 So probability, again, I'm repeating this is only for class 11th probability, it doesn't cover the complete J syllabus or for that matter engineering entrance exam syllabus because that would also include class 12th probability. So this is probability only meant for grade 11. Okay. Only meant for grade 11. Now, why I'm saying this is because after doing is doing this chapter, you should not start attempting some J level, you know, past year questions on probability thinking that now you are equipped enough to solve that question. You may not be able to solve many of the question. Okay, because the major chunk of the chapter comes in class 12. I already discussed that with you. So in class 11th probability, we are mostly going to talk about these features. Okay, so I'll just give you what all we are going to cover. We are going to quickly talk about types of events. Okay. These types of events basically again is a supporting a concept for your class 12th in class 12th you'll come to hear a lot of technical terms, like mutually exclusive collectively exhaustive, and so on and so forth. So in 11th we are preparing you so that you are used to those terms. Okay, so if you hear some strange terms coming out from my mouth with respect to probability, you should be aware that oh, we have already done this in our class 11th. Okay, so in 12th you'll be hearing a lot of technical terms with respect to types of events. So I'll be introducing you to these types of events in class 11th itself. So again in the major chunk of this chapter is the use of PNC, the use of PNC permutation and combinations in evaluating probability of an event in evaluating probability of an event. Okay, what we call it as P. So I hope everybody knows probability of an event is nothing but number of favorable events by the number of elements in the sample space isn't it this is the classical definition of probability. So this is the classical classical definition of probability. So in order to count the number of favorable events and the number of events in the sample space you need to be good in your counting. And for that you need to know your PNC very well. Okay, so when I was teaching PNC, I told you that this chapter is going to be very helpful in probability. Okay, and the time has come now that we start using the concepts of PNC I hope most of you are well versed with it by now. We completed it last to last week I believe. Okay, so good amount of practice has gone in and we'll see how good you are prepared or how well you are prepared with PNC concepts in evaluating probability of an event by using the classical definition. Last but not the least we are going to touch upon addition theorems of probability. Don't worry these addition theorems are very much rhyming with your cardinal properties that you didn't sets. The last part of the sets NA union B is NA plus and B minus and intersection B. So all those properties, they are going to come again in the addition theorems of probability so we are going to revisit it, however it will not take much of a time. So these are the three things that we need to address in class 11 probability. In 12th probability we'll be talking about several other things like conditional probability in combination to multiplication theorems. We'll be talking about law of total probability will be talking about independent events will be talking about probability distribution functions will be talking about you know the concept of base theorem etc. So there is there are a lot of things so let's wait for you to come in class 12 then we'll start those concepts. Is this fine is the scope of the topic clear to you is the scope of the topic clear to everybody. Okay, so this is what we are going to do can be easily covered in one class itself. So let us talk about types of events as our first topic. types of events. So what is an event. First let me ask that to you. What's an event. What's your birthday party. Marry ceremony. Red ceremony, they are events. Of course they are cultural events but very many talk about mathematics events are outcomes from experiments. Okay, so first I will talk about experiments a bit. So experiments are of two types. So when we perform experiments, our experiments fall into categories one is a deterministic experiment. What is a deterministic experiment experiments whose outcome or whose outcomes, whatever is the case here are well determined. Okay, or it can be predicted with certainty. Or can be predicted with certainty with certainty. Okay, so tell me an experiment whose outcome is known to you. So heating a water you know the temperature is going to rise. Okay, putting dilute SCL to zinc granules you know hydrogen bubbles are going to evolve. Throwing a ball up it has to come down of course let's say I'm talking about the situation in normal cases right. So these are some experiments whose outcomes are well determined or you can see it can be predicted with certainty. Okay, let's say you know you try to make fun of your teacher then obviously outcome will be you will be in your principal's office right. Okay, so these are deterministic experiments. We'll talk about that area. Okay, the next type of experiment which is our concern area are probabilistic experiments. Probabilistic experiments also called as random experiments. Okay, random experiments. What are random experiments. Experiments whose outcome or outcomes cannot be predicted with certainty cannot be you would be knowing total outcomes, but which of those outcomes will come that cannot be predicted with certainty. Okay, that cannot be predicted with certainty. For example, when you roll a fair die, you know the outcomes one, two, three, four, five, six, so these are the outcomes you know, in Mesa he could either right one to six some number will turn up, but which number will turn up that cannot be predicted with certainty. So this is a probabilistic experiment. So tossing a coin where head and tail both may come. I mean, you know the sample space you know either head or tail will come, but which one will come. We cannot predict with certainty. I'm talking about a fair point. Okay, not a biased one. Okay, so let us talk about, you know, probabilistic experiments in our discussion of this chapter we are not going to talk about deterministic experiments. So I'll be giving some example, like tossing a coin tossing a fair coin tossing a fair coin. Rolling a fair die, rolling a fair die. Fair means it is not biased. Okay, not like fair, fair and lovely, not that fair. Okay, rolling a fair die. Okay, and experiments could be anything. I mean, many of us think experiments are like, you know, the ones which we perform in physics lab or chemistry lab or bio lab experiment may be as good as you know, taking a step forward or backward, right experiment. I'm walking as an experiment, I may walk forward or I may walk backward. So these are the two outcomes. Okay, so experiment could be as simplistic as trivial as these. Okay. Now, let us talk about types of events which are associated with probabilistic experiment. Okay, types of events, types of events for probabilistic experiment, probabilistic experiment, or random experiment. Okay, few of them you already know, like show event. Okay, let's say I start with my show event. What's a show event. The show event is that if you perform the experiment, that event will definitely occur. That's the show event. Okay, so does that make that experiment a deterministic experiment, not really. Okay, so deterministic experiment, the particular outcome, right is always show to occur. So in a show event is a special case where if you perform it in a probabilistic experiment also that outcome is definitely going to occur. For example, getting a number lesser than equal to six or getting a natural number lesser than equal to six when you roll or die. That's a show event. It will occur. An example that occur has to be lesser than equal to six when you roll and die. Can you give me for a few more examples where an event are resulting from a probabilistic experiment is a show event. Any example that comes in your mind. Don't give the same example which I gave right now. Think, think, think, think, think. Nobody. Okay, so very good example given my Harsita. She says that there is a box full of red balls. What is the probability of getting a red ball? Of course, if the box is full of red balls, any ball you pick up would be a red ball. So this is a show event. Very good example. Anybody else who would like to add to that example. Okay, so please think of some examples and do let me know. So in short probability of a show event will always be equal to one. Okay, in general, we already know this. I'm not repeating this. There are a lot of things you have already learned in your class 9th and 10th. In general probability of any event lies between zero to one. Okay. Now talking about zero, zero is basically the probability of an impossible event. Impossible event. So an event whose probability of occurrence is zero. That means that event is impossible to occur. Let's say the same example which Harsita gave. What is the probability of picking a red ball from a box which is full of, let's say, green balls. Right, you cannot get a red ball. So that probability would be zero. Yes or no. What is the probability of you getting a number greater than six when you roll a die? Zero, it cannot occur. Okay, so these are impossible events. You are already aware of it, so I will not be spending time on these. Rather, let's start talking about something slightly serious. Mutually exclusive events. Mutually exclusive events. Okay, let's try to understand this. Events A and B are said to be mutually exclusive. Are said to be mutually exclusive. Exclusive. If or when they cannot occur together, that means their intersection is a null set. Their intersection is a null set means they cannot occur together. So these two events cannot occur at the same time. Okay, in the language of probability, we say for such an event, the probability of occurrence of A and B is going to be zero. That means it will be an impossible event of both of them occurring together. For example, I can give you several examples here. In fact, you will also give me some examples. I will start with the first example. Let's say getting one and six on the top face of a die. Okay, let's say a die is rolled. So getting one is let's say event A, getting a six is let's say event B. So you getting both the numbers on the top face of a fair die. Is it possible? No, they cannot occur together. So these events are mutually exclusive. Right. Getting getting head and tail both when you toss a coin mutually exclusive cannot occur together. Right. Can you give me another example? Can you give me another example? Getting an even number and odd number on rolling a die on rolling a fair die again mutually exclusive. You can't get an even and odds. A number cannot be simultaneously and even and odd both. Okay. Yes. It raining and not raining. Okay. By the way, let's not get confused with mutually exclusive and complimentary. Okay. So right now the example which I gave you in the second one, these two events are complimentary also. But the first example one and six are not complimentary, but they are mutually exclusive. We will come to that. Don't worry about what is complimentary. First, let us focus on. So yes, raining not raining is absolutely mutually exclusive events. Very right. Drawing a number 15 from a deck of cards is an impossible event. You have to tell me two events which cannot occur together. One event is not mutually exclusive mutually. The word mutually comes when they are two people right mutually more than two or more than two. So tell me two events which cannot occur together. Okay. That's fine. Any other example? Okay. I'll give you some examples in terms of numbers. So let's say you roll a die. Okay. So A is where you ended up getting one, two, three, four. Event B is where you ended up getting five, six. Okay. Are these two events mutually exclusive? Yes, because you will say that there is nothing in common between them. That means the probability of A and B is going to be zero. That means both event A and B cannot occur together because they don't have anything in common. Is it fine? Yes. Okay. If you pick up two days of a week and let's say one day is a Sunday and other day is a Monday. Right. Sunday and Monday, one day cannot be both of them. It has to be one of them. So they are mutually exclusive. Okay. So in short and simple words, let me conclude this concept of mutually exclusive. If two events do not occur together or cannot occur together, they are called mutually exclusive. That means their intersection is disjoint. A intersection B is a null