 So let us begin with a new chapter. The new chapter is probability Probability now just to quickly recap this chapter is not a new chapter because you have done this in class 11th Also in 11th, what did you do? Let me just quickly tell you in 11th you primarily covered two major aspects one was the use of the use of P and C in finding in finding The number of favorable events and your sample space Correct and using that you can find your probability of the event. This is what you learned correct second thing primarily you learned was the addition theorems addition theorems So what is PA union B etc. What is PA complement intersection B PA intersection B complement all those things You learned in the addition theorem. It's not that we are not going to use it anymore These concepts will be still carried forward because while you're solving questions, you'll be again using them and one more interesting thing That we did. I don't know how many of you remember geometrical probability Remember when there was a continuous sample space You learned how to find out the probability of an event for that kind of scenario also So that is actually called continuous probability or geometrical probability So in class. Oh when gets see the moment you said I wrote it so in class 12. What are we going to study in? Class 12 we are going to study the following. Okay, we are going to focus upon independent events So out of all the events that we Studied in class 11th. We are going to talk about independent events. Okay, we are going to take up some Problems directly on that Then we're going to talk about Conditional probability Conditional probability so calculating the probability under a certain condition Okay, so when we know certain part of the events How do we calculate the probability of a subsequent event because of that and Conditional probability will be studied along with along with multiplication theorem So multiplication theorem and condition probability go hand in hand Because you derive the result of conditional probability through the use of multiplication theorem only Then we are going to talk about law of total probability Let me write it down law of total probability and Finally, we are going to talk about The elephant in this chapter, which is the base theorem base theorem is actually a conditional probability But it uses this guy also Okay, so it's the conditional probability where there is a use of total probability And that is why this particular concept is very widely asked both in J exam and in your board exam because it uses all The concepts that you have studied in this particular chapter mostly mostly all the concepts Yes, this theorem is also called as the reverse probability Reverse probability and finally we are going to wrap this chapter with PDF PDF what is PDF? probability distribution function probability distribution function PDF Okay Normally we only deal with one PDF, which is your binomial PDF But I'll also talk about Poisson's PDF if time permits, but Poisson is not that important I've never seen questions coming on it. I don't know why some of the books mention about it So we're going to talk about binomial PDF and we'll talk about Poisson's PDF and To give you an insight about what is going to be in future coming for this chapter So when you go to engineering colleges, you will have a chapter called you will have a course called Probability and statistics where you'll talk more about PDFs only so there will be different types of PDF Rectangular PDF languages PDF a lot of PDFs will come to you Okay, so this is the course content for us in class 12. So we are going to start with independent events So let's get started Some of the brave hearts have joined despite having exam today. Well done guys proud of you Okay, so independent events independent events So what is the definition of a compound event Consisting of independent events. So we say two or more events two or more events are independent are independent if Accurance or non occurrence if occurrence slash non occurrence non occurrence of any one of them of Any one of them one of them Does not affect Does not affect the probability of probability of occurrence slash non occurrence Slash non occurrence of the other of the other Okay, some typical examples we can give here like let's say There's a bag Okay, let's say this bag contains This bag contains two red and three blue balls Okay, two red and three blue balls. Now you're drawing two balls One by one with replacement Okay, so two balls are drawn two balls are drawn are drawn One by one With replacement with replacement means you have picked up a ball You probably note down its color or something and then put it back and then again you pick another ball. Okay like that What are the probability that the second ball is a blue ball? What are the probability that the second ball is a blue ball? What do we answer be three by five correct? Does it depend upon what was the first ball? Correct, it doesn't depend what was the first one so the answer will be three by five Yes or no, right? So this event a and b here where a is picking up one ball b is picking up another ball Okay, they are both independent of each other because of this replacement factor because you replace them They became independent Right, so here they are example of this is an example of an independent independent events independent events Okay, but on the same hand if I change the question slightly and I say two balls are drawn one by one with replacement I'm sorry one by one without replacement What is the probability that? The second ball is a blue ball Now here you will have to see what was the first ball drawn Correct, so that would come under the concept of conditional probability or for that matter if I ask you Given that the first ball was a red ball What is the probability that the second ball is a blue ball that would come under conditional probability? Because your second event is will now depend on the first because you are drawing the balls without replacement Is this clear? Okay, so in the language of in the language of probability we say that events a and b events a and b are independent are Independent Provided this satisfy this condition pa intersection b is pa into pb Okay So for example, let's say another case I can give you what are the probability of you getting a head on tossing a coin and Six on rolling a die your answer will be simply half into one six Because both are basically independent of each other getting a head doesn't influence getting a six getting a six doesn't influence getting a head Okay, and this can be extended further so you can extend it further. So in general I can say in general I can say if events A1 a2 etc till an are independent are independent Then the probability of This compound event to happen that is a1 intersection a2 all the way till a n will be just the Product of will be just the product of probability of occurrence of AI Okay, now few properties that we need to look into when we are studying independent events Let us first look into that before going to a question Yeah, any query on whatever you have whatever has been done so far. Oh, yes, again, like it is Yeah, so properties of independent event properties of Independent events the first property is if a and b are independent if a and b are independent Then number one a and b compliment will also be independent Okay, and needless to say a compliment and b will also be independent Okay, and a compliment and b compliment. They will also be independent So some direct questions can be framed on this not in JEE probably for but in some regional entrance exam Most of you would be writing at least three four exams minimum, right? So let's say you're writing your VIT exam and this question came. Okay, so how can we prove it? See if you're claiming that a and b are independent Let me just prove one of them. Which one you want me to prove. I Will not prove all of them. Which one you want me to do? Let's go to the third one. Third one seems to be more robust see If you want to prove that These two events are independent We need to prove that p a compliment intersection b compliment is p a compliment into p b compliment Then only this event will be independent right because you have learned