 So good afternoon and welcome to the session on probability now probability for you in class 11th is mostly a Precursor of the probability that you are going to see in class 12 Yes, probability is there in both the years of your study in class 11th in class 12th, right? In fact, this chapter was also there in class 10 In fact, it was also there in class 9 Correct if I'm not mistaken you studied this chapter in 9th You study this chapter in 10th. Now you are studying it in 11th You'll be studying it next year also in 12th and to make the matter worse. You will be studying it in your first year or second year I don't exactly remember which year of engineering also, right? So there will be a course called probability and statistics where you will be studying this chapter again So every year the scope of this topic will become broader and broader, right? So what are we studying in probability of class 12th? Sorry, probability of class 11. So let me give you an overview of the topic here. We are studying three things primarily one how to use permutation and combination for counting Okay Mostly we all know that probability of an event is nothing but the number of favorable events by the sample space Right, this is your classical definition of probability. This is the classical definition of probability Okay So as you can see here We require to have a count of the number of favorable events and have a count of the sample space and for that You are going to use your permutation and combination skills PNC skills for counting because they will be You know so huge in number that you will not be able to count it by making sets out of it The way you used to do it in class 10 or probably the way you used to do in class 9, right? There if you remember used to count them and they were very small in number You could count them on your fingertips, right? But now in class 11th and 12th probability of course will be coming across such cases where The number of events favorable events or number of sample space would be running into lacks Lacks course also, okay? So there is one thing that we are going to see Second thing that we are going to learn is your addition and subtraction theorems addition subtraction theorems Okay, this is the next concept that we are going to talk about This concepts are very similar to what you have learned in your Cartesian property of sets So all of you remember your sorry cardinal property of sets, okay? So in the cardinal property of sets whatever theorems we had learned about NA union B NA union B minus C NA intersection B Compliment NA complement intersection B etc etc. They will be coming again back to you in the form of probability So that is what we are going to learn under addition and subtraction theorems and the third thing that we are going to talk about is continuous slash geometrical probability Geometrical probability where geometry would be used in order to calculate the probability because Basically, they are continuous in nature means you will not be able to count Because counting can always be done for discrete items counting cannot be done for continuous items For example, if I ask you how many points are there on this screen? Of course, you can tell the pixel, but there are infinitely many points, right? So in such cases, we cannot make use of counting principles. We have to go with geometry Okay, so we'll give you a I'll give you a brief idea about these cases as well Okay In fact, the main probability that you're going to study and the probability that is going to come for your Comparative exams and the board exams both of them will be basically based out of probability of class 12 So we sometimes call this as an introduction to class 12 probability. Okay All right, so let's get started. So the first thing that I would like to discuss with you is Events Yes, everything is happening around events, right? So what are the meaning of an event? What do you understand by an event? anybody Event is a outcome from an experiment Isn't it? So what's an event? It's an outcome from or outcomes from an experiment correct and When we talk about experiment experiments are of two types One is called a deterministic experiment and other is called a Probabilistic experiment, okay What is a deterministic experiment? In deterministic experiment by the way in any experiment you actually know what all outcomes are possible Correct. So whenever you try to perform an experiment, you basically have an idea. What kind of outcomes can I expect from there? correct So in all the type of experiments, we normally know the gamut of Or you can say you you know the all possible scenarios that can result from it, isn't it? But in deterministic experiment you can actually make out which of those possible outcomes is actually going to happen Okay, so those are called deterministic experiment where the outcome from that experiment is Well-predicted Okay, so depending upon how you are taking your experiment the outcome is well-predicted. I'll give you a very simple and funny example Let's say This is me. Let's say, okay. I have a ball in my hand I toss it in there Okay Now what are the possible outcomes of this experiment? Either the ball will go up and come down after a certain time or the ball will skip the Gravitational pull of the earth, right? It will escape the gravitational force of the earth Provided I have thrown it with the escape velocity and I'll go to the to the space But I know given the strength I have I do not have the strength of a rocket launcher Right, so given the strength I have this ball is going to go up Okay, and it's going to come down again. Okay, so this is a deterministic experiment You know what is going to happen when you're doing that? Okay So we are not going to talk about this because we are talking about chances We are talking about probability and for that will come to the probabilistic experiment What a probability exp probabilistic experiment those experiments whose outcome is Random see you know the total outcomes, but which of those outcomes will happen that is completely random For example tossing a coin. I know either a head or a tail will come out But which of the two will come out? Nobody knows it's a random When I roll a die, I know which of the outcomes is going to come It's somewhere between one to six, but which number from one to six will come out We cannot predict right. It's completely random. Okay, so in such cases our outcomes are random. We get random outcomes Okay, example tossing a coin tossing a coin Rolling a die Okay, so they can be so many probabilistic experiment picking up, you know Let's say there's a bag which contains balls of different colors Picking a ball from the bag and you know getting a particular color back where ball that is completely random Okay, you don't know which color ball is going to come into your hands. Okay, so this is the Interest area for us. We'll be not talking about deterministic experiment will be limiting our discussions to probabilistic experiment Okay, so we were talking about events. So what types of events are possible? Let us talk about Types of events that we can actually come across Okay Now events can be on a broader level classified as simple Okay, also called elementary or compound also called mixed Okay, so what is a simple event simple event is there There's one only one outcome is happening from that particular experiment that you're performing for example If a roller died Right, I can get a number one or I get a number two or I get a number three or four or five or six They are called simple events Okay But if I'm performing let's say at experiment where I am tossing a coin and rolling a die simultaneously So me getting a head on the coin and me getting a six on the die Right that together would be a compound event Are you getting my point the difference between simple and compound event? Okay, so there is only one outcome possible From that experiment and here there is a mixed outcomes possible Okay, so normally I you know when I am giving you a situation In fact, there are questions on your DPP's they will be asking you whether this event is simple or compound Simple is where you cannot break down that event into simpler, you know sub events Compound is where that event is made up of several simpler events Okay, another example of a compound event would be let's say Let's say I pick up I pick up two balls from the bag What are the probability that one will be of red color and the other will be of a white color Let's say they are red and white balls in the in the back only so that is a compound event Right because there are simpler events associated with that experiment okay, so the first type of Event that we are going to talk about is Assure event This is on a broader basis on a very broad scale event can be classified as Simple event and compound event. I'm going to talk about the different varieties of events. Okay Now this can be applied to both the types of events simple or compound. Okay, so what is a sure event? Assure event is an event which will definitely occur when you are performing that experiment For example, what is the probability of you getting a number or natural number less than equal to six when you roll a dice In fact, any outcome that comes will be less than equal to six. So that is a sure event So for such events, we say that the probability of occurrence is actually one So any event whose probability is one that is called a sure event. By the way, all of you should know that Probability of any event ranges between zero to one Okay, it cannot be more than one. The reason why I'm giving this even though it is known to you is many people There was a question I asked to one of the students His answer was some five C2 So probability you cannot have an answer like five C2 five C2 is 10, right? You cannot exceed one actually, okay So, um, let me hear out from you some examples where the event is a sure event Can you give me some examples on the chat box where you feel that event is a sure event any example? Front of guy three. Yeah, but can you think of some examples where the event is a sure event? Getting a one on a die is a sure event. That means whenever you throw a die, you should get a one guy three Okay, getting either head or tail on a coin When you toss it is a sure event. Yes, absolutely Picking a red ball from a bag of red balls very good example Pooja. Absolutely. So very good examples Rehan and Pooja any other example next is Impossible event. What's an impossible event an event which can never occur? Okay Yeah, getting a number more than six on a die. It's an impossible event correct or for that matter The Sun not rising what is the probability that the Sun will not rise in Bangalore tomorrow Impossible event it has to rise right. We are not living in Norway or you know The country is near to the pole right land of Midnight Sun and Those type of situation is not there in Bangalore. Okay, so these are impossible events Now these things are very simple, but now we'll come to the real situation Just a second guys Yeah, sorry Now next thing that we are going to talk about those things are very important from your 12th Probability also point of view. So please listen to this very carefully mutually exclusive events Mutually exclusive events, what are mutually exclusive events? So we say event a and b are Set to be mutually exclusive mutually exclusive exclusive If they cannot occur together that means their intersection is a null set Okay, that means that these two events have nothing in common are you getting my point? So you can say that this compound event will never occur This compound event will never occur a simple example. I can give you is let's say What is the probability that? When you roll a die you get an even or an odd number or even and an odd number Even and an odd number a number