set. That means A and B are disjoint sets. Okay. So probability of two mutually exclusive events of occurrence of two mutually exclusive events is always a zero. Is it fine? Any questions? We'll come back with more questions based on this after I am done with all the types of events. So now let us go to the next part, which is independent events. See, these words or these names look very, very similar, very, very rhyming to each other, but there is a lot of difference between them. Okay. What are mutually exclusive? Sorry. What are independent events? So event A and B are said to be independent events are said to be independent events. Independent events. When occurrence or non occurrence, both are applicable. When occurrence or non occurrence of one does not influence or affect the occurrence or non occurrence of the other occurrence of the other. Okay. A simple example that I can cite over here is that let's say you are rolling a die. Okay. And let's say A is the event that you get a six. Get a six on a fair day. And you are tossing a coin and you get, get heads. Okay. On a fair coin on tossing a fair coin. Okay. Let me ask you that if you roll, let's say there is an experiment where you are doing both the activities, you are rolling a die, you are tossing a coin. Okay. What is the probability of occurrence of a head on the coin and occurrence of a six on the fair die? What will your answer be? In such cases, you will say it is as good as you get a head on the coin and you get a six on the, on the die. That means half into one by six. Okay. So any two events with satisfy this criteria, they will be called mutually exclusive. Sorry. They will be called. Why I'm saying mutually exclusive. They will be called as independent events. Okay. Pardon the slip of tongue that I had. So it is called independent events. So any two events with satisfy this criteria. So many people ask me, sir, what is the three events? What is the condition for three events to be independent? Same thing. You just have to extend it to, you know, three events like saying PA intersection B intersection C is equal to PA into PB into PC. When such a thing is happening for our experiment or events associated with an experiment, those events would be called as independent events. Are you getting my point? So tossing a coin, let's say you're tossing a coin and rolling a die. You getting a head and you getting a six, of course, on the coin and the die respectively, they will be independent events. It is not that if you get a six on the die, it will, it will affect you getting us the head on the point. Will it, will it affect? No, right? And vice versa. Also if you get a head on the coin, it will not affect you getting a six on the, on the die. Isn't it? Now, many times by reading the question also, you can come to know that these events are independent. That means they are, you know, they're not affecting each other out. Okay. But many times, I mean, in higher versions of independent event questions, which you will do in class 12, but reading the question, it is not very obvious many times that the two events are independent or not. For that, you need to rely on this formula. Okay. But don't worry too much about it right now. In 11th, you will not be bothered with independent events, concepts. It is a subject matter of class 12th. Okay. So next year, almost at the same time, we'll be talking about, we'll be talking about independent events questions. Okay. Is this fine? Any questions? Any questions? So do not treat independent events and mutually exclusive events to be the same thing. Mutually exclusive events do not occur together. Independent events, they occur together, but one will not influence the occurrence of the other. Okay. I know one of them is an impossible event. So we cannot please be careful, please be careful with the use of the word mutually exclusive and independent. I have seen people in class 11th using them one for the other. No, they are quite different from each other. Next is exhaustive events. Or sometimes people call it as collectively exhaustive events. Exhaustive events or collectively exhaustive events. So what are collectively exhaustive events? So let's say we have E1, E2, E3, da, da, da, da, da, till E n. Let's say these events are collectively exhaustive. Collectively exhaustive. Collectively exhaustive. When their union is basically your sample space. Okay. When this union becomes your sample space. Okay. That is to say the probability of union of these events will definitely be one because probability of a sample space is always one. Okay. So events E1, E2, E3, da, da, da, da till E n are collectively exhaustive when their union is the sample space. For example, let us say you are rolling a die. Okay. Set A is the event where you ended up getting one, three. Set B is the event where you ended up getting five, six. And set C is the event where you ended up getting a two, four. Now, in fact, I can say one, two, four also. Okay. You need not be all disjoint. Let's say, let me put three here also. Okay. Now, if I ask you this question, are these three events collectively exhaustive events? What will you check? You will check their union. Are their unions giving me the entire sample space, which I think so it is correct. So this is your sample space. So if this is happening, you will say, yes, these events are, that is your A, B, C events are collectively exhaustive. Are collectively exhaustive. Is it fine? So if somebody asks you to check whether three events are collectively exhaustive or not, just see whether their union makes the entire sample space. Whether the union makes the entire sample space. Is it fine? Any question with respect to collectively exhaustive? So can I say events can be, these events could be mutually sorry, these events can be collectively exhaustive, like you getting odd number and you getting even number when you roll at I correct collectively exhaustive because the union will make the sample space. Head, tail, they are also collectively exhaustive. Okay. Because either you get a head or you get a tail, that means the union will be your sample space. Okay. So these all could be examples of collectively exhaustive events. Now, MCQ answers. Yes. All possible MCQ answers. Of course, you know, they can, if, if, let's say there are multiple options, correct. Also, they are collectively exhaustive. Correct. Now let me have now a combination of these two types. So I will now introduce you to the third, sorry, a sixth type of event, which is mutually exclusive and collectively exhaustive. This is one of the most commonly seen type of event in class 12 also mutually exclusive, mutually exclusive and collectively exhaustive, collectively exhaustive. Okay. So let me give it an abbreviation M E C E, me see. Okay. So I'll be using this word very commonly in class 12. I'll say these are MEC events. Okay. So MEC means mutually exclusive, but collectively exhaustive. Okay. So let me first define it. Satyam has a question. So this is the tossing off point example and getting either ahead of tail or you're talking about this case, mutually exclusive and collectively exhaustive. Yes, you are absolutely right. Satyam, you're getting ahead of a tail. They are mutually exclusive and collectively exhaustive. Very, very, very nice example he has given. Okay. So let me formally write it in mathematical jargon, not for the purpose of confusing you, but making it, making the idea clear. So these events are said to be M E C E are said to be M E C E when these two criteria are met number one criteria. If you take any two of them, they are disjoint. If you take any two of them, they are disjoint. That's a, by the way, if out of even E to E three Tilly and if any two of them are disjoint, will any three of them will be disjoint? Yes or no. I'm asking a very stupid question, but still I want to see whether if any two are disjoint, can I say any three of them will also be disjoint? Of course, yes. Nothing to think and their union. Okay, let me write in terms of symbols and their union would be a sample space. So the two words mutually exclusive and collectively exhaustive, I made the two criteria out of it. So events even E to E three Tilly and are said to be mutually exclusive when if you take any two of them, they cannot occur together. Okay, but, but the union of all these events make the sample space, but the union of all these events make the sample space. Okay. A very interesting example that I can cite for this is again getting an even number when you roll a die and getting an odd number when you roll a die. They're mutually exclusive because they can't occur together. But at the same time they're collectively exhaustive because together they make the entire sample space. Okay. Now a special case under this is your complimentary events. A special case under this is your complimentary events. Let me name it as number seven complimentary events. Okay. Now why I call this as a special case. Let me explain this to you. Evens E and E dash are said to be complimentary. I said to be complimentary complimentary when they are MEC when they are MEC. So if two events are mutually exclusive and collectively exhaustive, that means E intersection E complement is a null set. That is to say P of E intersection E complement is zero. But at the same time their union is the sample space. That is to say that P, E union E complement is one. Then the event E1 and E, sorry E and E dash. Okay. I purposely put a bar on the top of it. I could have called it as even E2 also. But I wanted to give them a special identification. So I called them as E and E dash. The same way as we do it in our set chapter. Remember compliment of a set. I think I still remember some of you had a hard time understanding it. So E and E dash are said to be complimentary events when simultaneously these two criteria are satisfied. Now a typical example of this is your head and a tail. Okay. So head and tail cannot occur together. And union of them is the sample space. So getting a head and getting a tail are complimentary events. Only two events are complimentary. You cannot say three events are complimentary to each other. No. Complimentary means opposite. Raining or not raining. Complimentary events because they cannot occur together. And either it will rain or it will not rain. Right. The union is the sample space. There is no third situation. Now many people ask me, can you give an example to clearly distinguish between complimentary and MEC. So complimentary are special types of MEC events. Try to understand it. Special types of MEC. So let's say I have an event like this. Okay. I'll give an example for this. Let's say you roll a die. You get one, two. You get three, four. You get four, five, six. Okay. You can see here that any two of them, if you take, you will get a null set. Intersection is a null set, isn't it? You take A, B intersection. You take A, C intersection. You take B, C intersection. And if you take their union, you will end up getting the sample space. So these are examples of MEC events. Correct. But you cannot say they are complimentary events. So complimentary is only applied to two events, not three or not to more than two. So as to say, are you getting my point? So can I say here A and B are complimentary? Can I say that? Can I say A and B are complimentary? No, I cannot say. They are mutually exclusive, but they're not collectively exhaustive. If you do A union B, will you get all the outcomes when you roll a die? No, right? So A and B, you cannot say they are complimentary events. You can say they are mutually exclusive events. Similarly, B and C, you can say they are mutually exclusive. Similarly, A and C, you can say they are mutually exclusive, right? But if I take an example like this, A is one, two, three, four, and B is five, six. Then can I say A and B are complimentary events? Can I say this? Can I say these are complimentary events? Yes or no? Yes or no? Yes or no? Correct? Because they satisfy both the criteria that their intersection is a null set and their union is the sample space. So A and B becomes complimentary. So if you want, you can write B as A bar or A as B bar. Do you get the difference between the two now? Right? In case, let's say you attend a KVPY interview. We never know. It may happen. Let's say you're writing it in 12th and the professor wants to know, hey, what's the difference between complimentary events and mutually exclusive, collectively exhaustive events? You should be very, very clear with your understanding. How is their union a sample space? Arya, I believe that the union will give everything which is going to come out from rolling off a die. Am I making any error in that? Don't say sorry. Say, are