that if any event two event a and b are independent P a intersection b or p x intersection y should be p x into p y Okay, now Let us start with our left-hand side What is p a compliment intersection b compliment in your childhood days you have learned something called the Morgan's law correct The Morgan's law, okay, so this is your de Morgan's law, which means it is one minus p a union b Correct. Now you had learned addition theorems in your class 11th Where we had learned that p a union b is p a plus p b minus p a intersection b correct so if you open the brackets we end up getting this and Also, note that a and b are independent events Since a and b are independent events you can write p a intersection b s p a into p b. Okay This is because a and b are This is because a and b are independent events because they're independent events Right now if you factorize this one minus p a take minus p b common again one minus p a So you can get one minus p a times one minus p b and this is clearly p a compliment into p b compliment P a compliment into p b compliment. This is your right-hand side and hence proved Similarly, you can prove the first and the second as well. I think this was the Heavier of the three. So I proved it. Similarly, you can prove your second and the third one. Is it fine? Any questions any Oh my god, what happened to your voice? It's echoing Let me know once you're done. Okay. Next is the complementation rule Next is the complementation rule Complementation rule so this says that if a and b are independent if a and b are independent then P a union b is nothing but one minus p a compliment b compliment Okay, this is very easy to prove all we can do is write p a union b as P a in a union b compliment minus one, correct Okay, and you can apply Morgan's law here Shotgun Morgan, okay, and then you can use the fact that if a and b are independent So will be a compliment and b compliment and there you go Okay, now our question may arise is this rule extendable. Yes, it is extendable so in general we can say that if a1 a2 etc till a n are independent events are Independent events So they will follow the complementation rule So complementation rule means p of a1 union a2 da da da da da till a n will be One minus p a1 compliment p a2 compliment The da da da da da p a n compliment Okay Not a very important rule, but yes It is a byproduct of whatever you have learned in the first property So you can keep this in mind in case it is required now. We'll take a few questions Somebody go up slightly Go up slightly Yeah, thank you sirs for one minute. This is all I've written on the page here for the reason a and b are independent now many people when they're asked to Check whether two events are independent. They try to check it from you can say non-mathematical point of view Oh, this is affecting this is not affecting it. So don't go into that apply mathematics Sometimes it may seem that the events are independent, but they may be dependent Okay, so use the property p a union the p a intersection these p a into p b to verify it Don't go like intuitively. It seems sir. It is independent or intuitively. It seems dependent. Don't do that Oh, I don't remember the examples exactly, but I've seen See many a times p a and p b would be values will be such that They'll speak volumes about it Okay, sir When you find p a intersection b You may realize that it is not equal to p a into p will take questions. In fact, we'll have questions based on that Don't worry about it. Okay, sir questions So Let's see. I'll start with a simpler one Yeah, let's start with this question Let a and b be two events P is point four pb is p P a union b is point seven then find the value of p For which a and b will be independent a very simple question I'll put the poll on very good All of you voted for the same option very good Okay, I'll close the poll in 15 seconds four three two one go Very easy question Everybody should have got it right most of you have voted for see see for Chennai. Let's check What is the answer for this? So p a union b is point seven. This would be equal to p a which is point four. Let me write it down directly Point four pb is p minus p a into pb because p a intersection b will be p into pb because the events are given to be independent Okay, don't assume p a intersection b to be always p into pb It is only when they have mentioned that they are independent you can write it like that. Okay So this clearly gives you point three is equal to point six p. So p has to be half option number C is correct Is it fine? Okay, now let's take a Question here a bag contains four tickets a bag contains contains Four tickets. Okay. These tickets have numbers One one two one two one one two two two One ticket is drawn at random from the back one ticket One ticket is drawn from the back Okay let II Equal to one two three denote the event denote the event that the I it Digit on the ticket on the on the ticket is two then Which of the following options are correct? So multiple options may be correct. Even and e2 are independent okay option Be 2 and e3 are independent Option C e3 and even are independent an option D Even e2 e3 all of them are independent Okay, our multiple options may be correct So please put your response on the chat box. I'm not running any poll for this Now Venkat reading at this will you be able to comment which events are independent which are not no, sir It is very difficult to predict in that way Many people they just read the question and they try to reason it out that it doesn't look like independent It doesn't look like dependent. So you don't do that mistake Let's go from a mathematical and logical point of view rather than from our intuitions Got it. Yeah, very easy question. I mean you can you have multiple option correct Aditya Shisti No Amala you will get part marks But Shisti you will get negative marks because you have mark an option which is not correct I mean subject to solving it. I'm commenting from the options. I see in the answer answer key So all of you know how to join the Crash course session, right? So I don't have to send any kind of instruction. All of you are already exceeding almost three four classes Very good Pratham correct Aditya correct Rippan. Okay guys, let's discuss this Yes, you see the answer here that the book says is ABC only now. Let us check So first of all, let us write down even event in how many ways can you get the first number as a two? So only one case is there. Sorry two cases are there two one one and two two two Correct. So the probability of E one will be two by four or you can say half E two will be what E two is second number is two. So I think it is one to one only and two to two only this is also This is also two by four one by two Even and E2 what is even any two means both the first digit and the second digit should be should be two Okay, so that is only happening in the two to two case So only one event will be there under this set. Okay, so this probability will be one by four because they are four Numbers one by four. So can we check now is This condition getting satisfied P even intersection E2 is P even into P2 Yes, very much because one fourth is equal to half into half. So option number a is definitely correct Okay, similarly, let us figure out E2 and E3 E3 let me write it down because not make take much time E3 will only contain I think two two two and one one two So that's also one by two only P3 is also one by two Okay, so I think this will also be correct Because E2 intersection E3 will only have two to two Correct E3 and even they will only have again two to two. So again, this is also correct But are all of them independent of each other that is slightly doubtful because P even intersection E2 Intersection E3 is one fourth, which is not matching with half into half into half So this option is not correct. ABC only are correct Okay Can I move on to the next question? Sorry if I'm going a little fast because Today's our last class and you have to cover a lot of ground la Okay, there are some questions which are based on your Understanding of okay, let's take this one Let's see Venkat if if it is required a License plate a license plate is three capital letters followed by three digits If all possible license plates