cannot be simultaneously even and odd both So these two sub events or you can say simple events are mutually exclusive. They cannot occur together So if I write a in the form of a set, let's say if I talk about a rolling a die rolling a die So if I talk about a a will have two four six if I talk about be it will have one three five There is nothing common between them if I take the intersection it will be a null set so you cannot get You cannot get an even number and an odd number simultaneously on a die So they are mutually exclusive mutually exclusive means one Excludes the occurrence of the other a Simple example would be head and a tail Another example would be any two number on the die on the head of a die only one number can occur at a time Okay, so these events are mutually exclusive events in the language of probability. We say that for such an event The probability of occurrence of a and b will be zero So as an event a intersection b it will become an impossible event if they are mutually exclusive Is this fine? So you can have multiple events mutually exclusive also So if I say if I say even e to etc till e and arm Exclusive events or you can so mutually exclusive events Mutually exclusive events that means the occurrence of any two of them Will be a null set or they are disjoint from each of them Okay, so for such cases you can always write down that the probability of their occurrences Intersection will be a null set for the point so mutually exclusive this word you are going to hear very very frequently in classes So if you have any kind of a doubt any kind of a concern, please do bring it to my notice right now Okay Any questions any questions? Okay, let me take the attendance next is Exhaustive events exhaustive events So what are exhaustive events? Huh many times you call it as collectively exhaustive events also sometimes it is called collectively exhaustive events So what are these events? What are these events? So events even e to Till let's say Ian are said to be exhaustive are said to be exhaustive if Their union becomes the Their union becomes the sample space Their union becomes the sample space. So yes, they're union should constitute the sample space By the way, do not get me wrong over here. I don't mean to say that they must be disjoint with each other They may have common elements that is fine But all I need to check is whether their union makes the sample space or not Are you getting my point? So a typical example that I can cite for this is let's say is an event is an event where you get a even number where you get an even number on on Rolling a die Okay, B is the event that you get a prime number prime number On rolling a die. Okay. I let's say C be the event that you get a number Less than four on rolling a die. Okay. So if you start writing These in the form of sets even number are two four six Prime number are two three five Number less than four will be one two three Okay, now they may have common elements in them, but when I do the union I Get the entire possible sample space when you roll a die, which is one two three four five six Which is your sample space. So these events these events would be called as Exhaustive events or collectively they exhaust the sample space Is the idea clear? So now having learned exclusive events having learned exhaustive events there could be a combination of the two also So you could have something called mutually exclusive and collectively exhaustive as well So let's talk about that By the way, any questions that you have from these four types of events that we have learned so far Any questions? Any questions any concerns any discomfort in understanding? Please highlight. So next we'll have Mutually Exclusive collectively exhaustive Collectively exhaustive in and short form we call such events as M E C E M E C E this will be used Quite a lot I'll write it down over here useful in Great 12th probability Especially in topics called law of total probability Okay, so I'm just giving you an idea up front or itself that it is going to be very useful and base theorem and base theorem base B A Y E Based okay, very very useful. So please understand this are very very carefully So what is mutually exclusive and collectively exhaustive events mutually exclusive and collectively exhaustive events as the definition itself is known to you? so We call these events as M E C as M E C if if number one their union Gives you the sample space Okay, and Intersection of any two of them is a null set Okay, when these two properties are satisfied when these two properties are satisfied then normally Normally what every time you will realize that the sum of the probabilities of such events will always add up to give you a one This is our biggest you can say Way to identify whether certain events are M E C E or not M E C events Probability will always add up to give you a one. Okay a simple example. I would take over here Let's say Even is the event that you get a 1 and 2 on the die? E 2 is the event that you get 3 and 4 on the die E 5 is the event that you get Let's say 5 and 6 on the die. Okay. Now ask yourself Are they collectively exhaustive? You will say yes because when I take their union I get the entire sample space which is 1 2 3 4 5 6 Okay, which is actually a sample space. Are they mutually exclusive? Yes, you take any of two of them That's even any two. There is nothing in common You take even an E3 nothing in common You take even and sorry E2 and E3 nothing in common Okay, so they are mutually exclusive also and now you'll see that when you talk about their respective probabilities They will add up to give you a one see probability of even to happen is 2 divided by 6 Probability of E2 to happen will again be 2 divided by 6 probability of E3 to happen is again 2 divided by 6 So when you add them you will always get a 1 Okay, this is the one of the biggest trademarks or biggest identification point for you to identify Okay, I'm not saying that I'm not saying that These events Cannot be complimentary. Now, that is what I'm going to talk about next. So first note this down first note this down Then I'll talk about complimentary events Any question regarding MEC now? What are complementary events and how it is? Similar or different from mutually exclusive events. Now. What are complementary events? Complementary events, please do not confuse it Complementary events are type of MEC events only but with only two events associated there Okay, so we say E and E bar are complementary events are Complementary events if number one, they are collectively exhaustive and Number two, they are mutually exclusive, right? So complementary events is a special case It's a special case of MEC events where only two events are, you know associated So we can say E and E bar are complementary of each other. We cannot have, you know, one event Complementary to five events. No only two events are can be complementary to each other For example, it raining and it's not raining. They are complementary events You getting an even number you getting an odd number Complementary events, but you cannot say one is complementary to two, three, four, five, six when you roll a die No, they're not complementary events. They are exclusive events agreed, but they're not complementary events So in short, all I want you to know is complementary events are those events Which are MEC events, but only two events are associated there E and E dash Are you getting my point? So it follows the same nature as what MEC events follow. That means the sum of their probability Will actually be a one. So many times we say PE complement is one minus PE Okay one minus Any questions? So many people, you know, they Interchangeably use these words That's why it is very important to get the definition of these type of events very very clear. There's something called Independent events, but those are subject matter of class 12. I will not discuss with you right now independent events Okay, so we'll be talking about independent events in class 12 So in class 12th probability will be making use of these terms while we are discussing certain kind of an event Okay, so I hope these all types of events are clear So we'll take one few questions on this one question one simple question on this. Yes, any question? Okay, no question So let's Take a question Which are one one small thing I missed out Equally likely events equally Likely events Now what are equally likely events? Okay, so we say even e2 Till e and r equally likely events Equally likely events if their probabilities of occurrence are equal If the probability of occurrence are equal Okay, now why it is important to say this is because I'll tell you a very small story of mine So When I was very young of course now also I'm young only We used to have a big campus right and in that big campus we had a small house Okay Now the campus was big enough so that we could actually make up a cricket pitch over here So we had a small cricket pitch. Okay, where me and my brothers used to play cricket okay, so Let me put some stumps Okay Okay, so we used to play cricket over here So many times when we used to hit the ball and when you should hit the ball in this direction Okay, so we were not sure whether the ball was inside the house or it actually crossed the house and fell in the Premises somewhere. Okay, because it was a blind spot for us, right? Now whenever we used to hit the ball in this direction My brother used to tell there is a 50-50 chance that the ball is within the house and 50-50 chance that the ball is outside that means 50% chance that it is within the house and 50% chance it is outside the house Now you tell me Did he use the correct statement 50% Was he correct when he said it was 50% chance that is inside the house and 50% outside. Do you think that? Charan says yes Okay, so Charan there is a 50% chance that tomorrow when you go out for jogging You will find a bag full of corrors of rupees and 50% chance is that you will not find it That means 50% chance is that you will be a corpathy tomorrow And 50% chance that you will not be a corpathy Is it a correct statement to use? The answer to this both these questions are no Because these events are not equally likely now whether the ball is inside the house or the ball is inside the you know Promises but not within the house is not equally likely It depends upon what is the area of the house within the premises For example, let's say this area of the house was 100 square feet Let's say 1000 square feet 100 square feet is too small for a house Okay, and let's say the rest of the premises was 2000 square feet Okay, that means it is twice more likely that the ball will be inside the premise than in the house Are you getting my point? It all depends upon the area covered by this house within the premises So smaller the area lesser is the chance that the ball will be inside the house Are you getting my point? So why I'm telling you this story is because I keep getting these kind of I would say Illogical arguments from the students and it is 50% chance that I will become you know I'll get these many I'll get a you know a gold coin or something in the in the in the in my house When I reach house, right? So it's not a 50% chance. It is not equally likely that you will get or you will not get your chance of getting a Bag full of course of rupees is like one in one million very very less chance Are they hitting up one? So don't say it is 50% chance that I will get 50% chance. I will not get no It is not 5050 9.9999 Probabilities that you will not get because that is that is Given by your historical data also how many times you have found a bag full of you know You know tons of notes when you're walking to your home or you know walking jogging in the in the in the park Very less, right? I cannot deny that it cannot happen. But what are