you? Okay. Is it fine? Any questions? Any questions related to whatever we have done, all the seven types of events. Okay. Now something very interesting, I would like to tell you beforehand, even though we are officially going to talk about it, when two events are complimentary. Okay. Something very interesting links their probabilities. This. This is always true. This is always true for two events which are complimentary. Now, many of you would be wondering, how do I get this formula? How did sir get this expression? See later on, I mean, many of you have not done this chapter in school. Am I right? What am I saying? Many of you, all of you have not done this chapter in school, right? I forgot you all belong to the same school. Have you done this chapter in school? Somebody please confirm. No. Okay. So you have not done. Okay. No problem. Okay. So later on that there is a special. Addition theorem. Very similar to what you have learned in your. Set chapter. Sorry. P a union B is PA plus PB minus PA intersection B. Okay. Yes. This is going to come up little later on. We'll talk about it little later on. In light of this, if I introduce A and B as. E and E complement. Okay. Let's say A is E. So can I say as per this formula, which as of now you have to assume it to be true. Okay. Okay. I get this. Now E and E dash are complimentary. Their union is the sample space, right? So probability of a sample space is always a one. Okay. So this left-hand side will become a one. This is PE. This is PE bar. And when two events are complimentary, their intersection is a null set. That means their probability of occurrence is zero. Okay. In light of this, can I not conclude that PE plus PE bar is going to be always a one. Okay. Hence the result. Okay. That is to say PE or PE bar is one minus PE. Okay. This is something which you will be using very, very frequently while solving problems. Okay. So please make a note of this and do let me know in case of any concerns. Next one is equally likely. What are equally likely events? That is something again you get to hear many a times while solving problems. So we'll talk about that as well. But before that, please make a note of it. Absolutely right. We'll talk about that. So please allow me to drag the screen. Oh, I've only written here something. Okay. I'll write it on top here. Last is equally likely events. Equally likely events. So events, even E2, et cetera. I mean, depends upon the question. Let's say if they are equally likely are said to be equally likely. Okay. When, when their probabilities are all same, that means they have equal chances of occurring. A simple example is let's say the numbers on the die. They're all equally likely. If it is a fair die, as you may, it's a fair day. Okay. And for a bias day, it may change for a bias day. Numbers are not equally, these events are not equally likely. So you may have, let's say, let's say that the die is heavier towards, you know, let's say a number six, then the opposite number will have a higher chance of occurring. Right. So in that case, your chances of occurring of these events are not equally likely. Okay. So for equally likely events, they have equal probabilities of occurrence. Is it fine? Any questions? Okay. Before I move on, I would also like to test you upon odd in favor of an odd against everybody knows this. Odds in favor of an odds against. Okay. You already know it. Okay. Good. So if I say the odd in favor of an event. Or odds in favor of an event. In favor of an event is a is to be. Okay. What does it mean? It means probability of occurrence of E by probability of occurrence of E bar is a by B. Are you getting my point? That means odds against occurrence of an event. So odds against occurrence of the event would be the reciprocal of it. That means it will be P E bar by P E. So this will become as per the formula B by. Okay. Now many people ask me this question, which I would like to ask you. If somebody gives you odds in favor of an event, what is the probability of that event? Simple question. Who will write it down? If P E by P E complement or E bar is a by B, what is the expert is the value for P E in terms of A and B? You are writing in terms of P only. Give me probability of an event in terms of only A and B. Your answer should only contain A and B. Nothing else. Yes. Come on. I thought I would get the answer by this time. Am I audible to everybody? Sometimes it makes me doubt whether you can hear me. Yes. So please tell me the answer. See guys, I've already told you that P E bar is one minus P E. So this is a by B. Okay. So let's let's call. Let P E B B X. Oh, sorry. I wrote X inside on me. Yeah. Let P E B X. So basically I've written X by one minus X as a by B. Find X now. Make X the subject of the formula. So B X is equal to a minus A X. Bring it to one side. Okay. So X, which is nothing but P E becomes a by a plus B. Simple as that. Right. So if somebody gives you the probability, somebody gives you odds in favor of an event as ratio, some ratio, or some fraction, please note that the probability of that event then becomes a by a plus B. Are you getting my point? Is this fine? Any questions? Any concerns? All right. Now, the main aspect of this chapter is, I'll come to that, but before we take some questions based on this. So let me begin with this question. In fact, it's a question having five sub parts. I hope it is legible to everybody. So question says discuss whether these events are equally likely. Mutually exclusive and exhaustive. So what they have done, they have given you the experiment. They have given you the events coming or events related to those experiments. You just have to say yes, no for each one of them. That means do you think A and B are equally likely? You have to answer with the yes or no. Do you think event A and B are mutually exclusive? You have to answer with the yes or no. And do you think event A and B are exhaustive or equally exhaustive? You have to answer this with the yes or no. So let's say a typical answer for, let's say, question number one could be like, yes, yes, no. Something like that. So you please give me your response for only the first one in the chat box by writing y, y, n or whatever combination you feel. So you just have to tell yes, no for these three. So you understood what is the way you have to respond. For example, the first one, you can answer something like this, no, no, no. I don't think so. They're equally likely. I don't think so. They're mutually exclusive. I don't think so. They're exhausted. Got the point. So this is how you have to answer your question number one. Okay. In fact, all the questions. So I'm waiting for your response. By the way, let me erase this. Else you think that question number one is already solved. No, it is not solved. I'm waiting for your response. Okay, Satya, please put the question number next to it. So that, you know, it is not lost. So Vihan. Okay. So Vihan has also answered. Okay. Okay. Okay. Okay. Maybe let me solve the first one. Then you'll get an idea. Only the first one. Let's answer her for the time being. Else I would lose track of your answer. So now let's say you're rolling a diet. You're throwing a diet, rolling a diet. Same thing. Okay. Dye means. Dye and plural is dice. I hope everybody knows it. So when you roll a die. Even day is getting an odd face. So odds are one, three and five and getting a composite. Now composite is four and six only. Right. So non prime numbers are composite numbers. Isn't it? So please don't include two. Because two is a prime number. Okay. So you getting a has a probability of three by six, which is one by two. You getting B is a probability of two by six, which is one by three. Are they equally likely? No. So first answer is no. Are they mutually exclusive? Do you think, do you see any common element between me? No, right? So are they mutually exclusive? Yes. Are the exhaustive? That means that does the union of A and B make the entire sample space? One, two, three, four, five, six, you'll say no, because any how to will be missing in both of them. So the answer to the first one is why? And why? Is it fine? Oh, Vishal has given all the answers. We're too fast. Is it fine? So answer to the first one is. And why? Okay. So first one is done. Let's move on to the second one. A ball is drawn from an urn containing two white, three red, and four green. Even is the event of getting a white ball. E2 is the event of getting a red ball. And E3 is the event of getting a green ball. Yes. Okay, Satyam. I think Harshita had already given one response. Okay, Harshita. Anybody else? Satyam, Harshita and Vishal so far. Okay, guys. Easy. Okay. Okay. Are they equally likely? That means are these balls present in equal amounts in these bag? In the sun? No, right? You have higher chances of getting a green ball. Right? Then you have red ball. Then you have the least chance of you getting a white ball. Right? So you getting, so let me just put a question number here. This is the first one. So you getting a white ball. The chances is two by nine. You getting a red ball. The chance is three by nine and you getting a green ball. The chance is four by nine. Are they equally likely? They are not. So the answer to the first one is no. Okay. Are they mutually exclusive? Yes. So when you're drawing one ball only from the bag, it can either be white or a red or a green. It cannot be all of them simultaneously. Right? So they are mutually exclusive. Okay. They are mutually exclusive. Are they exhaustive? Yes. They are exhaustive. Because there's only three possibilities. Either you get a green ball or a white, a red ball or a white one. There's no other third possibility. Okay. So the answer to the second part of the question is NYY. NYY. Satyam got it right. Vishal was wrong. Harshita was right. Arya was wrong. Okay. All right. Next one. Third one. Throwing a pair of dice is the event of getting a doublet or throwing a doublet. These event of throwing a total of 10 or more. I hope everybody knows a doublet. Doublet is basically when you get the same number on both the dice. Or both the dice. One, one, two, two, three, three, four, four, five, five, six, six. These are all called doublets. Okay. So let's wait for the answer. Okay, Satyam. Should we discuss it? Okay. Doublet. Doublets are one, one, two, two, three, four, four, five, five. So let me call them as event A, which they have also called it one, one, two, two, three, three dot, dot, dot. Let me write them all. Not many. Just six of them. Okay. These are called doublets. Now, the probability of you getting a doublet is the total number of events in A by sample space. Sample space is 36. When you roll a pair of dice, you get 36 pairs of outcome. Out of it, doublets are six of them. So probability of you getting a doublet is one sixth. Okay. The probability of you getting a sum of 10 or more. So four, six, six, four, five, five, six, five, five, six, and six, six. Okay. So if you check here, this also is one, two, three, four, five, six, six in number, right? So probability of occurrence of B will also be six by 36, which is one by six. Yes, they are equally likely. They are equally likely. Okay. So the first one, the answer is yes, they are equally likely. Okay. But are they mutually exclusive? I don't think so because five, five is occurring and six, six is occurring in both of them. So they have certain elements which are common to both these events. So A intersection B is not an ulcet. So if it is not an ulcet, they are not mutually exclusive. So this answer is N. Are they exhaustive? That means if you take the union of A and B, do you get all the 36 events that occur when you roll a pair of dice? You'll say no, sir. Not all of them are covered. So are they exhaustive? Answer is no. The answer for the third question becomes YNN. Very good. I can see Vihan, Harshita, all of them getting correct. But why only a handful of you are replying? I don't see like, you know, people, I don't see any point for not replying for such easy questions. Okay. I couldn't understand if the question becomes tougher. Next one. Let me just drag it up. Fourth one. Fourth one, I hope you can all see the question. Since I needed some space, I have just dragged it up. From a well shuffled pack of cards, a card is drawn. So one card is drawn. So even as the event of getting a heart, E2 is the event of getting a spade, diamond and club. I hope all of you know the suits in a card. Okay. So there is a heart. Okay. There is a spade. So heart looks like this. Okay. Spade, I think it looks like this. Diamond looks like this. And club, I think it looks like this. Correct, Arya? Arya knows it very well, Arya. Every day she plays. Joker and all I know it can be the part. Okay. So there are 13 cards of each type in a suit of cards. So 13 hearts are there. 