are equally likely the probability that a plate has either a letter palindrome or a Digit palindrome or both now when I say digit You can have zero also coming in the first phase don't worry about it So we can have a number of a car as C80 025 Okay So in number plates, we can have the first number as zero also doesn't matter Because it's just symbolic. It is just symbolic Letter palindrome palindrome. Everybody knows what's the palindrome palindrome means If you read it from left to right or right to left, they will be read as the same thing Let's have the pole on good. Good. Good. Let's close this in another 30 seconds Okay, five four three two one go Not many people have voted but Yeah, most of you have gone with a and b options. So and we have got equal number of votes Okay, let's check See first let us define the events let a be the number of cases where you get a letter Palindrome, okay, and b is the event where you end up getting the digit palindrome Digit palindrome. Okay. So what do you have to find out? We have to find out PA union be right Okay. Now one thing I can observe here is that the two events are independent of each other Correct. So can I not write this as this basically I am using my Law of or you can say complementation rule complementation rule Okay, now PA So in order to find PA, we have to find NA by NS. Let us find NS first So how many how many number plates can be made? First of all, that is the question. How many number plates can be made? 26 cube into 10 cube 26 cube into 10 cube See when you're trying to find out the The total number of ways in which you can make a letter palindrome, right? So why are we counting how many ways we can you know put numbers? Won't it be just 26 cube? Yes, yeah So what are the probabilities that when you make a 26 cube combination with these three letters you will end up getting a Letter palindrome, so NA would be nothing but see In letter palindrome, whatever you choose here that will automatically come in the last place So you don't have any option for the last place and this can also be chosen in 26 ways So letter palindrome the number of ways to choose is 26 square. So PA would be 1 by 26 Okay Now same for PB also will do NB by NS You should call it as SNS dash probably or you can say Yeah S dash so in this case S dash and S dash would be 10 cube because 10 numbers can come in any of the three places and out of which I want Palindrome to come so again palindrome can come in 10 way one way 10 way That means 100 ways or 10 to the power 2 so PB would be nothing but 1 by 10 Okay, so now the answer that I would get is PA intersection B will be 1 minus PA complement which is 25 by 26 into P B complement 9 by 10 Okay, a lot of things can be simplified. I guess 2 and this is 5 So 1 minus 45 by 52 will give you 7 by 52 which is option number Option number 8. Okay I understood 3 but if you're doing that, that means you are taking into consideration two cases I mean, you're taking care of the entire six digits simultaneously Yes, I'll be the same only Because the extra two zero three zeros will get cancelled because for the letter palindrome you can have 10 cube 10 to the power 3 combination and same with the digit palindrome Sir, even both are possible right. We're considering that also. Yeah. Yes, it does both are also considered another same thing. Don't worry. Okay Let's move on to the next question Let's take this one. If all the mappings that can be defined from set A to set B a mapping is selected randomly What is the chance the word here is chance not change Chance What are the chance that the selected mapping is strictly monotonic? Option is on not exactly to do with whatever we have done. It is just a testing of your normal probability skills Pole is on so please respond on the pole last 30 seconds five four Three Two One go Okay, most of you have gone with option number B B for Bangalore See is first of all when a function is monotonic It could be strictly increasing or it could be strictly decreasing Okay. Now if you have to make a function, which is strictly increasing Remember you have to have first of all one one mapping. Okay, so Let's say one maps to something two maps to something three maps to something and four maps to something in other words, you have to pick up four such numbers from here and There's only one way in which you can arrange them in an ascending order because one will map to the least Two will map to the slightly greater than that three will map to slightly greater than that and four will map slightly greater than that So what is what is trying to say that f1 should be Lessor than f2 should be lesser than f3 should be lesser than f4 in order to have a strictly increasing function Correct. This is for your strictly increasing function Right in other words, whatever number you choose whatever four numbers you choose There's only one way to arrange them in a ascending order. Let's say if you choose six seven eight nine So it's only one way six can go to one or six can map to one seven can map to two It can map to three and nine can map to four if you choose five seven eight nine five will map to one Seven will map to two eight will map to three nine will map to four So out of four five numbers you can choose any four of them in five C four way Can I say same will be true if you're looking at strictly decreasing function? So whatever four numbers you choose and map to one two three four They can be arranged only in one way in descending order. So that is also five C four So total number of ways is five C four plus five C four Okay, and what is the sample space sample space is the total number of functions you can form total number of functions that you can form Between set A and set B is your number of elements in the core domain To the power of number of elements in the domain. We had done this concept in our functions discussion. Correct. Co domain to the power of domain so five to the power four. So this will be your answer So answer is ten by five to the power four five to the power four. I can write like this So this is two by five cube five cube is 125 option numbers B is correct. Got it Clear, okay Let's take more question. These are all warm-up question guys Easy easy. This is what I hate searching for the right question So This you have done Okay, let's take this one This is touch and go guys touch and go I think if I put the poll you should be giving me the answer within 20 seconds the probability of Happening of three independent events are P1 P2 P3 find the probability of happening at least one of them Guys, let's close this in another 15 seconds five four Three Two One Go after all I mean most of you have given the vote to the right option But most of you have made it made mistakes also see guys. This is law of complementation or Complimentation rule what did I discuss with you if ABC etc are independent events We can use complementation rule Okay, that's one shot done So this is one minus. What is PA compliment if PA is P1 PA compliment would be one minus P1 one minus P2 one minus P3 Which option is correct option D is correct. Is it clear any any any problem in understanding this? I Don't know why you people made mistakes. Okay next question AI triple E 2005 question Let me put the poll on Good So they're they're trying to test you on two fronts mutually exclusive and independent And of course equally likely as well. Okay, let's close this in another 20 seconds five four Three two one Go, okay, most of you have voted with option number B So you guys first let us understand the difference between these three first of all equally likely when you say two events are equally likely It means PA should be equal to PB Okay, when you say they are mutually exclusive it means PA intersection be must be zero and Of course, when you know it is independent You know PA intersection B is PA into PB now. Let's try to figure out from this information. What is PA and PB first of all Since PA compliment is one fourth It means PA is three fourth. Okay second thing is PA union B will be 56th, okay, and We have been given that PA is three fourth PB I don't know Minus PA intersection B is one fourth. That means five sixth is half Plus PB. So PB will be equal to Five sixth minus this that is nothing