the chance of it happening very less? So there's 50% chance that the flight I take to travel to my home will crash So 50% chance is that I will die today when I take up when I board a aircraft It's not 50% chance. It's it's almost point four percent chance Okay, or point zero zero one percent chance that it will happen because these modern day Aircrafts are backed up with a lot of engines and a lot of support systems, right? So me losing my life when I board a aircraft is not 50% is not half Okay, so get this difference. So you losing the life or not losing the life when you board an aircraft is Not equally likely Are you getting my point? So when can you say equally likely? Let's say you are trying to solve a question Okay, four options are there. You are not sure how to solve it and you guess the answer Right. So all the options that are sitting there. They're equally likely to be chosen You do not have a preference towards any one of the option vis-a-vis the other Okay, unless until you have a special attachment to option B, which most of the people I think have Okay, so most of the people will go for option B. But actually in reality, it is equally likely to be chosen Okay, so these are things which you need to understand when you are trying to solve questions. So now we are ready to take a small Problem based on this Discuss whether the following are equally likely mutually exclusive and exhaustive events. Okay So all you need to do is tell me Yes and no for each one of them. So let's say this is your first question. You are throwing a die These are the two events that you're talking about. Are they equally likely? For example, if they're equally likely, you'll say yes. Okay, if they're not say no So please give your answer in terms of yes. No. Yes. No. Yes. No Okay, so for the first one, I would request you to give me a response on the chat box Okay, so I've got few answers Very good. Okay Let's discuss the first one so when you Throw a die you getting an odd number Okay, so you getting an odd number odd number can be one three five So chances of you getting an odd number is three by six, which is one by two Okay, and you Throwing a composite number. Okay. So composite number Are four and six Okay So probability of B will be two by six, which is one-third. Okay, since PA is not equal to PB They are not equally likely for the first one answer will be in Okay People who are considering two to be a composite number. God bless you all I can say that So if you don't know the difference between a composite number and a prime number in class 11 Bhagawan Malik hai aap logota I'm serious. Don't make these kind of stupid mistakes at this juncture of your life Are they mutually exclusive? Yes, there is nothing in common between these two So the answer for the second part is yes So all you need to see is for them to be mutually exclusive. Are they disjoint? If yes, they are mutually exclusive. Are they exhausted? No, if I take the union, I will not be getting two in that. So the answer is n So the first one the answer is why sorry n y n n y n I think nobody got it, right? Now, I'm sure second one. You should be getting it, right now when you're answering second one Please put the question number Please write two or please write that Roman number and then answer Okay, very good Arjun Very good. I think this is the most easiest one. Very good Gayathri Good Rehan, Pooja, nice Doublet is 112233 445566. Doublet means same number on both the dice. That is a doublet I'm sure you would have heard of this term. Doublet. In 10,000 also you had you would have seen questions where they asked about doublets Okay Yes, let us discuss the second one Second one You getting a white ball? Let's say I call even even is the name they have given also so even is the event of getting a white ball, which is 2 by 9 E2 is 3 by 9 E4 or sorry E3 4 by 9 So they're not equal 2 by 9 is not equal to 3 by 9 is not equal to 4 by 9 So they are equally likely? No, they're not equally like Okay, are they mutually exclusive? Yes Because a ball cannot be simultaneously white and red both Okay, so yes, they are mutually exclusive. Are they exhausted? Yes, they are exhausted because You know the union of even either you get a red ball or a white ball or a green ball that will make up your sample space So answer is n y y Most of you have got it right. Well done third one Third one You are throwing a die is throwing a doublet be throwing a total of 10 or more Total of 10 or more very good guy three Good run good Nice. Okay. Let's discuss So doublet is like you getting okay. Let me write down the doublet events first So doublet is where you end up getting 1 1 2 2 3 3 4 4 5 5 6 6 okay, this is a doublet, right? So what are the probabilities that you get a doublet when you roll a die when a roll a pair of die? Remember when you roll a pair of die, there are 36 outcomes and Out of which doublets are six of them So the chances of you getting a doublet when you roll a pair of die or a pair of dice is one by six okay now You getting a number whose total is 10 or more So you can get four six six four five five Five six six five Six six correct. These are the ways you can get 10 11 or 12. I don't think so. There's any other way Let's have a count of it one two three four five six. So again The probability of you getting a number which is you getting a sum of ten or more is six by 36 Which is one by six since P and P we are equal the answer to the first one is Yes They are equally likely Okay, there are people who have made mistakes here. Yeah, be careful. These are basics that I'm testing you on right Are they mutually exclusive? No, they are not because five five and Six six are common to both these events. So they have something in common. So they are not mutually exclusive This is no Are they collectively exhaustive or are they exhaustive you would say definitely no because there are so many cases Which is not covered where is two three? Right, there is enough over five all those I mean so many cases are there which are missing So the answer is again I know for it So the answer for this is why nn Sharon got it right. I think nobody else Okay, good, let's do the fourth one Let's do the fourth one. Please make sure you are writing your question number So that I identify which question are you talking about? This is quite easy very good urgent very nice very nice to job some of you are No participating in every question. That's the spirit. I'm looking for some of you are absolutely quiet That is not acceptable, right? I want people to be in action Good. Okay. Let's discuss. So when you pick up a card from a well-shuffled pack of cards Even is the event of getting a heart? Okay? It was the event of getting a spade if he's the event of getting a diamond and you for the event of getting a club card Okay, so as you all know a pack of 52 cards will contain four suits and In those four suits, there will be a heart spade diamond and club so 13 of each one of them are there So P even is 13 by 52 P e2 will also be 13 by 52 P e3 will also be 13 by 52 and P e4 will also be 13 by 52. So yes one fourth one fourth one fourth So are they equally likely? Yes Are they mutually exclusive? Yes. A card cannot be simultaneously a heart and a spade both Are they exhaustive? Yes, I don't think so. There is any other type of cards other than hearts spade diamond and trumps Okay, why why why? Absolutely correct next fourth one. I'm sorry. Just one first one From a well-shuffled pack of cards the card is drawn a is getting a heart and B is getting a face card now Face card, let me tell you the three face cards in every suit. What are they King Queen and Jack? Right, Raja Rani or Ghulam. That is what we say. Okay, King Queen and the Jack okay, so Let's answer are these two events equally likely Are these two events mutually exclusive? And are these two events Exhausted very good very good So let's discuss this out first of all you getting a heart When you pick pick up a card from a pack of 52 cards is 13 by 52 But you getting a face card is 12 by 52 because there are 12 face cards in a pack of 52 So they are not equal. They're not equal Right, so this answer is no, they are not equally likely Now are they mutually exclusive? No, there could be a face card, which is also a heart So there can be common elements, right? So no, they are not mutually exclusive. They have some common elements Are they exhaustive? No, there are other cards also Present which can be neither heart nor a face card Okay, so there could be other cards also So the answer for this is also n so n and n is the right combination and I think most of you who answer it Okay, any questions so this was just a simple exercise to let you know the type of events Now the next thing that I'm going to talk about is actually how do we apply PNC? application of permutation and combination for counting for counting for counting The favorable events and the sample space By the way, you all have done permutation combination. It's not a new chapter for us. So You all know how to apply it also. We have done enough number of problems in your DPPs also So let's directly jump to its application, right? So let's start with our question. Let me start with the question So let's start with this question question is n different books n greater than equal to 3 are put at random in a shelf Among these books among these books There is a particular book a and a particular book B okay The probability that there are exactly our books between a and B The probability that they're exactly our books between a and B is which of the following is Which of the following? I'll put the poll on I'll put the poll on Let's have around two and a half to three minutes for this Anytime starts now Almost two minutes gone. I've got just two responses One response per minute Okay, so I think the first problem needs a bit of hand-holding So just take your best shot, then we'll discuss it. Don't worry about Write a wrong person. Okay. So in another 30 seconds, we'll close the poll and we'll discuss the scenario Okay, five One. Oh, I can see most of you have gone with option number C Let's check See first of all, what is the What is the classical definition of probability? Classical definition of probability is nothing but the number of favorable events divided by the number of sample space Let's find each one of them out So how many ways can you arrange and different books on a shelf? I'm sure None of you would make a mistake in answering this it is n factorial, right? I hope nobody has made a mistake in this at least. Okay Now, what is a favorable event? Let's start to talk about your favorable event here. Your favorable event is between the book a and b Between the book a and b you should have our books Exactly our books. Okay, there should be our books our books okay, and There are total n books out of which are plus two is already occupied. So outside there will be Outside there will be let's say I will write and Minus r minus two books, right? So all together there are n books and you have to arrange them in such a way that These are books or they should be always our books between a and b So first of all, what will you do? What will be the first task that anybody will think of you will think? Let me first choose these our books. Okay, so if you bar your a and b books There are n minus two books So you can choose your our books in n minus two CR way, correct? Yes or no correct now having chosen these our books Use string method. What is the string method? We tie them as a unit So wherever these will go they will go as a unit Am I right? So all together if I say there will be one n minus r minus two plus one Okay, so total you can say there will be n minus r minus two plus one books, which you are trying to shuffle superficially So superficially these books you are trying to arrange