13 spade, 13 diamonds and 13 clubs. Okay. Now, are these events equally likely? That means, is it equally likely to get a heart, a spade and a diamond and a club? You'll say yes, sir. Because each of these cards, each of these probabilities is going to be one fourth each. Okay. So they are equally likely. Now, are they mutually exclusive? Yes. A card cannot be simultaneously a heart and a spade both. Okay. So the answer is yes for this as well. Okay. Now, are they exhausted? Yes. So when you're doing a card, it has to be from one of these four suits. You cannot get a fifth type of card, right? So they are collectively exhausted as well. So the answer to the fourth part is yes, yes, yes. For all the three. Is it fine? Any questions? So I could see. Yes, yes, yes. Being given by Harshita. Tanvi, Satyam. Awesome. Great. Fifth one now. From a well shuffled pack of cards, the card is drawn. Okay. A is getting a heart. B is getting a face card. By the way, all of you should be aware. All of you should be aware that there are three face cards. King. Queen. And Jack. These are called face cards. So every suit has got three face cards. King. Queen and Jack. Okay. What do you call it in Hindi? I heard some people playing cards in Hindi. They say. Right. Right. Right. Right. Right. Right. Right. Right. Right. Right. Right. Right. Right. Okay. Now, like that, this isn't it? Am I right? Yeah. Okay. So the, the first part of the question is. Getting a hard and getting a face card. Are they equally likely. So remember. Getting a hard, the probability is already one by four. I'll write it once again. Getting a face card, please remember there are three face cards per suit, so there will be 12 face cards in all. So 12 by 52. I don't think so it is going to be one by four because one by four it needs 30 so are they equally likely. No. Right they're not going to be likely. Okay. Okay, are they mutually exhausted, so are they mutually exclusive. So there could be a face card which is a heart, let's say King of Hearts, Queen of Hearts, Jack of Hearts. Okay, so they're not mutually exclusive. So the answer for this is again an N. Are they collectively exhausted that means is every card either a heart or a face card. No, there could be speed as well they could be diamonds as well they could be clubs as well. So this answer is no. So NNN for the third one. Excellent. Many of you have in fact have got this right. Is it fine any questions any concerns. Or good that we revised our, you know, suits of the cards as well. Should I move on. Okay, so the next thing is the use of permutations and combinations in evaluating the probability of an event. Now of course I don't need to teach any theory over here because I'm assuming the prerequisites are already taken care of you know your PNC, I know quite well. So better to start, you know, using solving questions where PNC knowledge may be required may be required. Okay. So let's start with this question. Now give me a response on the chat box by putting the proper question number against it. Let's read the question together about a bag contains eight red, five white walls, three balls are drawn at random. Find the probability that let's say I am talking about the first question, all the three balls are white. All the three balls are white. Okay, now before you start solving the question, let me ask you one question. Do you think these all eight red balls are identical? Or do you think all these five white balls are identical? What's your take on that? Do you think these are all eight identical red balls and five identical white balls? Right, Satyam. So Satyam is absolutely right. Unless until specified, you will not take them to be identical. Okay. If they are identical, the question setter will specify that in the question. He will say there are eight identical red balls. There are five identical white balls. So here is the take. Here is the rule that we follow. If not stated, it will be considered to be distinct. So these eight balls will all be distinct from each other. Yes, here it will be distinct, Satyam. All these five white balls will be distinct from each other, whether the distinction is caused by their difference in sizes or maybe by difference in their hue. So white also, a lot of fuses are there. Creamy white, off-white, whatever white, sparkling white, phantasaline white. So they are all different hues of white. Okay, now proceed. Okay, another thing I would like to, sorry to interrupt you. I wanted you to be very clear while you're solving the question. When they say three balls are drawn at random, what does it mean? Anybody can explain this to me in simple English. Three balls are drawn at random. Okay, now Satyam, what I mean to say when I say, please explain to me. One after the other keeping, let's say I draw the first ball. Okay, then do I put it back and then draw the second ball? And then I put it back and then draw the third one? Or is it like, I take one out, keep it with me. I take another out, I keep it with me. I take third one out and keep it with me. Does it mean that? Or does it mean like in one shot, I put my hand and draw all of these together? How am I doing it? Okay, so those who are saying drawing them all at once or drawing them all together, you are absolutely right. Okay, now here is the distinction that you should all appreciate between the terms that we are using here. Sometimes you draw with replacement. Drawing with replacement. Okay, later on in the questions, you will see drawing with replacement. Sometimes you draw. Let me not write drawing. It appears to be as if I'm doing some sketch. Draw without replacement. And sometimes we draw together. Okay, or draw at once. At once. Now, please answer this question of mine. Drawing with replacement is very obvious. You draw a ball, you notice something about it. Maybe you notice its color, whatever you want to notice about it and you put it back before you draw the next ball, isn't it? So that is called, let's say I'm drawing two balls with replacement. What is the meaning of that? You draw one ball, observe its color, size, whatever you want to do it, put it back. Then you draw another ball and do whatever required. When you say I'm drawing two balls without replacement, what does it mean? I draw one ball. I notice the color, but I don't put it back in the bag. I keep it with me. I retain it. And then I draw the next ball. Of course, the sample size now will vary for the second round because one ball is already out. But what I want to ask from you, what is the difference between these two processes? Drawing together and drawing without replacement, is there any difference between them? Mathematically, what do you think? Are these two processes different or same? Mathematically, different or same? What do you think? Write on your chat box, write DIFF for different and SCME for same. I'll come to know whether your understanding is. I never say here I'm ordering the balls. I'm drawing the two balls, let's say, and I'm saying I'm drawing without replacement and I'm saying I'm drawing them both together at once. Mathematically speaking, are these two ways of drawing the ball different from each other or same? Up till now, I have got just one person saying same. Okay, now there are two people who are saying it is the same. Anybody else? See, it's just about knowing your opinion. Please don't refrain yourself from answering thinking that, oh, what will sir think? Let it be wrong. I just want to know your opinion. Let's have an interactive session. Okay, anybody else? Okay, now the verdict of this is they're actually same. Okay, now see why it is same. When you're drawing two balls one by one without replacement, what are you doing? Let's say there are two balls A and B. How do I show? Maybe I don't have a small object with me. Maybe let's say I have these two pens with me, let's say. Okay, let's say this two pens. Okay, I hope you can see these two pens on the screen. So they're kept somewhere and I'm drawing one pen. I'm retaining it with myself and then I'm drawing another pen. Can I say when I'm drawing them together, it is like I pick one and within a fraction of a second other one is picked, thereby giving you a sense that they have been picked together. But mathematically speaking, when you're picking them together, there is a time lag of let's say infinitism in between them. So it actually becomes picking without replacement. So mathematically speaking, drawing two balls together and drawing two balls with replacement does not make any difference to the answer. It is going to be the same process mathematically. Whether you say sir, English wise it is different, but mathematics doesn't distinguish between the two case. Are you getting my point? So these two process are actually the same process. Now, let me give you a live example. Let us say there is a bag. This bag contains three green balls and let's say two red balls. Okay, now, let's see the question is you are drawing two balls without replacement. What is the probability that one is a red and other is a green? Okay, so question is very simple. I'll give you two questions. You are drawing draw two balls, two balls from this bag, from the bag without replacement, without replacement. Okay, find the probability that one is red and one is red and other is green. See, what you did was you just, you know, you put your hand, you got a ball, again you put your hand, you got a ball. Ultimately, you should have a red and a green. Okay, so if you're drawing the balls from the bag without replacement, what is the probability for this? And the second case is if you're drawing the two balls together. If you draw two balls together, okay, then what is the probability? Now here if somebody tries to solve this question, he will get the same answer for both of them. See how. So if you want the two balls to be drawn without replacement and at the end of the experiment, you want to see a red and a green in your hand. So it can either happen in this way, you ended up getting a red first and then a green or a green first. So what is the answer coming out from this case? It is going to become six by 20, twice of six by 20, which if I'm not mistaken is going to be three by five. Now if you solve this question by using your idea of picking together, you want to pick one red out of three, sorry, one green out of three green and one red out of two red. Out of a total number of ways is five C2 because two balls from five distinct balls can be picked up in five C2. So this is your favorable number of favorable events. This is your number of sample space. So if you solve this, it becomes three into two by five C2 which is ten, which is again a three by five. So as you can see the answer doesn't change because of the situation that one is drawn without replacement and the other one is drawn together. So mathematically speaking, these two cases, the last two cases are identical. They are not, are same things, not different things. Of course, drawing with replacement is a different thing. Okay, how about the same question if I have to solve with replacement, how would you solve it? Okay, now you're drawing two balls with replacement, with replacement. So if you're drawing two balls with replacement, you'll say okay, I could either end up getting a green first and then a red or a red first, then a green. In this case, your answer will become two into six by 25, which is 12 by 25, drastically different from the other two answers. Is it fine? Okay, now, I thought it would be, you know, good to actually know this concept beforehand only else people do a lot of mistakes on these type of questions initially. Now, the actual question. Now let me invite the responses for the actual question. First one, all the three are white. What's the answer? Please give me an answer in a raw form. So five C three, that's the favorable event, total balls are 13 and you have to draw three out of them. So the first answer is five C three upon 13 C three. Okay, I'm not interested in knowing the exact values of them, but thank you for letting me know that. So the first answer is five C three into 13 C three. Sorry, five C three divided by 13 C three, what am I saying? Is it fine? Any questions? Any questions? Now here, many people ask me this question. Sir, if I solve this question from a very layman's perspective. What is all this question? Very simple. You say okay, the first ball should be white whose chances is five by 13. Second ball should also be white whose chances is four by 12. I'm sorry, four by 12 because the number of balls will be reduced. And third also is white that chances is three by 