but five six minus three six, which is our two third Which is sorry two sixth, which is one third Okay, now is PA equal to PB. I do not think so So they are not equally likely Not equally likely Second thing is PA intersection B is not zero. So they cannot be mutually exclusive Correct. So cannot be mutually exclusive Not equal to zero. That means not mutually exclusive Now are the independent let's check so PA intersection B is five by six is it equal to three fourth Which is PA into Into Into one third. No, I don't think so. Have I calculated? Yeah, so more than one options will be correct in this case. I guess How can they ask in a no only one option can be correct because it is asked in the AI Tripoli paper Oh PA intersection B is given to us one fourth. Sorry not five sixth. I Was wondering none of the options will be correct. Yeah into one third. Yes, it is equal So you they are independent They are independent events So let's check which option is correct. So mutually exclusive and independent that cannot be correct Independent but not equally likely bees correct equally likely but not independent incorrect equally likely and which is incorrect So option number B only can be correct. So B is the right option Okay Let's have one more question and then we'll go to conditional probability The question is one second. So let's say a and B Are two people who are participating independently in a shooting competition. So they shoot independently So a and B Shoot independently Until each hits his target until Each hits his Target Okay The probability of a hitting the target the probability of a hitting the target in At each shoot is three by five Okay, probability of B hitting the target at each shoot is five by seven at each shoot Not the whole event at each shoot So one shot that they take The probability that a will hit is three by five and the probability that B will hit is five by seven Okay, then the probability that B will require more shots than a is Then the probability That B will require we will require more shots than a is Option a five by twenty one Option B six by thirty one Option C option C seven by forty one option D nota So I hope the question is clear. So a and B both are aiming at their respective targets Okay What are the pro and the game will continue until each hits the target So both of them have to hit the target. It's not that who hits first It is both of them have to hit that at their respective target So what are the probability that B will require more shots to hit than a what is on? Almost one minute gone. I'll give you one and a half minutes more think about it Then answer no need to be in a hurry last 30 seconds five four three two one go now Just 11 of you have voted but out of 11 seven of you have voted for option B. Okay, let's check See guys, this is actually two summations undergoing in this particular process. So let's say a takes a takes our attempts to hit the target our Attempts to hit the target finally Okay, that means B has to take more than our attempts to hit it Okay, so here can I say if a has taken our attempts means He will first fail for our minus one and Then he will hit it in the last of his our attempt. So our minus one attempts he'll fail Remember the moment he succeeds you'll have he has to stop shooting Okay, so he would have failed R minus one time Then the Rth time he would have succeeded. So PA compliment to the power R minus one into PA and remember B has to fail more than that. That means B comp P compliment R into P B Okay, but when I say more it need not be Just one more than what a takes it may be two more three more four more five more and so so on So can I say one thing? For each of this are attempts at a takes there are following possibilities that we can take He can fail for our succeed in the R plus one it he can fail for our plus one it succeed in the R Plus to it and this can go on and on for infinity this will go on and on for infinity and This whole thing we have to sum up for R equal to 1 to infinity because you never know how many attempts will a take Right So there is a infinite summation going on and whatever you get there is an infinite summation going on on there also Are you getting my point? So let us put the values here So P not taking are not hitting the target. The chance is 2 by 5. So it's 2 by 5 to the power of r minus 1 Okay, this is going to be 3 by 5 Now here we have a geometric progression whose first term is this Whose first term is this and the common ratio is P B complement Okay, so first term is this means we can write it as P B complement to the power of r So B complement how much will it be B complement will be a 2 by 7 to the power r into 5 by 7 by 1 minus P B comp P B complement P B complement 1 minus P B complement is P B only so you can write 5 by 7 here also Okay, so thankfully this will get cancelled off Okay, and you're left with this summation of summation of 2 by 5 to the power of R minus 1 PA you can put it outside and P B complement That is 2 by 7 to the power of r Okay, and R equal to 1 to infinity Again, this can be very easily we dealt with you just bring that 2 by 5 to the power minus 1 outside So minus 1 means recipe procl of that so it will become 3 by 2 and then you have summation of 2 by 5 Into 2 by 7 to the power of R going from 1 to R going from 1 to infinity Okay, so this is going to give you 3 by 2. This is an infinite GP whose first term will be 4 by 35 Okay, yeah, and common ratio is also 4 by 35 correct Anything I missed out do let me know So it's 3 by 2 into 4 by 35 and this will be 35 will go up and you'll have 31 down 6 by 31 So your answer will be your option number option number B Well done. This was not a you know simple question at least to solve But well done those who solved it 11 of you seven of you got it, right? Good Okay Yes, sir I didn't understand that part which part This one see for every R attempt at 8x we can take R plus 1 R plus 2 R plus 3 R plus 4 Dada Dada Dada Dada infinity. Yes, I got the R itself is changing What if R is 1 B will take 2 3 4 5 6 7 R is 2 B will take 3 4 5 6 7 till infinity R is 3 So like that there's a summation here and for everything there is a summation going on. Yes, okay So this is a GP. I use GP formula infinite GP formula and then whole thing is a GP. Oh Got it. Oh, yes And sir another doubt. Yes, sir. Tell me What if just could we like take some two three minutes more and think for what will happen if three shots are required and all Three shots are required means like a and B have to shoot for three times each and then Find the number of ways to find the probability that B requires more shots than a So let's say a shoots in the first go B will require two or three shots If a takes two shots B will require three shots And if it takes three shots then there is no chance because he himself has taken maximum number of shots No, no, no, sir I meant each of them have infinite shots and until both of them hit the target thrice like a hits the target thrice And it need not be narrow and then he stopped and then B hits the attempt to hit the target like once then miss miss miss Then twice and then miss hundred times and then thrice like that Okay, so you are trying to say that a will hit the target three times and The total number of attempt that he has taken to complete that three attempts of hitting is more than that of B Right is less than is less than that of B, right? Anything yes, sir. Okay. So in this case what you have to do is in this case you're last You have to first figure out it's you have to apply binomial That is the first two shots that a has successfully taken can occur in any of the are Shots that he has already taken before he finally shoots it So R-1 C2 you have to you know make it into the system and multiply it with PA square That will undergo binomial process now See who can I explain it to you? Let's say our attempts are taken by a to hit the target, right? That means in the arid attempts must have taken his third shot And in our minus one he should have taken the two other shots Among our minus one right to all right So what will happen the first guy over here will change to our minus one C2 PA square and Again PA because in the RS shot. He's again hitting it. Oh Are you getting it and