Among each other so that can be done in r minus n minus r minus one factorial way But what about intra arrangement? So your our books can be internally arranged in our factorial way and Your a and b can shuffle its position into factorial ways So this will be your total number of ways in which you can do the favorable event. So I'll again repeat why this Why this? This is to enable the choice of our books, which you are going to put between a and b. So choosing Our books to be Put between a and b Okay, why this? Why this expression? This is the number of ways to superficially arrange So you're treating this as a unit Right, so one unit and n minus r minus two so together will be n minus r minus one Units that you are arranging right because total units. I'll write it down because the total Units now will be n Minus r minus one. Okay. Why this? Why this? It is because you are trying to arrange the our books between Between a and b. Okay. Why two factorial? Because you are trying to arrange a and b among each other arrange a and b You know among each other or between each other Are you getting my point? So your sample space and your favorable events have now been Figured out. So let's use our formula Ne by ns, which is nothing but n minus two cr Into n minus r minus one factorial into r factorial into two factorial By n factorial. Let's try to simplify this. This is n minus two factorial by R factorial n minus r minus two factorial Other terms I'm just copying Okay, a lot of cancellations will happen. First of all, I can see r factorial r factorial going off Second thing I see is n minus two factorial if you cancel out with this This will be this will be left with n into n minus one and If I cancel this term with this I'll be left with n minus r minus one So your answer finally will be two factorial, which is actually a two by n n minus one This is going to be your answer Is there any option which is saying so I think it's option number B It's option number B Let me check the poll again. Oh my goodness people have gone with C First question clean bold first ball clean hold. Okay, so there's nothing to teach here There's nothing to teach here. All you need to do is brush up your skills of counting once again Right Very good, but I turn Any doubt in this explanation any concerns with this explanation all set Okay, let's move on to the next question Okay, let's read this question. The question says there is a guy Mr. A who lives at the origin Okay, so his house is at the origin and he has his office at four comma five So the office of Mr. A is at four comma five Okay, his friend's house is at two comma three. Good Mr. A can go his office traveling one block at a time either in the pocket or in the pocket One block at a time either in the positive y direction or in the positive x direction That means he can either go right or he can go up He cannot go left. He cannot go down So when he's going towards his office, he can either move a unit right or he can move a unit up Are you getting my point? For example, one of the ways he can reach his office is he took your unit up up right right Right up up up right. So that's one of the parts that he can take. Okay. I'm just giving a sample example If all possible parts are equally likely Then the probability that Mr. A passed by his friend house while he was going to his office is which of the following So favorable event is where he is passing through his friend's house while he's going to the office Sample space is where he can reach his office in any part by taking any part Okay, so I'm just giving a hint to start with let's have around three minutes To solve this very good very good. I can see a lot of responses coming in Think carefully We have solved such kind of questions in probability and permutation combination chapter Last one minute Okay, let's wrap this up in another 15 seconds five four three two one Most of you have gone with B Let's check. Let's check B has got almost 50% votes. So many people have not voted. Let's check See, what is the sample space here sample space is how many ways you can go to the office, right? Now to go to the office. He has to go Four units to the right and five units up Okay, that means he has four R moves five you moves What is our R is right you is up, okay? So in order to reach his office From zero zero to four comma five. He has to go four times, right? Khabi to Jaya four times light in some stage of his journey Total number of right move should have been four only and Total number of up moves would have been five only then only really reach office So the number of parts that he will take to reach his office Will be the same number of ways in which you can make a nine alphabet word from our RRR you you So how many ways can you make a nine alphabet word from these, you know repeated alphabets? Your answer would be nine factorial by four factorial five factor Isn't it? Let's figure out about these values. So nine factorial I can write it as nine eight seven six By four factorial is twenty four So this is two fourteen in two. This is one twenty six. Okay Now favorable events favorable event is that event where he first goes from Origin to F So O2f and from there he goes to his office. So F2p Right. So how many ways can you go from O2f? So O2f basically has to take two right moves and three up moves Then only he will reach two comma three So two right moves and three up moves just like we did the sample space can be taken in five factorial by two factorial three factorial It's like saying how many ways can you arrange? Five alphabets where there are two hours and three years You've already done this in permutation combination chapter, right and means multiplication and means multiplication and means multiplication Yeah How many ways can you go from F2p? So F2p if he wants to go he has to take two right moves and two up moves So four factorial by two factorial two factorial Okay, which is very easy to calculate the first one is a 10 the second one is a six. So it is 60 So your probability is nothing but Ne by ns That is nothing but 60 upon 126 Does it simplify yes 10 upon 21 Which is option number be well done Janta well done good It's all a good problem. I'm proud of you Okay, any questions any concerns any discomfort Please do ask Yeah, remember I took a question based on this when I was doing a chessboard question I said that there is an ant which is sitting at one diagonal and it wants to go to another time, right? So it's based on that similar concept Shallow with this will move on to more questions. This is all about practice. This is all about practice So here is a passage based question. So let's read the passage carefully This question says a is a set containing 10 elements okay a subset P of a is chosen at random and Set a is reconstructed by replacing the elements of P Another subset Q of a is now chosen at random. Basically, it's trying to say that you are making two subsets out of a okay They write it in such a way that as if they're giving some very important information They should have just written they are making two subsets out of a P and Q Okay, 27th question says What is the probability that P union Q? Will be the set a itself Okay, good question Please think about it. I'm giving you two minutes time. I'm putting the poll as well What is the chance that when you make two subsets out of a? Their union will make the set a itself. Okay Let's solve this Just 27th one you have to answer 28th and 29th we'll do after about okay We are nearing about two minutes only two responses. I've got I can give you 30 more seconds And this is a very simple question because If you think it in the right direction, it'll just take you 20 to 30 seconds to solve it one What what would only eight people have voted all right? So confuse response Almost equal number of fours to a b and c. Okay. Let's check it out. See Set a has got 10 elements right a 1 a 2 a 3 a 4 a 5 da da da da till a 10, okay now Every element that you have over here Every element that you have over here has got four options Right, so every element every element has four options What are they number one? That element. Let's say every element. I'll name that element as let's say a I Okay, either a I can go to P and It can go to Q Or it can go to P and not go to Q or It cannot go to P and goes to Q Or it doesn't go anywhere. That means it stays in your set a itself. That means it is not used at all to make a subset Yes or no So every element has got four options. What are those four options? It can go to both the sets P and Q So when you're formulating a subset Right when you're formulating a subset that subset can contain both the subsets can contain that element a I okay, or Only P will contain Q will not contain or Only Q will contain P will not contain or It is not contained by both P and Q that means that element is not used in making the subject Okay, these are the four possibilities any element can undergo. Okay now If every element has got four four four four four four choices The total number of samples space is undoubtedly four to the power of ten It will put four to the power of ten other Total number of possibilities that can happen when you are making subsets P and Q from elements of setting Is there any doubt regarding this? Is there any person who is not happy not convinced with this argument? Everybody's happy sub cushion. Take it. Let's move forward Now what is your? favorable event favorable situation is where your P union Q should make your set a When will this happen? This will happen when every element either goes to P or Q or both of them That means only this situation is ruled out Because if let's say an element doesn't go anywhere No matter whatever P and Q you formulate its union will never make a a Anna So every element should go either in P or Q or in both of them. It cannot stay back in your set a agreed So for favorable case all elements should have only three options Yeah, so bhaiya don't know may job P and Q man Yeah, so P may job Q. Mehman job. Yeah, so Q. Mehman job P. Mehman job Yeah, it cannot stay back in your setting So your favorable event would be Three to the power of ten because now you are providing every element just three options for option fourth option is ruled out And you can't stay at home. You have to go out You have to go to P's house or Q's house or you can go to both the houses, right? Then only their union will make the setting Agreed no, so your favorable event divided by the sample space will lead to three by four to the power of ten So your answer is actually C Right so out of the three options, I think C got the least in fact D got the least which is zero But from the chosen option C got the least are you getting my point? So this is not a rocket science It's all about your counting skills and nobody can actually make you better than you yourself Okay, so this is your 27th question answer. Now. I think this is a good enough I don't hint for you to work out the 28th question So what is the probability that? P Junion Q will have exactly four elements Exactly four elements. I'm relaunching the poll this time. I'm expecting better response and accurate response Don't disappoint me. Yeah Again, two minutes is good enough for this as well Last 10 seconds. I've only got one response Very few people have responded. So we'll wrap this up in another 15 seconds five four One I just tell you one thing Any exams are going on for you? Have you some missile exam started? Not now, right? Hsr and YPR people Feb 15 will start Okay, okay So after Feb 14 celebration So exams are there oh Febs 26 Okay, okay. Good. Good. Good All right, so let's discuss see first of all The sample space is not going to disturb sample space is going to be 4 to the part in like no difference in your sample space Okay, now favorable events favorable events is where? You