11. Are you getting my point? If you check this and this, the answer would actually be the same. Let's try to do that. So let's try to figure out these two values. What is 13 C three by the way, 13 into 12 into 11 by six, 286. So just 10 by 286 or to be precise, five by 143. And you can see that this answer will also be five by 143. So this is how this is how a person who knows PNC will solve it. And this is how a person who knows basics of probability, he will solve it. Okay. Now, in this case, this approach seems to be okay, but when you start talking about complicated cases, something like the third case where one is red and the two are white. This approach will be lengthy. This approach will be lengthy to solve it. So if I know PNC, I would prefer this. So I prefer this approach over this. Okay. Now you will also start preferring it once you start spending more time doing it in a very, you can say, you know, basic way or very fundamental day. Okay, so second one answer is very straightforward. Yeah, 8c3 by 13c3. Okay, or if you want to solve it in a more fundamental way, you will say, okay, 8 by 13, 7 by 12, 6 by 11, right, both the answers will be the same. Now coming to the third one, here, here is something where you will learn why PNC is more important, why PNC method is preferred. So if you want to have one red and two white, right, 8c1, 5c2 by 13c3 straight way. But let's say somebody wants to do it in a very, very fundamental way. See how he will do this question. So I want one red and two white. So he will say, okay, there are following cases I can do red first, then white, white, white, white, white, white, red. Now this person will sit and evaluate each one of these cases separately and add them to get his answer. So let's do that and try to, you know, match our notes. So first getting a red is 8 by 13. Then getting a white is 5 by 12. Again getting a white is 4 by 11. Okay. Now this case is over. This is first case. So second case, let's focus. White will be 5 by 13. Red will be 8 by 12. And again, white will be 4 by 11. Then white 5 by 13, 4 by 12. And this will be red means 8 by 11. Okay. Now let's try to add these three and see what do we obtain. Let's add. So you can see here that the figures on the numerator will always be 8, 5, 4 and denominator will always be 13, 12, 11. And they will get multiplied with three times because you are writing them thrice. Isn't it? Okay. Now see everybody. If you evaluate it, what are you going to get? 8. Let's try to cancel things out. This and this will get cancelled with this by a factor of 2. So there will be 2 on the top. Okay. So it will be 40 by 143. Am I right? Am I missing out anything? Okay. So with so much effort, you got 40 by 143 here. Let's try to check what happens over here. This is 8 into 10 by 13C3. 13C3 is 286. So cancel it out. 40 by 143. Okay. Which now the verdict is quite loud and clear. The first method, this method, the one which I did it in white, this was much faster as compared to this. Yes or no. This was slightly slower. So the choice is yours. Whether you want to stick to your basics or you want to start using PNC to solve your question. Is it fine? Any questions? Any concerns? Anybody? Okay. Great. Let's take the next question. Three integers are chosen at random from the first 20 integers. Find the probability that their product is even. Find the probability that their product is even. I'm launching the poll for answering this question. So please leave your response on the poll. Okay. Almost two minutes gone. I could just get two responses so far. Okay. Satyam, we will see. First 20 integers. I think they mean 1 to 20. Okay. They should have written it properly. First positive 20. Okay. So we'll stop the poll in another 45 seconds. So those who want to answer, please do so in the next 45 seconds. Okay. 5, 4, 3, 2, 1, go. Okay. Just 10 of you have responded. Less than 50% of the class. Now let me ask somebody. Arnav Pai, which option have you gone for? Arnav. If at all you have solved it. Okay. Okay. Arnav says A. But I know most of the people have gone with option C. Okay. Let's check. See, if you want the product to be even, at least one of these three numbers must be even. Let's say my numbers are A, B and C. At least one of these three should be even, right? If you want the product to be even. So why don't we do this question in this way? So let me figure out that the case is where the product will be odd. So if you want an odd product to come out, you must pick up A, B, C, all of them to be odd. So what is the number of ways in which I can get the sum to, sorry, the product to be odd. It's the same way as you pick up the number of odd numbers out of 1 to 20. There are 10 odd numbers. So if you pick up any three of them, it will be odd. The sample space will be 20 C3. So probability of you getting a product of odd will be 10 C3 by 20 C3. Am I right? Am I right? So 20 C3, let's calculate. This is going to be 120. 20 C3 is 20 into 19 into 18 by 6. Okay. So if I'm not mistaken, this goes by a factor of 2. So 2 by 19. So can I say a probability that the product will be even will be 1 minus the probability that the product will be odd because they are complementary events. So answer is 17 upon 19, which is option number C. So people who have said C, absolutely right. Is it fine? Any questions? Any concerns? Now, if you want, you can also solve this question by taking cases. That means let's say exactly one of them is even. Two of them are odd. Exactly two of them is even. One of them is odd or all of them is even. If you add them all also, you will end up getting the total number of cases in which you get even product. But that's a longer way to solve the question. I believe that. Can we have the next question now? Let's take another one. Let's take this one, a simpler one where maybe a lot of PNC is not required. A bit of find the probability that the two digit number formed by 1, 2, 3, 4, 5 is divisible by 4. Reputation of digit is allowed. I'm relaunching the poll. The way to solve any probability question is getting these two. What are the number of favorable events and what is the number of samples placed? So find these two by using your knowledge of probability if needed. If needed. I mean, not for all the questions you will need it. So there is an option for that as well. I will not say which one else people will get. So I've got two responses in little less than two minutes. You should always, actually, a mistake which commonly people make is that they turn a blind eye towards the options, which is not actually good. Now, looking at the option and finding, looking at the option while reading the question is very much recommended. Many people believe in, okay, revelation, I solve it and I finally get to know explore which option is matching. So look at the options while solving the question. Okay, I can give 30 more seconds, not more than that. Okay, okay, Satyam. All right, so five, two, one. I can see this time the response has gone slightly higher. 52% of you have voted. By the way, most of you have gone with option number D, none of these. Okay, let's check. So first of all, sample space would be total number of two digit numbers that you can make by using these digits where reputation is allowed. So five ways to fill this five ways to fill this. Correct. So there are 25 such numbers. Okay. Now, your favorable events are those events where your number happens to be a two digit number divisible by four. So what are the options you can have a number like 12. Correct. You can have a number like 21. Sorry, not 21, 24. Okay, you can have a number like 32. Correct. You can have a number like, like, like, like 52. Correct. Yes, I know. Any number which I'm missing out 44. Because repetition is allowed. Right. So altogether, the number of favorable events is only five. So your probability of this event to occur is five by sample space, which is 25 answer is one by five, which is not there in your options other than none of these. So exactly option number D is the right answer. Well done. So did not require much of a PNC here, but yes, your counting skills is definitely needed. Is it fine. Any questions. Right. None of these is not one of the answers normally. That's why many people have this element of doubt. Okay, let's take another one. Let's take this question. Mr. A lives at origin on the Cartesian pain. So this is the house of Mr. A. Okay, this is the house and his office is at four comma five. Okay, so this is the office of Mr. A. His friend lives in two comma three, which is already shown in the diagram. Mr. A can go to his office traveling one block at a time either in the positive Y or positive X direction. Okay, very similar to what we had discussed the other day in the permutation combination topic. If all possible paths are equally likely. What is the probability that Mr. A passes by his friend's house while going to his office. Okay. So what is the chance that Mr. A will choose a path which is passing through his friend's house while going to his office. I'll put the poll on. Again, go by the basics. You need sample space. You need favorable events. So find both these out by using your PNC skills. Okay, I got one response so far. Oh, only two people responded two and a half minutes gone. I hope all of you are aware that the upper limit to solving any PNC question is just three or four minutes max to max, because just a matter of click. Okay, I should do. Okay, last 30 seconds and then we'll close this poll. Five. Four. Three. Two. Okay, very few of you responded. I think only one third of the class responded and those who responded maximum Janta said option B. Okay, let's see whether B is correct or not. See, first of all, what is the sample space in how many ways can he travel from his home to his office. If you recall, I had basically explained this concept by trying to see the situation like you have four rights. Or four alphabets which are ours, and five alphabets which are used. Correct. See if he has to go from his home to his office, he has to travel four units right. That means positive exertion and five units up. So the number of ways in which you can arrange these four identical hours and five identical use among themselves will define how many parts he can take. The number of parts that he can take will be nine factorial by four factorial five factorial. Okay, let's compute this. So nine factorial by four factorial will give you nine into eight into seven into six by 24. Okay, so this happens to give you 18 into seven. We'll compute it later on if required. Okay. Next, what is the favorable event now understand this fact. In order to know the favorable events first he must go from his office. Sorry, home to his friend's house. And from friends house to his office. How many ways can you go from O to F again O to F he has to take two rights and three ups. So can I say it will become five factorial by two factorial three factorial. And means multiplication from his friends house to his office he has to take two rights and two ups, which is nothing but four factorial by two factorial two factorial. So it will become 10 into six, that is 60. Okay, so now let us come back to our calculation of the probability so it will be 60 by 18 into seven. I think this will go by a factor of six. So that is nothing but 10 upon 21 option number B is right. So Janta is absolutely right. B is the correct answer. Right. Any questions any concerns. If you have any questions. Any queries with respect to any of the steps. Please feel free to ask. Else we'll move on to the next one. Okay, let's move on to the next one. We'll come to this question later on I think we'll take. Since the response is slightly on a poorer side I'm giving you an easy question so that everybody gets it. Let's take this one up all is on question says if the letters of the word as size in are written down in a row. What is the probability that no two SS occur together. Easy question based out of your basic understanding of permutation combination. I will not name the method, because I will give you a hint. Okay, Satyam, I think you're the one you have answered on the poll as well. Okay, two minutes gone. Only two people have responded. Okay, five, six, I can give one more minute maximum. Okay, five, one. All right, I can see again seven of you are active and five of you say option C. Okay, let's check whether C is right or not. Again, the basic classical definition is of a probability of an event is number of favorable events by sample space. Okay, so first of all, what is the sample space. If you use all the letters of the word as I said and try to make a word out of it. How many such words can be created. As you can see, there are total eight