he's failing in the remaining so you'll have to do our minus Sorry, how many attempts he's failing? He's failing in our minus three attempts, right? So this figure will come in the first part here where I've underlined it here this part Yes, sir. What the point now? Yes, sir. Accordingly you have to change your be also Oh, okay, sir. Right. Yes, we have to make him take our first one attempt right the figure Our first two attempt right the figure and then apply a summation to infinity So two summations will go Now with this we are going to talk about conditional probability Conditional probability that is the main thing we are going to study in this chapter conditional probability Okay Now many people have a very wrong understanding about conditional probability Conditional probability just a second. Yeah conditional probability. I'll just give you a you know first some examples to connect to see, let's say I'll take a very funny example if let's say two people are running the race let's say Aditya Aditya Manjunath And Mr. X okay, they're running a race hundred meters Right What is the chance that Aditya will win the race? Equal now is saying zero. He'll not win the race. Sir. He's fat, sir. He'll not Yeah, Aditya will kill me. Yeah, half-half, right? Yes, I know Okay, so there's half chance equally likely. I know some Aditya might have Superpowers Wait, wait, wait, as of now when you don't know anything about it There are equal chances of them winning it right both are competent as right 50-50 chance Sir, but then still you have other factors. I have not mentioned other factors right now now. I will mention. Oh Now Aditya is running with Mr. X which happens to be Mr. Usain Bolt What is the chance of Aditya winning the race? You'll say zero Definitely Usain Bolt will win because he is going to be the fastest person on the earth, right? So now given the information about X You change your answer, isn't it? This is what we are actually going to study under this concept When you know a condition in that event Your answer is actually going to fluctuate Okay, so an additional information is going to change your probability answer Let me give you another example, which is not as funny as this. Let's say Let's say I told you there is a family which has got two children Okay, child one child two elder and the older sorry elder and the younger What is the probability that both the children are girls? What is the probability that both the children are girls? What will you say? Both are girls What will you say one fourth? Correct because there are four possible gender distribution you can do boy boy boy girl girl boy girl girl So out of that both are girls will be one and total number of cases were four So one by four correct But if I ask you the same question in this way find the probability that both are girls Given I know The younger one is a girl The younger child is a girl Right now your probability will increase Because you know that there is an additional information discharged at you or given to you that the younger Child is known to be a girl So now your answer will become half because this child is a girl so they can be boy girl or girl girl So only two possibilities out of which one is the possibility that both are girls So your answer has increased Because of this information given this information your answer has changed. It is no longer one-fourth Are you getting my point? So given that person X is Hussein world the chances of Aditya winning becomes negligible unless until he Has some super power on that particular day. Okay, so this is what we are going to study So I'm just explaining this in a very very you can say layman terms, right? So let's try to discuss this in more formal sense So when you have a compound event occurring, right? When you have a compound event occurring When these two are not independent And we are not independent what happened to my spelling are not independent Okay, so when we write the expression for this We can either have a occurring and then be occurring given a has occurred or we can have be occurring and Then a occurring given be has occurred now this term that you see is no longer PB unless until they are independent This is no longer PA unless until they are independent because it depends upon it Basically, this is the given condition to you that a has occurred before it So let's go back to our the bag example, which I gave you So there are two red balls three blue balls Okay, you are drawing two balls out of which out of this without replacement Without replacement if I ask you what is the probability that the second ball is a blue given the first one is a red Then this answer will be no longer When you were trying to find out the probability of getting the second ball red Probability of getting a sorry is getting the second ball blue if it was done with replacement Correct. So here your answer will be as you told three by five, but here the answer will be three by four because it will be now changing So the first information given to you that the first ball was a red basically changed your mind or your influence your Probability of getting a blue ball in the second trial Okay, so this will be basically what we will be studying under conditional probability So there will be something given to you about that event which will change your answer Are you getting my point? So this thing that I have written in front of you in the very first line. This is what we call as multiplication theorem so through multiplication theorem, I'm going to derive the expression for P B given a has occurred or P given a P a given B has occurred by the way this is pronounced as as as Probability of occurrence of B occurrence of B Given an information is given to you that a has occurred Given that a has occurred right Now many people ask me sir. Is it P B by P a It will be P B by P a but provided B is a subset of a B is an event which is subset of a Else not Else not so please do not treat it like this every time unless B is a subset of a I'm not denying that it cannot be P B by P a It can be if B is a subset of a Okay, because if you look at this formula It tells you that P B by a would be Let's let's divide P a to the other side of course P a is not zero in this case So if you divide by P a P a is not an impossible event So it is not equal to zero You can see that this is the actual formula And if B happens to be a subset of a then P a intersection B will be P B only Isn't it if B happens to be a proper subset or subset of a then P a intersection B will be P B only Isn't it then in that case it can become equal to this the first formula over here Else not Are you getting my point So this is the formula for the probability of occurrence of B given a has occurred Okay, similarly, you can write probability of occurrence of a given B has occurred is P a Intersection B by P B. Of course P B should not be zero Okay, so we have written this in light of Multiplication theorem. So what is multiplication theorem? I will write this theorem in general So multiplication theorem states that If they are if there is a compound event occurring that means all these events a 1 a 2 a 3 etc occurring together You can write it as P a 1 into P a 2 given a 1 has occurred into P a 3 given a 1 and a 2 have occurred into P a 4 by a 1 intersection a 2 intersection a 3 has occurred da da da da Till you reach P a n Given a 1 and a 2 and a n minus 1 has occurred Okay, this is the more general version of The multiplication theorem what I have taken is a very small case where there are only two events in that compound event Is this clear? What is conditional probability? First of all, that is clear How multiplication theorem is helping you to get the expression for conditional probability that is also clear and Under what case can I write it like this because most of the people have a habit of writing it like this So under what case we can write when you know your B is a subset of it Is that fine and Under what condition will P b by a will be P b when a and b are independent Under what condition P a by b or P a given b has occurred is equal to P a when a and b are independent Okay, so what could you repeat that? See what I'm trying to say is P b given a has occurred is P b only when you're