want exactly four elements to go to Exactly four elements to be in P Union Q So now first of all which four elements Correct, which four elements? Yeah, poll results are B by the way Okay, B is what people have said B for Bangalore. So which four elements first of all you have to choose that So out of A1 and A10 out of A1 to A10 which four elements will you want to you know go in P? Union Q so first of all out of 10 choosing four is 10 C4 Okay, now let us say let us say you chose a 1 a 2 a 3 and a 4 correct Now these four elements that you have chosen should either go to P or Q or Both of them Correct that means these guys have three three three three options, right? Rest should not go anywhere Rest should actually stay back in the set a that means these all will have one one one one one option That means do not go to P. Do not go to Q stay home Correct because you want only four elements to come in P Union Q Right, so only the chosen ones the ones which you have chosen they will have three options Either go to P's house or go to Q's house or go to both of these houses Then only the Union will make those elements appear in the answer and rest of the element should not go anywhere Correct that means three to the power four is the total number of ways If you want to be more accurate, you can write three to the power four into one to the power six doesn't make a difference It is still three to the power four. So this will be your favorable events So your probability will be nothing but ten C four By three to the power four divided by four to the part ten ten C four is two hundred and ten Three to the power four is eighty one this you can write it as two to the part twenty also So out of this you can cancel off or two So this will be two to the power nineteen and I think this will give you eight five zero five By two to the power nineteen, which is your option number D my dear friend and let me tell you How many people have voted for D Just two people have voted for D Okay But I'm sure I'm still hopeful that you'll be able to get the 29th question. Correct So 29 questions is what is the probability that? P and Q have no common elements P and Q have no common elements your poll starts now Yeah, I'm Pucka that you will get it right Most of you should get it right very good panel Guys come on three people have voted P and Q should have no common elements. Okay five Four two One Go again Same old story eight people have responded which is like anyways only eight of you are only active in today's class and 50-50% to A and B see again sample space is not going to change sample space is still four to the part in right Now what are the favorable events if you want an element? To come in both the sets It should actually go in both P and Q But if you want it to only come in one sets it can either go in P and not Q Go in Q, but not P or it could also go in neither of the two So now if you want your P and Q to be disjoint Then you should not allow any element to go to both the sets They can go only to one of the sets or don't go anywhere and a either you go to P or Or you go to Q or you don't go anywhere Then only your P intersection Q will have no common elements So now every element will be given three choices again. So your answer is going to be the same as what you got for 27 So your probability of the event will be 3 by 4 to the power of 10 which is option number B So half the Janta or four of you who said option number B well done Are you getting my point see it is not a very difficult concept that you are you know giving us a confused response So if you want an element if you want the Intersection of P and Q to be blank to be null to be disjoint you want P and Q to be disjoint sets You should not allow any element to go to both of them, right? They can exclusively go to one of them or don't go anywhere. So the first case will be ruled out Right in the option list only second third and fourth that means three options every element will have Okay, so three into three into three into three ten times three to the part and will be a favorable events. Is it fine? Let's move on to the next question. Okay for a change. I will take an easy question Let's see if I am getting a response for this Consider a function f of x which has got zeros as four and nine So four and nine are the zeros of a polynomial function. You can say Okay. Yeah, this is a polynomial function, which has got zeros as four and nine Given that Mr. A randomly Selects a number from minus ten to ten Then the probability that Mr. A chooses a zero of f of x squared F of x squared So again, let's have a poll. It's a very simple question. I'm sure most of you will get this right very good last 30 seconds Guys, this is super easy Don't miss out the opportunity not to answer this question Okay Seven of you have responded and all of them have given the same response good five Four three two one Go so nine responses I've got out of which eight are saying a and one of the person is saying see Okay, see let us try to understand this if f of x has a zero four or nine Can I assume that f of x is actually this function? Correct. What are the meaning of zero zero is basically that number which you put in place of x so is that the entire function An entire polynomial becomes a zero So this is the best choice of the polynomial, right? You can have some constant also doesn't matter So what is f of x squared f of x square will be just you replace your x square with a sorry x with an x squared So what are the zeros of this zero of this will be plus minus two and plus minus three Okay, so what are the chance that if you pick out a number from this it would be a zero of this function So remember the favorable cases will be four numbers, which is plus two minus two plus three minus three And how many ways can he actually pick up a number from these numbers? There are by the way 21 numbers isn't it? Minus 10 to 10 there you are passing through zero also. So that is 21 numbers So the answer is pretty loud and clear in front of you four by 21 option number a any questions Come on. I thought I would give a easy question to boost of your confidence But people have not been able to capitalize on it. Let's take this question Three integers are chosen at random from the first 20 integers Okay, three integers are chosen at random from first 20 integers What is the probability that their product is even? Let me put the pole Two two and a half minutes is maximum. I can give you for this Time starts now. Actually, I don't understand the meaning of the word first over here. So let's say 20 integers from 1 to 20 actually the answer should not change if you pick up any consecutive 20 integers So, yeah, okay, that's fine Guys two minutes over three responses one more minute. I can give you I'm sure you have done tons of such questions when you are doing PNC so in PNC the same question would have become in how many ways can you make their product as even okay? So that would be the you know synonymous question analogous question in PNC Okay, five four three One okay Again nine people are active that I already told you out of that nine Okay, so I tell people are active out of that most of you have gone with C Okay, C for Sharon C for Shit night. Okay. Let's check First of all, what is the sample space sample spaces in how many ways can I pick up three integers from 20, right? So we know this from our childhood days. It will be 20 C3 Okay Now favorable events Favorable events is basically those events where at least one of the numbers should be even because if you want the product to be even At least one of the number should be even correct now Can I say it is total number of cases where all minus all of them are odd See, I could also take one even two even three even that is fine, but that is slightly longer So I'm going by a smarter route. So what I'm saying is the total number of cases, which is 20 C3 Minus those cases where I'm picking all the three numbers which are actually odd correct now out of 120 we all know that From 1 to 20 there are 10 even numbers and 10 odd numbers Correct. So I want to pick up three numbers and all of them should be odd So the number of ways I can do that is 10 C3 Okay, you can also do this by taking cases three cases. What are the three cases one of them is even Two of them is even all the three is even but that is as good as saying Total cases minus all of them are odd Okay, so your probability of the event will be 20 C3 minus 10 C3 by 20 C3 Which is actually one minus 10 C3 by 20 C3 Okay, so let's evaluate this 10 C3 if I'm not mistaken that will be 10 into 9 into 8 by 6 20 C3 will be 20 into 19 into 18 This will be half this will be 2 this will be 4 This will be going again by 2 so it's 1 minus 2 by 19 that's 17 by 19 Is it there in my option list? Yes, it is definitely their option number C Janta was correct. Well, Jandanta very good. Nice Good work a cube with all six faces Colored Okay, so there's a cube Who's all six faces have been colored fine is Cutting to 64 cubicle blocks of same size Okay So just like you cut onion So 64 cubicle blocks have been created out of that cube find the probability That two randomly chosen blocks Have two colored faces or have exactly two colored faces So which face will not be colored or this side of the cube will not be colored the one which is facing the interior of the cube Okay, because you are you are coloring a cube from outside and then you're cutting it with a knife Right. So what are the chance that you pick up two cubes? What is the probability that when you pick up two cubes that two cubes will have exactly two of its faces colored I'll run a poll for this. Let's have around two and a half three minutes and as you solve I'm making a cube Okay, I just tried my best to Make some set of cubes Just one person has answered. Oh my goodness Alright, we will stop the poll in another 30 seconds okay, five four three one So 11 people have voted out of which five say option number B. It's a good amount of polling better Response than the previous few questions. See first of all, let's try to understand. What is the sample space? Sample space is you are picking up two cubes from 64. So which is 64 C2 Everybody agrees to this Now when you cut the cube into 64 pieces, which cubes will have exactly two faces colored So I'm shading the ones which have exactly two faces colored this too Of course, I cannot face shade the within do these two these two These two These two These two these two These two These two Okay, so the ones which are in the center two of every edge Correct. So how many such cubes are there? One two three four five six seven eight only are here the same eight will be on the backside which is hidden eight plus eight sixteen and 17 18 19 20 21 22 23 24 24 cubes are there which will have exactly two faces colored and Let's say your favorable event will happen when you pick two out of those 24 So your probability of the event will be 24 C2 upon 64 C2 Let's try to figure it out. How much is it 24 into 23 by two factorial two will get cancelled off so I'll not write it and 24 into 64 into 63 So I think this goes by a factor of Eight you can write this as a three and this is an eight Okay, and this will go by a factor of 21. So answer will be Answer will be 23 upon 168 23 upon 168. Yes option number a is the right option Chanta Again, not a correct response. Most of you went for B. Is it clear why 24 cubes will be there? Which will have exactly to I hope you can imagine it which ones I'm talking about So these two these two these two these two these two these two these two these two these two again Down something will be there backside. They'll be there. So I know I will not be able to shade all of them Okay, so out of those 24 if you pick up any two those will be your favorable events Divided by the sample space that will give you an answer. Okay Moving on to another question in an objective paper There are two sections of ten questions each For section one each question has five options and only one option is correct For section two each question has four options with one or more than one correct and Marks in each section is awarded only if he ticks all the correct options Okay, good. So as of now the story is there are two sections one and two This is a single option correct question. This is a multiple option correct question And they are ten ten questions in each one of them Okay And what next marks for each questions in section one is three. So this is a three marker. Oh, sorry Section one is one and this is a three marker. Okay, good And in multiple option correct, you have to mark all the right options in order to get the full marks Okay, and there is no negative marks. Thankfully no negative marks. Good There's no negative mark in the question I mean in these questions not in the entire question So this may be a negative mark question also So there's no negative mark as per the situation given to you in the question Now 30th question says if a candidate attempts only two questions One from section one and one from section two What is the probability that he will score in both the questions? That means what are the probability that he'll get both the questions right? Your time starts now. I think two minutes is good enough for this DPS people have exams. Where is Aditya? I have not seen him for two classes And you have anybody has an idea about DPS exams running on Akash Does DPS have exams? 