alphabets here. And out of those eight alphabets that two is which are repeated and there are four s's which are repeated. Am I right? Okay, so this is your sample space. Now, for your favorable events, you must have to count those cases where the two s's no two s's are together. So here we will use gap method. Anybody speaking here, Neil, I think your mic is on. Yeah, so we'll be using gap method over here. So what I'm going to do, I'm going to put the two a's, I and N. And as you can see here, within these alphabets, five gaps are created. So first of all, how many ways can I arrange A, A, I, N? A, A, I, N itself can be arranged in four factorial by two factorial because two of them are repeated. They call your PNC concepts. And then you have to choose any four gaps out of these five to place the s's. In short, this becomes your favorable events. So as for the formula that I have used here, so let's write down. So probability of the event would be four factorial. In fact, this we can just write it down as 12 into five divided by this eight factorial again two into 24. So let's try to simplify this. So 24 you can, in fact, 24 and five can go. And this will be eight into seven into six. So this also can be removed by a factor of four. This also can be removed by a factor of two. One by 14 is the answer in this case, which is clearly option number C. Well done. Is it fine any questions? Okay, now for the last 20 minutes of our break time, we will, before we take a break, we'll take some complicated questions, the type of questions which normally is asked in J exams. Okay. Sorry, if anybody had any questions. Oh, no problem needed happens. No need to apologize for that happens. Next question is read this question carefully. Right, it's, it's a, you know, question which requires you to be good with your PNC. Okay, this question says 10 different books and two different pens are given to three boys so that each get equal number of each gets equal number of things. The probability that the same boy does not receive both the pens. Now I'm just asking you to give it a try. I'm not very hopeful that you will definitely get it right but just give it a try. Just see how you can solve this question. We will discuss this out. Meanwhile, the poll is on. I can give you around four minutes for this question. Think it out how to do it. You can do it. Hint is just use your fundamentals of division of distinct objects into groups. If you feel there is no option correct. Okay, then don't mark anything in the options. Okay. You give me the response in the chat box. If you feel there is no option correct. Okay, three minutes gone. No answers received so far. If you want, I can give one minute from here on also no issues. But something positive should come out from this. Some learning should happen. Okay, very good. Excellent. Should I discuss it? I mean, I'm asking to those who are trying hard. Should I discuss it? Okay, so poll is off. No use. I think only one person has responded and that too on the chat box. Okay, so let's nullify the poll. See, same formula. I'll begin with number of favorable event by sample space sample spaces. The total number of phase in which you can distribute 10 different books and 12 different pens. Sorry, two different pens to three boys. So now 12 different items are there. You have to distribute it between three boys equally. Okay, that means four, four, four. Okay, so it's a equal size group. Okay, and is the order important? Is the order important? Yes, because there are three different boys orders important. My know her shit order is important. Because the same objects and change hands between the three boys. Isn't it? So the number of ways to do this is so factorial by four factorial, four factorial, four factorial and three factorial because they are equal size groups. And into three factorial because order is important. Okay, in short, these two will get canceled. And you will end up getting this as your sample space. Number of elements in the sample space. Any doubt with this so far? No doubt. Okay. Now is our event where let's say the same boy does not get both the pens. So what I will do I will come from a complimentary angle. So let's say a particular boy, let's say a boy, not I should not use the word a particular particular means already chosen one. So let's say a boy gets both the pens. Okay. So how many ways can it happen? So first of all, you'll say, let us select the boy who you want to give both the pens. So what I'm trying to do I'm trying to find out the number of ways in which I can end up giving both the pens to a single guy or to a single boy. Right. So first I need to select which boy I want to give both the pens. There are three boys. So I can select one out of those three. Am I right? Yes or no. And once I give him both the pens. Remember I had to give him four items. Two pens I already gave him. That means I now have to give him two books also. So you choose the two books out of 10 books which you want to give him. Yes or no. Right. So now four objects have already been given to this guy. How many objects remain? Eight objects. And now you have to give now these eight objects. You have to distribute between the remaining two guys as four four. So this is again a case of equal size group where order is important. So this will become eight factorial by four factorial four factorial two factorial into two factorial because these two guys can change hands. Not the guy who has already been given those two pens and two books. Right. He doesn't, he will not change his hand. There is only accounted for here. Okay. So that three C one is only accounted here. Okay. So this is going to become your answer to the number of number of ways in which a boy gets both the pens. So let us find out PE bar. PE bar will be any bar by NS in the sense that you will have three C one into 10 C two into eight factorial by four factorial four factorial divided by NS divided by NS. Okay. A lot of terms will get cancelled. This, this definitely gets cancelled. Okay. And if I'm not mistaken, three C one is three 10 C two is 45 eight factorial will get adjusted from here. It will become 12 11 10 nine. Okay. I think this will also get cancelled five. This will get cancelled two. Okay. Anything that I'm missing out, please immediately bring it to my notice. Oh, there's a four factorial on top also. Yeah. Thank you. This and this will also go. So can I say it gives you three by 11. Okay. If PE compliment is three by 11. What is PE then you say PE is one minus. PE compliment. Right. And you are absolutely right. I think I should tell us the person who got it right. Eight by 11 is the answer and that is why I said in the beginning, if you feel none of the options is correct, then please give me a response on the chat box. Because I knew if I'm going to change the question somewhere, you will understand that that option is the right option. That is why I did not write it. Is it fine. Any questions. So this was a good problem where we, you know, recalled our concept of division of distinct objects into groups. Anything here that is bothering you. You can ask me. Your concerns or doubts. Is it clear. This thing is clear. Why this, why three C1, why 10 C2, why eight factorial by four factorial, four factorial, two factorial into two factorial. Everything is clear in your mind. Okay. Great. Next. Okay. Let's take this question. There are two vans each having numbered seats, three in the front, four at the back. There are three girls and nine boys to be seated in the vans. Find the probability that the three girls sitting together. The probability of three girls sitting together in a back row on adjacent seats. So you have to find what are the probability that these three girls sit together. In the back row. In the adjacent seats. All is on. Okay. So if you're wondering what are these two, these are headlamps and lights. So that, you know, this is the front. Lights and headlight. All right. Light is important. How will you know it is the front or the back? Very good. One person has dared to response. Who is that brave heart? It's a easy question only. I don't get scared. Okay. So it's a certificate, right? Okay. If you're good with your PNC, all these questions will appear easy. Don't worry. Okay. Arvind. Okay. Yes. Last 45 seconds. I can give. Okay. Okay. Five. Confused and lost. Okay. Never mind. I'll help you out. See, most of you have gone with option D. However, just five of you have voted. So less than one fifth of the class. Now see here, again, let's go from our basics. Sample space is first thing that we'll focus on. So first of all, how many ways can you make? 12 people. Sit in two vans having 14 seats. Tell me. At least this you tell me. I'm expecting answer from Harshita. Harshita, since you're saying you're confused a bit. Tell me 12 people, 14 seats. How many ways can you make them sit? No, that you're only choosing the seats on those chosen seats. You can shuffle. No, the people of Seaton. So your confusion is in NPR and NCR. So that'll go only with more practice. Okay. So this is your, this is your sample space. Now what is your favorable event? Favorite event is those cases where the girls are made to sit on the back row of every bus. And that two together. Okay. So let's say these girls, of course, we all know that, you know, girls like to travel together because they will take selfies and all that stuff. Right. All right. Harshita, you will do that. Of course, I know a lot of my friends who do that. Okay. So let's say these three girls, they choose van number one. First of all, they have to choose one of the vans because if they want to be seated together, they have to choose one of the vans. So two ways to choose the van, either van one or van two. Let's say they choose van number one. In van number one, they want to be seated in adjacent seats. So they can sit in the following way. They can either occupy four, five, six or five, six, seven. Yes or no. Then only three people can sit in adjacent way. So two ways they can sit in the adjacent seats. Now having made the girls sit, the girls can inter-arrange in three factorial ways. Let's say four, five, six is occupied. G1, G2, G3. Then it could be G2, G1, G3, G3, G1, G2, like that six ways. Now remaining how many seats? How many seats remain now? If you check, 11 seats will remain. Out of 11 seats, you have to choose nine and arrange the boys on that. So 11 P9. So this is your favorite element. What is confusing and lost about this? So this is going to be two into two into six into 11 P9. By the way, 11 P9 will be 11 factorial by two factorial. 14 P12 will be 14 factorial. Again, two factorial. Two factorial, two factorial will happily cancel out. And this is going to be two into two into six. 11 factorial will get adjusted from here. This will go off. Seven, one by seven into 13, which is one by 91. No, it doesn't look easy. You have to break the event in your mind. You have to do it slowly. There's no direct formula. Most of you are searching for some direct way to solve it. Choose the van, choose the seats, make the girls sit, then the boys will sit. This is a way to crack the favorite element. This is all simple. Now, let me give you another question where the further testing will be done. One more question I will give you and then we'll go on to a break from there. Any questions anybody has? Okay, let's take this question. Okay, so before you start the question, let me make some small correction here. A and B are randomly chosen from this set with repetition. Okay, or with repetition allowed with repetition, with repetition allowed. That means A and B can be same numbers also. Find the probability that this expression AX to the power four BXQ, A plus one X square BX plus one has positive values for all real values of X. That means this expression is always positive for all real values of X. Of course those X's should be only from this one, two, three, four, five, six, seven, eight, nine. All is on. So the question setter has tried to mix probability along with quadratic or by quadratic, whatever you call it, inequality. Okay, so this is how amalgamation of concepts are done in competitive exams because they don't have enough number of questions to test you on all the topics. So they mix two concepts in a problem. A beautiful example of that. What, sir? Beautiful for them, but for us it is hard. Okay, Satyajit, I'll put the poll on. Yeah. Okay. Thank you. Okay, last 45 seconds. Okay, Harshita. So none of them is matching. Okay, fine. All right, so should we stop it now? Five, four, three, two, one. Go. If you want to vote, please vote now. Okay. So just three people and three people, three different responses. A, B and C. Okay. Nobody