A and B are independent Okay, and of course P a should not be zero Okay, and same thing with P a by b also is equal to P a when a and b are independent Is this fine Sir, yes sir. Yes, sir. Sir. Can you repeat that multiplication theorem? Yes multiplication theorem says that the probability of occurrence of this compound event a 1 intersection a 2 intersection Is P a 1 now again, you can start with anything you want You can start with a 2 also and then whatever you choose you have to write probability of occurrence of that given a 2 has occurred da-da-da-da-da like that See, it's just coming from our common sense It is just coming from our common sense See if I say what is the probability that? You are drawing let's say there are five white balls three green balls What are the probability that? Just a second guys Yeah, sorry So now let's say if you're drawing let's say three balls out of which one by one without replacement So three balls are drawn one by one without replacement Okay, what are the probability that the first ball is a white second one is a black and the third one again is a black Right, so what will you do first ball white you will draw three five by eight Now second ball blue black you will draw given that the first one is a white So now your white ball has become lesser isn't it so it'll become why am I writing green over here? Sorry, three black Yeah, so this black will be three by four Because you know the first draw is a white and again this black would be given that you have already drawn a white and a black So you will be left with one less white one less black. So it's four and two over here. So this will be two by I think I made a mistake three by seven. Yeah, it'll be two by six Are you getting my point? Okay, yes, the simple thing is basically written and a Complicated language like this. See guys everything comes from our common understanding common sense understanding We just complicated through the use of mathematical formula. Let's take questions That's a sir So what did you write about that independent thingy the first? Where first first A and B are not independent sir they can be independent So then it'll just become zero and stuff and then you'll divide and become zero We can say A and B may not be independent. That is fine Oh, I got it sir. Thank you, sir Because then it'll become a special case, right? Yes, sir So you can also say see what is the probability of you getting a six on a die given that you got a Head on the coin which was tossed along with it In this case, even if you write probability of getting a six given the head as a card, it is just probability of getting a six Yeah, oh, yes, sir. Yes, sir Okay, let's take this two dies are thrown What is the probability that the sum of the numbers appearing on the two dies is 11? Given five appears on the first try so read this if has given So you have been given an additional information about this experiment Please put your response on the chat box Okay, now my job is also to tell you how to present your answer in your school and semester exams or board exams Don't solve any question in probability without defining an event It's a very wrong practice. I have seen among all the students. They just write a formula They just write down the values While it will not hurt you in your competitive exams, but it will hurt you badly in your board exams So you'll always should start with let a be the event that the sum of the numbers Some of the numbers appearing Some of the numbers is 11 Okay, b is the event that five appears in the first day five appears in the first day in the Or on the first day Okay, so what are we finding here? We are finding the probability of a given b has occurred Okay, now as per As per our multiplication theorem, we can write this as pa intersection b by p b correct Yes, I know Now what is the chance that both of them occur that you get a sum of 11 also? And you get a five on the first die also that is only one way, right? When your Yeah, five and six happens So this is one upon 36 Total number of cases Is 36 so one upon 36 cases will be you get a five on the first and the sum is 11 And how many ways can you get us five on the first? So if you fix this to five, then this can change in six ways So this will be six by 36. So your answer will be one by six Now through this example, I would like to highlight one more thing For both these the sample space is the same. So when people write it They normally ask me sir, can we just write it like this? Yes, you can write it like this because ns ns will get cancelled from both numerator and denominator So you can directly say one event will be where five and six Uh, sorry, uh, first number is five and some is 11 and six such cases where the first number is five So one by six is the answer Now let me ask you an additional question. If I change the question to what is the probability that The first is a five given that the sum is 11 What will be the answer? Now I flipped it. What is the probability that The first is five Given that the sum is 11. Come on guys. Very easy How same Because the other should be six I mean I didn't apply formula What is the probability that your first number is five given that you get a sum of 11 Sum of 11 may be also obtained if you don't have the first number as five Oh, oh that matters Correct. Ravi Kiran has got the answer Yeah, see Numerator will not change it will be b intersection a or a intersection b whatever you call it because it is competitive operation by n a Now how many ways can you get the first number as five and the sum as 11? That is only one way And how many ways can you get? The sum of 11 itself Two ways only five six six five I don't know how any third way to get it. So the answer is one by two. Are you getting my point? So see how knowing one thing changes the other Got the point. Is it clear? Does it make sense? Sir, but If you tell that It's given that the first number is in the first example Five is on the first time. Yeah, that means b of b should be one only, you know, sir Because it's pukka five only then pukka means one only No, you can make that as your sample space For example, you can make the events where a is pukka occurring as your sample space. So five six Six five will your sample space out of which one event is where your a is also occurring b is also occurring So that will come as a first event But you can't say it is a pukka event because the event itself is a bigger event But you're telling it's given I never asked you what is the probability of you getting a sum of 11 given that you have got a sum of 11 Have I asked you that? No, no, sir. I was talking about p of b only see I said b is has pukka happened But have I asked you the probability of b? Oh, no, no, no, okay, but you're asking in denominator Uh, how do I explain this in a simpler language? Let's say you have tossed a coin three times Okay, sir What is the probability that the last Uh, uh Thing that the coin shows is a tail Given that the first one was a head This is this is the type of questions we are trying to address Because even if it is head this can change its place, right? Yes, sir. Yes, sir. Are you getting my point? Okay So I'm asking you the similar type of question over here given that your sum is 11 What is the probability that you are getting a five on the first? So even if you know your sum is 11 There are two cases in which you can get 11 and out of it one of them is where you get a five So it's half half the chance No sir, I was talking about It helps you to reduce it helps you to reduce your sample size But it doesn't make it the entire thing as one Oh, okay, in the this one I understood p of b by a I understood I was talking about p of a by b sir Ah, what is the question there? There only given that five is already on the first day Then then why are we telling p of b is six by 36? That yellow six by 36 there. Yes, sir Around Ah that one that one Again, this is a way of solving it. You can actually consider that this to be a sample space also That is what this formula says this becomes your sample space And then you can treat in how many of those sample space have both occurred that is also a way to solve it This is coming from the formula perspective Oh, okay, sir. Oh, okay Getting