15th. Oh, okay Where is Aditya? He's taking a long leaf to prepare Normally he doesn't miss a class. Sick or something? Oh, I'm sure he's not well This is not he's a very tough guy. He'll never miss a class At least not for preparing for school exams Good Two people have responded See he sure that is attempting one question From section one one question from section two and he wants to get it correct both of them He wants to get this correct. What is the chance? Let's close this in another 30 seconds Okay, five four one Most of you have said C C for Chennai. See guys, that's a very simple question Before I start solving it, I would like to ask you two questions one If I'm attempting an MCQ question, which is single option correct Right, and there are five options to it What is the chance that if I take a guesswork? I will get that correct. You will say one by five Correct because they're all equally likely options. So you getting a right option is a chance of one in five But if I'm solving a Multiple option correct question Then what is the chance that when I attempt it, I will get it correct now see in multiple option correct question a Particular question can be answered in 4c1 plus 4c2 plus 4c3 plus 4c4 ways Correct and one out of it will be correct See because we don't know know how many options are correct. So it may have one option correct It may have any two of the options correct It may have any three of the option correct or it may have all the four options correct Correct So if it has only one option correct then 4c1 if it has any two options correct then 4c2 Any 3 4c3 any 4 4c4 correct Right. So in short Out of 15 possible answers that I can give only one possible answer is correct So the chance of me getting it right is one in 15 That is why multiple choice. Sorry multiple options questions are three times more difficult Then a single option correct question. So if somebody says how difficult is to get a Multiple choice multiple option correct question, right? The chances is is even 30% of what chances you will get So you should never waste your time doing multiple option correct question first Right, you should always invest doing, you know the single option correct first Okay, because the chances of you getting correct is very high Okay So you getting both the questions correct is one in 75 chance, which is option number D Hey Janta, what is wrong D only two people answered Yeah, what is happening? Okay Alright, let's go with 31 Time starts now if a candidate in total attempts four questions What is the probability of scoring 10 marks? He's just attempting four questions This should be very easy guys. You should solve this in one minute. Look at the options options will tell you everything Last 30 seconds I can give you because I was expecting everybody to get this right within one minute five four three two one Go just five response Out of that most of you have gone with B Okay, if you are attempting if you are attempting four questions and you want to score a 10 It can only happen when you attempt one question from section one and Three questions from section two and you get all of them correct So what are the chance that you attempt to one question from section one and you get it correct one fifth? Yes or no What are the chance that you attempt three questions from section two and you get all of them correct? It's one 15th one 15th one 15 because your chance of getting One section two question correct was one 15 and now you're attempting three questions So your chance of getting all of them correct is correct correct correct, which is this Okay, so your total probability will be one fifth into one 15th cube, which is clearly option number a That's why I said look at the options. It'll tell you how to solve it Okay any questions Any questions here 32 let's attempt 32 32 should be again an easy one again 90 seconds for this time starts now very good Question says what is the probability of getting a score less than 40? less than 40 Anything less than 40 the lease. I think you can get is a zero So zero to 39 you can say very good Again, look at the options options will tell you everything Correct pun up. Okay. Let's solve this five four three two One all right Maximum people who have voted have said be which is 71% of the world See 40 is the total marks Right, if you if he attempts all the questions of section one and gets everything right you'll get 10 marks If he attempts all the questions of section two and gets everything right He will get 30 marks. So 10 plus 30 is 40 marks so if he's getting if he's getting less than less than 40 That means he's getting at least one question wrong correct so The chance of him getting at least one question wrong is one minus Him getting full marks correct So one minus the probability that he gets full marks will be your total answer So one minus him getting a full marks means he getting all the ten one option correct question and all the ten multiple option correct question correct So which is actually one by 75 to the power of ten Option number B is correct. Most of you have gone for the right option. Well, that's up So with this I'm now going to begin with Addition slash Subtraction theorems Now mostly we use these theorems on Compound events we apply them to we apply them to compound events Remember I discussed about compound events where multiple, you know elementary events or simple events are going to happen So this particular theorem is exactly based on your cardinal properties of your sets Okay, remember in your Set chapter we had learned something like and a union B. What was that and a union B? By the way, a union B is a compound event Okay, and we may be simple events. They may be compound in themselves also So and we will first take as a compound event So if you remember we had done this formula in our sets chapter, correct the very same formula if I divide by If I divide by The sample space the very same formula will actually take the shape of The very same formula will actually take the shape of a probability Okay, so PA union B is PA plus PB minus PA intersection So this is a one of the very very basic very very commonly used addition theorems in probability Of course, there are many more to come By the way, one thing I would like you to understand here is you should know how to read this many people don't know Know how don't know how to read this How would you read this? You would read this as probability of occurrence of at least One of a or B are you getting a one So when you convert your language of English to language of sets You should be knowing what a particular phrase in English means in the language of sex So occurrence of at least one of a or B or occurrence of either a or B is All read as a union B. I'm sure you don't need that kind of knowledge right now because you have already gone through your Set chapter so what I'm going to do is I'm going to give you few more Properties or addition theorems, but you would realize that you already know them because you have learned them in your sets chapter For example probability of occurrence of at least one of a B or C or at least one of a B and C is PA PB PC Minus PA intersection B minus PB intersection C minus PC intersection a Plus PA intersection B intersection C same formula that you have learned in sets Okay, if I say what is the probability of occurrence of a and not to be So it is PA minus PA intersection B So your when diagrams the best thing is use when diagrams So instead of that, you know NA you use PA instead of that Okay, so let's say if I want to get this formula from when diagram So what I'll do is I'll make your set a and set B. This whole circle is PA circle This whole circle is PV circle So PA intersection B complement is this zone Okay, so this zone is nothing but PA union B or you can say PA minus This part which is your this part you remove this part. Okay, so of course I should not draw it Else you'll get confused. Okay, so this is how you get this formula So use your when diagram as much as possible because I may not be able to give you all the theorems Okay, this is just a guiding principle on to how these theorems are basically formulated Okay Okay, so if somebody says what is the probability of occurrence of exactly One of A and B so basically they're asking you the sum of these two so it's PA plus PV minus PA into sorry minus two times PA intersection B Okay, so as you know, we normally call this as a delta being set So a delta B is what a delta B is exactly one of them occurring exactly one of them occurring is this zone and the zone Right, so you can also formulate it like saying PA union B minus PA intersection B and PA union B is already known from our first formula. So which is PA. I'll write it down here Plus PV minus PA intersection B and there's another minus PA intersection B So that makes it PA plus PV minus two PA intersection Okay like this Okay, there's some inequalities also that you should be knowing you should always be aware that Any event A or B can never exceed their union probability. This is a very very important rule. Please note that The probability of occurrence of A or B can never be more than that means it can be less than equal to PA intersection B very very important theorem Okay, similarly PA or PB will always be Greater than equal to PA intersection B That means intersection can never exceed the probability of occurrence of A or probability of occurrence of B Are you getting my point? And of course other things that you should always know that any probability Should always lie between zero to one that is anytime given now this list can continue on and on and on I cannot give you everything you have to exercise your own discretion You have to exercise your own intellectual capabilities When you are applying these kind of a theorems just remember one simple statement of mine when diagram When diagram is where you can you know? Rely on and that will help you to solve most of the questions So let's have questions. Let's have questions I'll start with a very simple one By the way, it's not an MCQ question. It's a subjective question PA If A and B are two questions two events says that PA union B is three-fourth PA intersection B is one-fourth PA complement is two-third find the