went for D, by the way. See, first of all, favorable events or sorry, a sample space. You have to choose two numbers A and B from a collection of one, two, three, four, five, six, seven, eight, nine, nine ways to choose this. And since the repetition is allowed nine ways to choose this as well. So denominator will be either 81 or something which is a factor of 81. Okay. So there's a possibility of A, B, C and D all the four as of now, as of now. Okay. Now favorable events, favorable events. I want to choose those values of ABC. A B for which this expression is always positive. Okay, now let's do some analysis of this expression first of all. Okay, if I write it like this. I'm just breaking this term as x square plus a x square. Okay. I want this to be always positive. By the way, I can take an x square common from here. So x square plus one times a x square plus B x plus one. Okay, this is always positive. Now this term is anyhow positive. That means if you want this term to be positive, that means this quadratic guy must be positive. Okay. Now all of you recall, you know when you're doing inequalities and I told you a quadratic expression is positive. It can only happen in when two situations are met when these two situations are met. So A should be positive. And number two, the discriminant, the discriminant should be negative. Do you all recall this. You remember this. That means the quadratic graph must be opening upwards, why it should be opening upwards, because if it opens downwards, it will not remain positive for all values of x. Yes or no. And why should the discriminant be negative. Why should the discriminant be negative discriminant is negative because even if it opens upwards, it could open upward like this cutting the x axis that means at some part it may become negative also. So it must be hanging up in the air like this means no real root should be there. That is why both these conditions must be fulfilled. But now since A and B both are picked up from a set containing positive integers this condition is any helmet. Now I have to ensure that this condition this fellow this fellow should be met. That means discriminant should be less than zero. Now let us go to the side of the page. If you want your discriminant to be. What is the discriminant in this case B square minus four. A is a but C is one if I'm not mistaken. Yes or no. Yeah, C is one actually. So you want to choose those values of A and B for which this is negative that means B square minus four a should be negative. Are you getting my point. That means you want your B square to be lesser than four a. Okay, now let us take cases over here case one. If I take B as one. So what could be the possible options available for a you'll say sir a can take anything from one to nine. Isn't it because if you put a as a one one square will be less than four one square will be less than eight one square will be less than 12. So there are nine possibilities coming from here if B is one. Case two if you take B as a two. Then what are the possibilities with a can a B one. You say no because four is less than four is not true. So a has to start from two to nine. So two to nine there are eight possibilities. Am I right. Anybody has any doubt in the third in the second case. If these three, then this is already nine. So I cannot have as one I cannot have as two I can have a from three to nine. That means I have now seven possibilities over here. 3456789 count seven will come. If B is four. Then this is only 16 so I cannot have one I cannot have two I cannot have three I cannot have even four also so I have to have five onwards yes so a can be five to nine. Five to nine total possibilities is five five six seven eight nine count on your fingertips you can count it. If B is five, then the left side is already 25. So a cannot be one a cannot be two a cannot be three a cannot be four a cannot be five a cannot be six, it can be seven eight nine. So a can be from seven to nine which is only three possibilities. Okay. If B happens to be six. That means this is only 36. So even if you take nine as a value. Yeah, so there is no such a for this so for BS six. Nothing zero case. Okay. BS seven also. Nothing zero case BS eight also. Nothing zero case BS nine also. There is no A possible for this. Okay. So all these cases, I'm just writing them down for case or for the sake of making a note sort of it. Yeah. So all together the total number of cases will be nine plus eight 1724 2932 32 cases will be favorable for us out of 81 cases which we had already written over here I believe. So your probability of this event will be I'm writing it big. 32 by 81 32 by 81 matches with option number C. So only one person got it right. Is it fine any questions any concerns. Okay. So we'll take a break as of now. It's I think 626 will meet exactly at 641 p.m. after the 15 minutes of break. On the other side of the break I will take up addition theorems on probability. Okay. See you on the other side of the break. So the next concept that we are going to discuss is your addition theorems of probability. Not to be very frank. These theorems are all most if in fact I would say exactly the same as what you learned in your cardinal property of sets. The only difference is instead of and you will be using P now. Okay, I think you had a chance to see one of those addition theorems when I was explaining you the complimentary events. So this is exactly the same as what you learned in your cardinal property of sets. So if you recall your cardinal property of sets, the first thing that you would have that we learned was PA union B, sorry, NA union B is NA plus NB plus, sorry, minus NA intersection B. Okay. So NB are two events which are basically coming out from let's say an experiment. So the same, you know, cardinal property of sets, if you divide both sides of the equal to sign by the sample space. Okay, the same thing takes the form of a probability. So this, this whole thing will now be called as probability of a union B. Okay, this whole thing will be called as PA. The whole thing will be called as PB. And this whole thing will be called as PA intersection B. Okay, so this becomes one of the addition theorems. Similarly, you can start making probability theorems or addition theorems from all the cardinal number properties that you have seen. So we can have another one like PA union B union C like this PA plus PB plus PC minus PA intersection B, B intersection C, C intersection A plus PA intersection B intersection C. So there's no difference in the cardinal number property formulae and the addition theorems and probability because you are just dividing throughout by the sample space. And then your N by N, that is N of the event by N of the sample space will start becoming probabilities. Okay. Yes or no. All right, so let me ask you a few questions, if you could remember. What is the probability of occurrence of A and not B? How would you write it? Of course, give me one of the representations. Right, it's PA minus PA intersection B. Okay, so recall your, you know, sets chapter. This was A, this was B, intersection B complement is this zone, right. In terms of, you know, your sets. So this part is what P NA minus NA intersection B, just start calling NSP now that becomes your distinct. Yes or no. Similarly, what is your A complement intersection B? What is it? PB minus PA intersection B. Okay. What is the probability of occurrence of exactly one of A and B? What will you do? Basically, you take the union of these two. Right, so you'll say PA plus PB minus two PA intersection B. Okay. And not only these properties having just starting dot dot dot for it, depending upon, you know, what cases you get, you can also write inequality properties which you have learned in your cardinal property. For example, please note that probability of A union B will always be less than equal to one. Okay, and probability of A or probability of B will always be lesser than equal to probability of A union B. Okay, and probability of A intersection B will always be lesser than equal to probability of A or probability of B. So these properties also will be equally valid. Just like you used to say A union B will always be a subset of the universal set in the same way. The synonymous form for that is your PA union B will always be lesser than equal to one. Right, so I hope all these things, you know, make sense. So we will understand these through more and more problem solving. So this is the meaning of exactly one of A and B. So let's say there are two events, you want exactly one of them to occur so it will be either this zone, or this zone, exactly one of them is occurring. Okay, so you can say it is PA union B minus PA intersection B also which is going to result into the same expression. Absolutely Satya, this zone is actually called A delta B, symmetric difference. Perfect, you have recalled it exactly correctly. Okay, so let's see problems, from problems we will learn more tricks of the creator. So let's get started with this question, very simple one. A and B are two events such that PA union B is three by four, PA intersection B is one by four, PA complement is two by three. First one, I would request you to write your response for the first one. Very good. First one is very simple. PA is nothing but one minus PA complement, that is the basic formula that arises from your complementary relation, which is again coming from your PA union B is PA plus PB minus PA intersection B concept. Okay, so this is nothing but one minus two by three, which is one by three, absolutely right. Second one. Second one. Okay, so for the second one we can use PA union B is PA plus PB minus PA intersection B. PA is one third, PB I don't know so far, PA union B is three by four, PA intersection B is minus one by four. So from here I can say PB is take this one by four to the other side and subtract one by three. That's nothing but two by three. Absolutely right. A third one. Third one. Yeah, so PA intersection B complement as I already discussed with you it's PA minus PA intersection B. PA is one third, PA intersection B is one fourth so that becomes one by twelve, absolutely right. Next one. This is PB minus PA intersection B. PB is two third. Okay, so this is going to become five by twelve. Is it fine? Any questions, any concerns with anyone of the four? Basic ones. I've just taken the basic ones. Seven by twelve for the last one. Why? Why should it be seven by twelve? No, no, no, no. Don't call it silly. Silly word is to be dropped. What did I say? I've been telling this again and again. Silly means you are giving yourself a soft corner. Okay, let's not be soft with these mistakes. Okay, so if you give yourself that excuse, you'll always keep making that mistake. Okay, next question I will relaunch the poll for this. A, B, C are three events so that PA is three by four, PA intersection B intersection C complement is one third and PA complement section B intersection C complement is again one third. Find PB intersection C. Okay, let's do this. Anybody? See, take the help of the graph, the graph, sorry, the Venn diagram. The Venn diagram is going to make it very, very clear. So don't shy away from using a Venn diagram. Okay, so think as if this whole area is PA, sorry PB, which is given to you as three by four. Okay, what is A intersection B? A intersection B is this part. Intersection with C complement and this part is your, this part is your A intersection B intersection C complement. Okay, let me write complement like this. Correct. What is A complement intersection B intersection C complement? So A complement intersection B. So A complement intersection B will be this part. This is your A complement intersection B. Okay, now intersection C complement that would do away with this part. So this white region that you see, this white region is your A complement intersection B intersection C complement. Okay. Now, what are they asking us? They're asking us PB intersection C. B intersection C is this blue part, right? Isn't it? Now, from this diagram, it is very obvious, by the way, I'm stopping the poll. Yeah, from this diagram, it is very obvious that these three are mutually exclusive and collectively making the set B. In other words, PB is nothing but PA intersection B intersection C complement, PA complement intersection B intersection C complement. Yes. This is only three by four. This is only one by three. And I think this is also one by three and this is something which we need to find out. So this becomes three by four minus two by three, which is nothing but I think one by 12. One by 12 is the answer. Any option will say one by two. Option number eight. Correct. Absolutely right. So don't forget your Venn