so if this is like telling p of b occurs in chakthi two days scenario Just take one day and then see how much yes What you are trying to say is that you are making the Cases where b occurs to be your sample space because you're saying only that is accurate Oh, okay. That is basically taken care by the second formula Oh, got it. Okay. Nice Got it. Got it. All right So let us take first, uh Few properties of conditional probability few properties of conditional probability Then we'll take more questions the first property is Probability of occurrence of a given the sample space has occurred is actually pa Right in this case, you would realize that since a is a subset of s you your formula will work But anyways, I'll do it in a proper way So pa by s is pa intersection s by ps Pa intersection s will be pa So this is the one which I was talking about to you the other time when you know a b is a subset of a Then you can write it like this Okay, else not else. This is what you should always write If a comes to be a subset of s then only it'll become pa And this is nothing but pa only because ps is anyways one Not a very useful property, but note down just to maintain the Exhaustiveness of the situation But ps by a will be actually one right Why because ps by a Or ps given a has occurred is ps intersection a by pa And ps intersection a is pa only so pa by pa will cancel each other out and you'll get a one Okay, s is the sample space Let's take few more useful ones the probability of occurrence of a union b given c has occurred Is given by a probability of occurrence of a given c has occurred plus probability of occurrence of b given c has occurred Minus probability of occurrence of a and b given c has occurred How do we prove it? How do we prove it? Let's look into this So if you start with the left hand side probability of occurrence of a union b given c has occurred as per our multiplication theorem is probability of occurrence of c intersection a union b by pc correct Now again, we can use our distribution of intersection over union correct Now treat this as x treat this as y And you have written like p x union y. We know our addition theorems. Let us apply that So which is p x plus py minus p c intersection a intersection c intersection b upon pc correct So this is p c intersection a plus p c intersection b. Now if you look at this term very carefully You can write it actually like this Okay, and if you individually divide so let me individually divide each one by pc What does this term represent? What does the first term represent the first term represents p a given c has occurred What does second term represent it represents p b given c has occurred And what does the third term represent the third term represents p a intersection b given c has occurred Hence proved. This is what you wanted to prove Is that fine any questions any concerns? No doubt if you want to okay The last property, uh, the fourth property. Let's take it over here. Fourth property is very, uh useful P a compliment given b has occurred is just one minus p a given b has occurred Now what is the proof for this Again, very simple if you start with your left hand side, which is your probability of a compliment given b has occurred by the very definition Or by the multiplication theorem. I can say this Okay, and this is nothing but Now if you remember your, uh addition theorems What is a compliment in section b? This is the zone right a compliment in section z. Can I say it is actually p b minus p a intersection b? Correct. So the numerator part will be this So if you divide by the denominator individually it becomes this And the second term is clearly p a given b has occurred Which is your right hand side Can be in the common okay now very good question. So now, uh Aditya is asking what will what will happen to the formula if we are finding p a given b compliment So in this case, what do you have to do aditya? Follow the definition everything is coming from the definition Okay, now this is nothing but as per the addition theorem. This is p a minus b, which is p a minus p a intersection b Okay, by p b Sorry b compliment. So b compliment is one minus. Let me write it in the second step So this is p a minus p a intersection b by one minus p b. This will be your formula Got it aditya Yes, so thank you sir. Yeah So only when you're uh The top event is complimented then only the previous formula can be used else. You have to use this formula Okay, let's take a question on this Very simple question just to test your basics poll is on done Let's close this in another 10 seconds five four three two one go There you go. Most of you have got it option number a let's check. Let's check option number a is correct or not See what has been given to us p a has been given to us as one fourth P b has been given to us as one sixth. Sorry one tenth And p a intersection b has been given to us as one tenth Oh my bad when one six was correct My mistake. Yeah one six So as per our multiplication theorem, this is p a intersection b or p b intersection a whatever you want to write it So this is going to be one sixth upon p a p a is one fourth Is that fine? Oh, sorry one tenth upon one fourth one tenth upon one That's nothing but two by five option number a Is that fine any questions? next Given p a is point three p b is point four p a complement intersection b complement is point five find p b given a union b complement Zero. Okay, let's check Okay Okay, now see everybody please pay attention. So p b given a union b complement Can I say it is p b intersection a union b complement By p a union b complement correct As per our distribution formula, this will be p a intersection b intersection a union b intersection b complement By this now Now see here This is a null event Okay, so this will be nothing but p a intersection b Now let us figure out these these values so What is p? A intersection b we can get it from here. This is nothing but p a union b complement Which is nothing but one minus p a union b p a union b is p a p b minus p a intersection b Yes, I know So point five is one minus p a p a is point three minus point four Plus this factor which is not known to us, which we'll figure out from here Yeah, so this will give you uh One point two which is point two so point two is it will be your p a intersection b So the numerator will be point two What about denominator? We'll figure it out. Let's figure it out p a union b complement Can I write it as p a Plus p b complement minus p a intersection b complement Okay, this is nothing but p a this is nothing but one minus p b And this term is nothing but as per our addition theorem. It is p a minus p a intersection b If you open it up if you open this up p a p a will get cancelled and you'll get up one minus p b plus p a intersection b Which is nothing but one minus point four plus point two Which is point eight So denominator is point eight. So your answer has to be one upon four Has anybody got one upon four Oh trippan got aiyush got very good. Well done guys well done good Is it fine slightly lengthy problem? But everything was well within our uh understanding Which which part you want me to do by when diagram? So the a union b complement I mean that's effectively doing the entire thing See a Union b complement b complement is the outside part Yes, sir Okay Union the a union that with a means outside Yes Union with a means this part will also come into picture. No Oh, yes That's my bad Next I think we have Let's take this one Easy question. I'm putting the pole on Very good. Let's stop this in another 10 seconds five four three two one go All right, I'm seeing c option coming from most of you those who voted. Let's check Yeah, so this is nothing but p a complement intersection b complement by pb complement And as per the de Morgan's law, this is p union b complement And this is one minus p a union b upon pb complement Option number c is correct. Is it fine? Let's take one word problem also Three coins are tossed Three coins are tossed are tossed If one of them shows tail If one of them shows tail shows tail Find the probability that all three coins show tail Find the probability that All three coins Show tail Simple very simple question I'm not able to take a lot of questions on