following. So let's start answering them First one, what do you think is the answer? Could you go to the previous slide once? Oh sure Let me know Charan once you're done done Charan First one So first one is super easy if you know PA Sorry, if you know PA complement What is PA? He is one minus PA complement, isn't it? Isn't it one minus two by three which is one by three? So the answer for the first one is one by three Okay Very good second one PB So PA union B is PA plus PB We have just done the addition theorem minus PA intersection B PA union B is three-fourth, PA is one-third, PB I do not know right now, but the law PA intersection B is one-fourth Okay, so if you bring it to the other side, it will become one minus one-third one minus One-third which is PB, PB is equal to two-third. Is that fine? Any questions? Any questions any concerns? Very good. What is PA intersection B complement? Now again, they have written it in a very formal language But this is not how you may expect it to be written every time They may say what is the probability of occurrence of A and a not B So A and a not B is A intersection B complement Okay, so what was the formula that we had learned for this? It was PA minus PA intersection B correct So PA was one-third, PA intersection B was one-fourth. So this is going to be one-twelfth This is one-twelfth. This is two-third Tell me for the fourth one fourth one PA complement intersection B PA complement intersection B Isn't it PB minus PA intersection B? So that's nothing but two-third minus one-fourth Five by two Is it clear how to use these theorems to get these, you know answers any question regarding that? Okay, let's move on. Let's move on time is less time is less Let us move on time is less Okay, let's do one question On inequality Let two events A and B If odd against A by the way, everybody knows the meaning of odd against If somebody says odd against A What does it mean? It means P A complement is to PA Okay, if somebody says odd in favor of A Odd in favor of A what does it mean? PA is to PA complement Okay, so I hope you everybody knows how to read odd against an odd in favor of So odd against A is two is to one and those in favor of a union B is three is to one find the range of PB Find the range of PB Let's take two two and a half minutes for this Any time starts now No response Okay We'll close this in another 30 seconds So those who want to answer please do so within the next now 20 seconds five four two one Excellent excellent. So out of seven of you have responded most of you say option number C C is the most safest choice But that may not be the correct choice. Okay. See first of all, what is PA itself? Now when you want to get PA from this information, please note that If you this is two is to one, right? So if you want PA you can always use PA as PA divided by PA plus PA complement Because the denominator will always become a one in this case. Correct. So there's nothing but One by one plus two So one third is PA similarly PA union B will be Three divided by three plus one, which is three fourth. Okay That is very obvious that P B can never exceed A union B. That means the upper Or you can say the upper restriction that PB will face will be three by four Right. So as you can see here option number B And C will be ruled out. So people who said C Absolutely wrong. That is not the right answer But let me complete it. What is the lower restriction that it will face now see PA union B is PA plus PB. I'm so sorry PA plus PB minus PA intersection B Correct. So from this expression, can I say that PA intersection B will be equal to PA plus PB Minus PA union B Now as I told you Your probability will always be a quantity more than zero more than equal to zero. Correct. So even this will be more than equal to zero Correct. So this is one third. This is I don't know. This is three fourth. So this should be greater than equal to zero That means PB should be greater than three fourth minus one third Which means PB should be more than five by 12 In other words, if you combine these two If you combine these two Your PB will be a value between five by 12 and three by four Okay, so that is option number A and not C. So A is the right option So I hope you're getting an idea of how these addition principles are helping you to solve these questions The inequality ones are slightly tricky and Jay loves asking those, you know inequality based questions Okay, so don't be like always, you know, knowing those Equality ones know the inequality ones as well Let's take another one. Okay, let's take this question If A and B are sorry A, B and C are three events says that P is three fourth And P A intersection B intersection C complement is one third And P A complement intersection B intersection C complement is again one third. What is P B intersection C What is P B intersection C? Pole is on And I'll also like to give you a hint Go with Venn diagram All the theorems have come from Venn diagram. They have not come from Skype Okay Right. So use Venn diagrams. You'll be able to Get the answer Sorry for that crooked circle All right. So almost almost almost almost Two minutes gone No response Oh one response I've got. Sorry Guys, go with the Venn diagram Same thing I'm saying see Okay, let me stop the poll first. So in another 15 seconds, I'll stop the poll So if you want to put forth your response, if you want to take a guess work, also you can take it Five Four Three Two One Go Okay Almost equal votes have gone to each one of them as you can see A B C D almost same number of votes just they differ by one one. Anyway, see What is the given to you P B is given to you P B is three fourth See P B is what this full circle think as if this full circle area is three fourth Okay, this circle area is three fourth Your probability of that event will be the area of that circle thing like that correct Next what is given a intersection B intersection B is this zone Intersection C complement that means only this zone this zone area as per the question is one third Okay, so the meaning of this is This part is one third correct Everybody's fine with that next A intersection B complement intersection B complement is this zone Okay, intersection C complement intersection C complement means this zone Correct me if I'm wrong I'm sure most of you know your Venn diagram very well Even this area is one third as per the given question Okay Now, what are they asking us? What are they asking us? They are asking us to find P B intersection C B intersection C is this zone Ye chahiyein ko correct Now this is a child's work You know the whole circle is three fourth You know this guy is one third, you know this guy is one third So whatever is left is your answer. So three fourth minus one third minus one third That will be your answer So how much is it three fourth minus two third? That's one by 12 option number a Right and most of you have gone with a very good, but most of you have done mistake also Right So this is regarding your addition theorems. There can be several other questions I will be sending you a DPP on it The last part of our discussion will be on On geometrical probability, I will not take much of your time. I'll just take 15 minutes and then we'll have a break Geometrical probability Now there are many such situations where You will not be able to count your sample space. You will not be able to count your favorable events because they are not discrete They are continuous physical quantities For example time space Right if I say how many points are there on a line? Will you be able to count it? How many points are there in space? Will you be able to count it? Remember one of the questions I one of the instances I gave about my childhood that I was playing cricket in my premises of my house so There I cannot count the number of coordinates in my house and number of coordinates in my entire premises That will be practically impossible to solve because we are dealing with such quantities which are continuous in nature They are not discrete. You cannot count them on your fingertips. Neither can you count them by using your pnc concepts Right so for those type of cases you need geometrics And that's why we call these probability as geometrical probability. Many times we call it as a continuous probability Now I would try to explain this with a simple example because there is no hard and fast theory related to it So I can only give you a simple example to understand Let's say Let us say Let us say Two of your students in the class. Let me name them Who? Let's say Arjun and Arjun and Charan Okay, Arjun and Charan. They are Very good friends and they say, hey guy, let's meet up. Let's meet up for a coffee today Right, let's meet. They want to meet for a coffee or a coffee Okay in ccd Not that's coffee capital ccd Okay, so these two guys they decide to meet over a coffee. Okay, so there's a date between them Charan will meet me when I reach school Okay, yeah, so let's say they want to meet up in uh ccd for a cup of coffee And they decide that we will meet between Uh, they decide to meet between Uh, let's say let's say let's say 4 p.m. To 5 p.m. Evening time Okay, so how will they meet sir? Centrum classes go on Okay, let's say there's a non-centrum day and they decide to meet but the condition is Whoever comes first Whoever Comes first Will Wait for 20 minutes And then leave Okay, so they have not exactly decided what time to meet but they said somewhere between 4 to 5 Right depending upon how free you are or what is the traffic? Okay But whoever comes first He will not wait more than 20 minutes and he will leave What is the probability? What is the probability that they will meet? Right, okay now you don't solve it question. I'll just tell you how to solve this question, right Now see here we are dealing in terms of time And between 4 to 5 the time Of course, you know, it's one hour only or of course, you know, it's 60 minutes or of course, you know, it's 3600 seconds But there is no limit to how much I can break it I can go I can divide time to you know micro second millisecond Pempto second whatever units in physics, you know, which is smaller smaller, right So it is such an event where you cannot count Right, you cannot count the sample space here because I can arrive at any time between 11 to 12. It's like It's like, you know, um, you know counting how many points are there between 4 and 5 How many points are between 4 and 5? No, the gap is one. I know that but how many points are there? Nobody can say there infinitely many points Isn't it? So what time these guys will come that itself is a subject matter that you are dealing with a continuous You know physical quantity which is time in this case Okay, so how do I solve this question? See every probability question will need a sample space And it will need a favorable event, right? So now what I plan to do I plan to use geometry for that. How all of you please pay attention As you can see on your screen, I have made X and Y axes X axis is the time After 4 p.m. Let's say this is your 4 p.m. Time and let's say this is your 5 p.m. Time Okay, similarly Y axis is the Charon's time 4 p.m. 5 p.m. Okay, so let's say if I choose a coordinate over here Let's say the coordinate is 20 comma 35 Okay, so basically what I did was I broke this time of one hour into 60 minutes Now most of you would be thinking sir. You only told that it's not necessary that it'll be 60 minutes See it's all about what unit you are choosing So if I choose 0 to 1 my coordinates will accordingly change Remember it is the ratio. So the unit will not matter here Okay, so if I'm choosing 3,600 even this coordinate will be to that level Okay, so it will not make any difference to your answer Depending upon what unit you are choosing your answer has to come out to be the same Okay, irrespective of whatever unit