diagrams. They are going to be helpful here as well. Okay. Because everything is inspired from your idea of events, events are nothing but sets. Is it fine any questions? Any questions, any concerns? Sorry about that. I meant to drag it. Yeah. Let's take a question here. Very beautiful question. A is a power set of a given set X containing an elements. Okay. A is a power set of a given set X containing an elements PQR are taken from a if PQR are disjoint sets. Find the probability that P union Q union R is your set X. Find the probability that P union Q union R is your set X. Would you, would you like me to run the poll for this? Okay. So I'm just launching the poll for the first question only. Only the first question you need to solve. Don't answer for anything else other than the first question. Any idea how to do this? I mean, anybody has a vague idea how to solve this question. If you want, I can help you with the first one so that you can, you know, solve the other two. Do you want me to help you with the first one? Or anybody wants to try out the first one? Nobody is replying. That means most of you are trying it out. Okay. Okay. One of you have responded after four and a half minutes. So it's better. I should discuss the first question at least. I'll give you opportunity to solve 19 and 20. See, let us read this question very carefully. I think with this power set and all they're trying to confuse you. They're just trying to say that from a set X, you are trying to construct three sets. Three subsets. Okay. So these are subsets of X. Okay. So how can you make three subsets out of X? Since that they are disjoint in nature. And at the same time, their union gives you the complete set. Of course. It is not always possible that if you make three subsets, their union will be collectively exhaustive, isn't it? Okay. And they're mutually exclusive also. So how can you ensure that? So let us try to understand a simple logic here. Let's say your capital X set has got these N elements. So as the question claims that there are N elements. So let's say these are your N elements. Okay. And now your subsets are like buckets. Q R. Now you have to start picking an element and try to decide which bucket you're going to place this in. But remember the only difference from actual scenario of putting into bucket is not here. An element has a power to go in multiple buckets simultaneously. That means if you wanted to, it can go in P also Q and Q also, right? Unlike we, we have it in our objects. If an object goes in one bucket, of course, you cannot make it go in as some other bucket because there's only one such object. But here these elements have the power to replicate themselves and go into any number of buckets that they want. Okay. Now, just think from X one point of view. How many options X one has? Think carefully. Think carefully and then answer how many options does X one have? Just give me a second. Children, they will shout like anything. Three options only. Why? Why three options? Think again, Arya Satyam Satyajit. Anybody who can answer this part, how many options does this guy X one has? Seven options. Okay. Now Satyam has changed his answer to eight options. Why seven options? Arya. Okay. Satyam. Why two to the power three minus one. Exactly. So this can be put in any one of the three or any two of the three or all the three. So if you see it has got three plus three plus one, seven options, my dear. Are you getting my point? Why I'm saying seven options and why not eight option is because when you're making a power set, when you're making a power set that element, that element is to be used. I mean, you cannot see what are you doing? So you have to get P Q and R from the power set of a. Right. Are you getting my point? So in our power set, you have all possible subsets of elements of X. It cannot happen that X is not participating at all in making a subset. If X were not put in any of the subsets, then of course there will be eight options in this case, this X one or we have to have all possible subsets by using all the elements of your set X capital X. So this guy has got seven options. This guy also has got seven options and this guy will also have seven options altogether. Your sample space will be seven into seven into seven n number of times. That means seven to the power n options will be there. Correct. Now, this is your sample space. What is your favorable event now? Let me write it below here. What is my, what is the favorable event now? Favorable event will be those events where a particular element goes only in one of the sets. Right. It cannot go in multiple sets because then it cannot be disjoint. So it either goes in P or it goes in Q or it goes in R. That means there are three possibilities. So now for favorable event, every element will have three, three, three possibilities. That means your answer will be three to the power n as the favorable events. So your probability of the event will be three to the power n by seven to the power n, which is three by seven to the power option number C is correct. Now does it give you a fair bit of an idea how to approach this question? Yes or no. Now, will you be able to now crack the 19th question? 19th question says P intersection Q intersection R contains just one element. Then the probability that P also contains just one element is I'll put the poll on for this as well. Yes, only two people have responded so far. Okay, let's discuss this out. See, read the question carefully. By the way, three of you have said option number C. Let's check. If P intersection Q intersection R contains just one element, what is the probability that P also contains just one element? Okay. First, let us ask our self. In this case, what is the sample space? In the sample space, there is one such element which is there in all the three sets P, Q and R, isn't it? Yes or no? Am I right? So in how many ways can you achieve this sample space? That means only one element is common to all the three sets. How will you do it? Now here, see everybody please pay attention. First of all, you choose one of the elements which you want to put in all the sets. Let's say you chose X1. Okay. And you put X1 in all the sets. Okay. Now please understand here, you are distributing X2 to X1 in such a way that PQR are now disjoint. Now we don't have to worry about them being collectively exhaustive, but they should be disjoint. So now tell me how many options does X2 have? How many options does X2 have? How many options does X2 have? Okay. Now X2 here will have, now listen to this. X2 here will have now four options. What are the four options? Either it can go to XP or it can go to Q or it can go to R or it could stay back in X. I need not use all the elements. So this guy will now have four options. X2 will have now four options. So let me make a fresh diagram over here so that you understand what I'm trying to say. So X1 is already used up. Okay. So this guy will now have four options, my dear. What are those four options? To go only in P, only in Q, only in R or to not go in any one of the three. Correct? Because you want the remaining n minus one elements to be distributed in such a way that they don't have anything common other than X1 itself, which you have distributed to all of them. So likewise, all of these guys will have four, four, four four options each. That means four to the power n minus one options will be there. Am I right? Any questions, any concerns here? Do let me know. Now what is the favorable event? Favorable event here is an event where P contains just one element. So P only contains X1. Okay. That means if I want to distribute X2 now, it should not go in P. It should either go in Q or R or stay back. That means that means one way, NC one way to choose that element which you are trying to send. Okay. And the remaining elements will have n minus one choices. Sorry. The remaining n minus one element will have three choices. Sorry for that. Yes or no. What are those three choices? Three choices are. Let me write it down. So you choose that element which you want to be common. Correct. And now let's say X1 is that element. This will have only three choices to go into Q or to go in R or to not go in any one of them. So this will become your total number of, I mean, favorable events, which I've already written on the top. So your probability of this event will be any by NS, which will be three by four to the power of n minus one, which I feel is not there in the options. Or maybe they wanted to write this as three by four to the power n minus one, but they did not write it. So anyways, I don't blame you here because none of the options seem to be matching. Okay. Sometimes they forget to type properly. So this is a typo error, I believe, but the logic is clear to everybody, right? Is the logic clear to everyone? Any questions, any concerns with the logic? Okay. 20th one, I would request you to try on your own as a homework. Okay. So let me, don't worry, I'll be sharing this sheet with you. So you can always see which one was the homework. Don't worry. Don't start copying it. In fact, we'll take one more question before we call it a day. Let's take another one. From a pack of 52 cards, face cards, and tense are removed and kept aside. Then a card is drawn at random from the remaining cards. If A is the event that the card drawn is an ace. Okay. H is the event that the card drawn is a heart. S is the event that the card drawn is a spade. Then which of the following holds? Then which of the following holds? I'll relaunch the poll. This is an easy question. You should all be able to do this. Ace is the number one card. I hope all of you have seen the pack of 52 cards. There is a king, queen, jack, and then there is an ace. Ace is the card whose number is one. We call it as Ika. This is a super easy question. You should be able to get this. Okay. I can give 30 more seconds. S is not the sample space here. S is the spade. See, they have mentioned it. Okay. So don't assume that S means sample space always. Okay. Now given S is the event that the card drawn is a spade. Okay. Five, four, three, two, one. Okay. Let's stop this. Most of you have said not most. I would say two of you have said A. One of you have said B and others voted for D. Okay. See, 52 cards, you remove the face cards. So there are three face cards. And there are four 10s. I think, correct. So how many cards you have removed altogether? 16 cards you removed. Correct. So three into four. And these are four 10s. So altogether you removed 16 cards. So there'll be only 36 cards left. Correct. Now the question is, is the event that you're drawing an ace? Now remember, there are one ace in each. So four by 36. So one ninth is the chance that you get an ace. Okay. What are the probability of you getting a heart? Now remember, now remember 36 cards. Nine will be hot. Nine will be spade. Nine will be club and nine will be diamond. Isn't it? So you getting a heart is one by nine. Oh, sorry. One by four. Nine by 36. One by four. Okay. So now the question setters says nine PA. Is it equal to four pH? I think so because one nine into one by nine is four into one by four. Yes, it is correct. So I would definitely mark option number eight. Okay. Second part of the question, let's see why others are wrong. PS. PS is again one by four. Is it four times PA intersection H? Now see a intersection H means it should be ace of the spade. Sorry, ace of the heart. Okay. There's only one ace in the heart. So that is one by one by 36. Correct. So this is PA H. Intersection H means it should be ace also as well as heart also. So there's only one such card, which is ace as well as heart. So out of 36. One case is like this. Okay. So for PA intersection H is one by nine. Does it match? Definitely not. So this doesn't match with PS. So this option is not right. Okay. Similarly, let us see why option C is wrong. 3PH is three by four. And PA union S is PA plus PS minus PA intersection S as per our addition to the formula, which we have seen. So A is one by nine. S is one by four. And A intersection S. A intersection S will be again a one by nine. Right. Ace of the spade. Only one such case. Sorry. One by 36. One by 36. One by 36. Correct. So if you solve this, it basically becomes 13 minus one by 36, which is nothing but 12 by three. Sorry. One by three. Okay. So the question says three by four. Is it equal to one by three? Definitely not. This answer is not correct. And since one of the answer is right. None of these is also not correct. Is it clear any concerns over here? So keep practicing. I mean, this is a chapter where more and more practice makes you better, especially with respect to your PNC concepts. Okay. So I'll be not going any further.