this one. 81 divided by 81 into one plus 96. So how much is this? 81 by 177. Okay. I think it'll go by a factor of 27 because this is three. Three, it'll go right. So three by 59. Sorry. 27 by 59. Okay. So this is the answer to this question. All right. Now this question is not a difficult one. Once you learn the base theorem, actually base theorem is a special type of conditional probability where you cannot count these events. Okay. So here you have to stay with the formula. This comes through the idea of total probability. So I should have taken this question after that, but I thought I would give you an idea of the taste of total probability also. So don't worry. This is a concept where you are using total probability plus base theorem, which I will take up tomorrow's session base theorem. Okay. We'll take this up tomorrow. When I say counting means you, you want to find out the total number of ways. Okay. You want to find out the total number of ways of doing that activity, which is associated with multiple events. You can count it. I'm not, I'm not denying it, but I don't want to use that method because I can use my probability itself. See, the problem will come is the problem will come because of these, you know, these terms like one by six, two by nine, one by nine, et cetera. Why do we need even any two for counting? See, because Aditya, your event of getting N1, N2, N3 depends upon which type of, you know, die you are using. So if you're using a fair die and you getting N1, N2, N3 in that file in this particular, you know, fashion that that condition is satisfied. That is a separate way of getting that, you know, event E done. Another way of it is let's say your initial type phone was a biased one and then you're getting it. So there are two ways in which you are accomplishing the task. So you have to add the total number of ways that you're trying to do the same activity because it is happening via different, different means. Tomorrow when we do total probability, you know, you will understand the logic behind this. I will come back again to this question and try to explain you, you know, how is this working? Correct. Okay. Anyways, I think this was an overstressed class of by 12 minutes and I didn't give you a break also. So enjoy, take a break, relax. Yes, I saw the plan. The plan was very good actually. Okay. Thank you. How I take care. This could be a JEE main exam. This, this could come in a JEE main exam. Yeah. Thank you. Thank you everybody. Yeah. Yeah. Really. Thank you everybody.