you are choosing not depending upon whatever unit you are choosing irrespective of whatever Unit you are choosing your answer will come out to be the same So what I'm trying to say is that Let's say 4 o'clock is at origin Okay, and your x coordinate basically represents The minutes after 4 p.m At which arjun arrives at ccd Right similarly y coordinate. I have assigned it to charan That means whatever is the y coordinate of that point that point tells The minute after 4 p.m At which Charan will arrive into ccd. So basically if I choose a point 4 comma 35 Sorry 20 comma 35. What does it mean? It means arjun has come to ccd at 4 20 p.m And charan has come to ccd at 4 35 p.m. By the way, are they going to meet in such situation? Will they meet in such situation? Yes, because the time gap is less than 20 minutes. Absolutely. Are you getting my point? Okay Now let me choose another point Let's say I choose a point 55 comma Let's say this is 5 What does it mean? This point indicates that arjun has come at 455 And charan had already come at 405 Do you think in this situation they meet? No, because the time gap is more than 20. So here they will not meet here. They will meet. Are you getting my point? so basically When you start choosing all possible points in the square region Okay, that will be your sample space Are you getting my point? So all possible points that you can choose in the square region will basically be the possible times in which arjun and charan will come to the ccd, correct? Now, of course, I cannot count the total number of points here. I can only count the area So I will consider the area here to be the sample space, which is 60 into 60, which is 3600 Now a question will arise in your mind, sir. Had you taken 3600 in both x and y axis, your answer would have been 6600 3600 square, agreed But that will not make any difference to my final answer because when you're taking the ratio All those, you know Things will be taken care of don't worry about it. Okay now So if I consider any point, let's say I it is these two because I don't need them Okay, so if I consider any point x comma y in this particular space, right The meeting will happen only when the difference in the x and the y is actually less than equal to 20 agreed So the feasible Sorry, the favorable situation will be when I'm choosing such a point or I'm choosing such set of points Which will satisfy this criteria in other words I have to locate a region in this particular square, which will be satisfying This criteria x minus y should actually lie between minus 20 to 20. By the way, all of you know your modulus modulus inequality What was modulus inequality when you had when you learned to solve this? What was the situation? We basically wrote it like this, right? Same way. I'm done for this also Okay, so now it is equivalent to saying two things Your x minus y should be greater than minus 20 and simultaneously Your x minus y should be less than equal to 20 Now I'm assuming all of you know your Inequalities in two variables that you have already done in your school, correct? So if I have to plot this zone This one So first I will plot x minus y equal to 20 x minus y equal to 20 is actually a line like this Okay And when you're trying to solve the inequality x minus y less than equal to 20 Please note it will be one of the regions Whether to the one side of the line or to the other side of the line. So this side of the line will be your answer So in order to know that you locate the origin And see whether origin is satisfying this inequality or not So as you can see 0 minus 0 is less than equal to 20. So it is satisfying the inequality So you will shade the upper part of the region. Are you getting my point? So x minus y less than equal to 20 will be the upper part of this line Any problem in understanding the region which I'm shading over here Okay, similarly If you plot this you first have to plot x minus y equal to minus 20, which will be a line like this Okay, I'm just drawing a line x minus y equal to 20 Now when you're solving x minus y greater than or let's say less than equal to minus 20, there will be one region coming up So how will you decide the region again? You see the origin is the origin satisfying this condition 0 minus 0 is greater than minus 20 Yes, it satisfies. So you have to shade below part Let me shade it in Orange Okay Now the overlapping the overlapping see this is an intersection symbol the overlapping will be your favorable event or favorable region. So remember Do not entertain anything outside the sample space. So I'll just erase this guy I'll erase all the external lines that I have made That is not going to be useful Okay And I have to see the common region common region will be my dear students the one which is in between This region will be the common region right, so I hope you are able to see a Hexagonal kind of a structure being created over here Okay so only when Arjun and Charan come at a point or come in such a time For which the x and the y coordinates happens to lie in this zone, then only their meeting will happen What will happen if they don't say decide they will not meet wherever Are you getting my point? All right, the highest probability is where the The difference in their time is the least which is the same time Okay Okay, so what is my answer for the favorable events is basically the area of this particular Hexagonal feature. So how will I find the area of this hexagonal feature? Just take this triangle Okay, by the way, this triangle is 40 by 40 Okay, and take this triangle 40 by 40 and remove it from the total square So from 3600 remove two times half base into height Which will actually give you if I'm not mistaken 2000 Okay, so your probability of Arjun and Charan meeting is 2000 upon 3600 Which if I'm not mistaken is five upon nine So five by nine chances that the meeting will happen four by nine chances that the meeting will not happen Is this clear how this works? Is this clear how this works? Okay, any questions So this is the most interesting and challenging part of probability which unfortunately is not covered in the school exams Which is unfortunately not covered in the school exams But before I give you a break, I would just do one question and then we'll take a break after the break We'll start with the 3d geometry The question is very simple There is a wire of length l Okay, so there is a wire whose length is l Okay, this wire is cut into three parts This wire is cut into three parts Okay, find the probability Find the probability that The three parts Will form a triangle So when you connect it They will form a triangle getting my point So there is a wire You cut at two points You can start combining their ends. Okay I hope you have done these kind of activities in your l k g u k g montessori Okay, and you try to make a triangle out of it So what is the probability that you will be able to make a triangle? So what is the chance that a triangle will be formed? See, I'm sure everybody knows that it may not be able to form a triangle in every case Isn't it? Yes or no? Okay, so let's say if you choose If you choose a condition like this where the third one is a smaller one like this Okay, then you'll not be able to form a triangle. So in how many ways How many ways what is the probability that? The three pieces that you have formed will form a triangle Okay, uh in the interest of time, uh, I'll also contribute here Let's say this length is x this length is y now some of you are saying Or some of you would be thinking let's take this as z but actually you don't require a z because There there are no three degrees of freedom right x y will automatically decide the third one. Okay So there's no need to choose a z. Okay, so 2d is sufficient for you to solve the question Now what is the sample space your sample space will be basically Uh, the fact that x should be greater than equal to zero. Okay y should be greater than equal to zero And of course l minus x minus y should be greater than equal to zero. Correct So these are the possibilities that you can actually you know have in cutting a wire. Correct Now all of you please understand here This last condition is very important. That means x plus y should be less than equal to l. Okay So these are the three conditions that you need to follow while you are choosing your sample space Now many people make this mistake They tell me sir My x and y values can take anywhere from l to l Right, so I should have a square as my sample space Please note no If you choose your square as a sample space means you are saying You can actually take a point like this. Let's say if I take a point here Which is almost let's say 0.9 l and 0.9 l Can you extract 1.8 l from l? It's not possible, right? So many people make this mistake. They think that oh x can take any value from zero to l y can take any value from zero to l. So let's have a sample space, which is a square of dimension l by l Which is wrong Which is wrong That's where this third guy comes into picture Right, so people who consider this please be very very careful very very vigilant while you are choosing your sample space Okay, just don't get carried away by the previous example of charen and arjun. Anyways x plus y less than l is basically a line which is Or basically an region which is below this Okay, so x is positive y is positive and x plus y is less than l is basically the area within this triangle I hope everybody is fine with that So this is your sample space Okay, so sample space is half l into l Now this area is your sample space Now what are the favorable events? remember for a favorable event In any triangle the triangle inequalities must be taken care of that means Some of the two must always be greater than the third So I can say this I can say this And I can say this Okay, so basically what I've used the sum of two sites Must always be greater than the third site this is the Very very important prerequisite To you format for you to form a triangle Correct So this first condition gives me x plus y is greater than l by 2 Here I will get y is less than l by 2 And here I'll get x is less than l by 2 So a combination of these three restrictions will give you the favorable Say favorable event or the favorable region in that particular sample space So let's first talk about the last two first x less than l by 2 Now l by 2 is here If I'm not mistaken x less than l by 2 is the region on this site Okay Why less than l by 2 so l by 2 is this zone y less than l by 2 is everything below it Okay x plus y greater than l by 2 now all of you please pay attention x plus y equal to l by 2 is this line Everything greater than this will be above this So everything greater than this will be above this Okay, so where do you see the overlap happening? Where do you see the overlap happening? If I am not mistaken the overlap will happen in this small triangle that is formed over air I hope you can see that red triangle Let me shade it also So this part is satisfying all the three conditions. What three conditions? x should be less than l by 2 That means left y should be less than l by 2 means below And x plus y should be greater than l by 2 means above the inclined line So out of this Let me erase other things. Yeah So out of this sample space only this area is what I need Okay, so your favorable event area would be half into Base into height by the way, this will all be l by 2 l by 2 Okay, so your probability of the event will be n e by n s Which is nothing but half into l by 2 into l by 2 upon Half into l into l which is one fourth Which is one fourth Okay, so let me tell you these are not the easy questions. They require a little bit of practice Okay, don't worry. I'll be sending you a lot of worksheets on this As of now, let's take a break because I'm sure all of you are hungry So with this we close the chapter