this because we have to wait till we do law of total probability And in base theorem again, we are going to deal with conditional probability because base theorem is actually a case of conditional probability So read this as given one of them shows tail given one of them shows tail What are the probabilities that all show tail? I think it is at least one of them This should be written as at least one of them. Yeah So what are we trying to find out we are trying to find out this Okay, so as per our multiplication theorem, this is what we are going to write You can write it as n a intersection b by n a Right. So in how many ways can you get At least one of them showing a tail and all showing a tail see since all showing a tail is a subset of This this is as good as n a Sorry B is a subset of a sorry not a so all showing a tail is a subset of this will be n b by n a so in how many ways In how many ways can you get? All of them show a tail only one way Tail tail tail and in how many ways can you get at least one tail? Is just remove all the head scenario from eight cases. So answer will be one by seven. Are you getting my point? So we'll come back again to more questions based on this But before that we have to learn something very important which is called the law of total probability because Law of total probability is going to be one of the most important thing That we are going to use for such cases Okay, so it is not that i'm finishing off this condition probability condition probability will be again revisited In our base theorem base theorem is actually a condition probability But just to add more material to the concept we have to use Law we have to deal with law of total probability So law of total probability. What is this? law This is actually a common sense thing but written in a very very you can say complicated fashion like formula See i'll just ask you a simple question Okay There are three bags Bag one bag two bag three All of equal shapes and sizes Bag one bag two bag three This bag has got five green balls two red balls Three green balls five red balls One green ball two red ball Okay Now You are a person who is blindfolded Okay, so somebody has blindfolded you You are like Which was that character in Mahabharata Father of Duryodhan Gandhari Blindfolded as Gandhari blind or Oh So you are like Gandhari. So yeah, you should be female. Okay Right. So you go and you put your hand in one of the bags. You don't know which bag you are putting your hand And you get a ball Yeah, okay. What is the probability that? Venkat the Gandhari will get a green wall The way they made raman. Oh my god father is blind father cannot see mother knows it, but she is She has blinded herself against the, you know Faulty things of his of her son Very cleverly written story, you know Symbolizing the present scenario I If you want to enter into a one hour lesson with him then only He is very learned when it comes to all these things Yes, so how will you solve this question? You don't need a formula to solve this question. What will you say? Okay, there are three cases in which I can end up getting a green ball Okay, I said no one thing one one small thing. I would like to say why not the answer is Total number of green balls divided by total balls. Why not the answer is nine upon seven plus eight plus three Okay, first of all anybody thinks this is the answer One by two because there are nine green balls out of total 18 balls. Is this the answer? Yes, no, maybe Okay, this cannot be the answer because This is this particular thing is derived from the fact that There is an equally likely chance of a ball being green and a red Right, so there's an equal homogeneous mixture that you have prepared out of it But actually it is not it is basically Uh, you know, since in the language of physics, I can say the potential is different in all the surfaces over here Okay, so your chance of finding A charge when there is a sharp corner is more as compared to a blunt part of the object, isn't it? So when you divide it in this way, you are basically dividing The green balls into different potential zones. So potential of finding a green ball is high here Potential of finding a green ball is less over here. Are you getting my point? So they are not equally distributed Also has to say that there's not they are not homogeneously distributed Are you getting my point? So how will you solve such a question? So you'll say, okay, let my hand go in bag number one Now given your hand has gone into bag number one The chance itself That it'll go in bag number one is one third And now that your hand is in bag number one you coming out of a green ball you coming out with a green ball will be five by seven correct Or your hand goes in bag number two whose chance itself is one third and you And given that your hand is in bag number two the chance of you getting a green ball will be three by eight correct Or your hand goes in bag number three whose probability itself is one third and you Putting your hand in bag number three getting a green ball will be again one by three Okay So this is a total ways in which you can pick up a green ball And that's why the name of the theorem comes as law of total probability So this is the chance of you getting a green ball Yes or no Yes or no So this is from a very common sense way that you figured out what is the chance or what is the probability of getting a green ball Now the very same thing I'll complicate it with a lot of mathematical jargons So let us say Let us say i'm generalizing these things. Let's say e1 e2 etc are mutually exclusive mutually exclusive And collectively exhaustive events The word collectively is optional. You can drop it also or exhaustive events Okay What are the meaning of mutually exclusive and collectively exhaustive m e c? What are the meaning of that? It means that if you take any two of them their intersection will be a disjoint set or a null set Okay, so they will be disjoint sets and their Union will be your sample space Okay, when such a thing happens when both of these simultaneously are satisfied then we normally say p e1 union e2 etc till en Will actually result just into p e1 plus e2 plus e3 till p In other words, one will be equal to p e1 plus e2 till Yeah, so any events which are mutually exclusive and collectively exhaustive will always satisfy this condition. Please note that Okay, now let me complete the theorem. The theorem is not completed so let's say if even e2 en are mutually exclusive and collectively exhaustive events events and let's say there is an event a And there's an event a that occurs with occurs with these ei events Occurs with these ei events. Okay, then we say that Then we say that probability of occurrence of a would be nothing but probability of occurrence of e1 into probability of occurrence of a given even has occurred Plus probability of occurrence of e2 into probability of occurrence of a given e2 has occurred And so on till the last now How does this formula come about? Let us try to understand to a simple Bend diagram Let's say this is a sample space Okay, and let's say even e2 e3 are your certain events which are disjoint Like this So this is your even. This is your e2. This is your e3 This is your e4 da da da da da till your en Okay So they are disjoint, but yet they make the entire sample space And let's say there's an event a which occurs along with these all events So there's an intersection of a with these events. So this is your event a let me use your local Okay So what are the probability of occurrence of a If you look at this diagram, it will be some of this area this area this area And so on Till this area correct. So the first area will be P e1 intersection a Second area will be P e2 intersection a Correct, and it goes on till P en intersection a Now as per our multiplication theorem, we can write this as P e1 into P a given even has occurred Similarly, this is P e2 Into P a given e2 has occurred And da da da da till P n Into P a given en has occurred Yes, this will be your pa That is what formula I wrote over here. Is this fine? So if you compare this with our example, which I gave you then your even e2 were like picking a bag one back to bag three So if you compare with this example Okay, this example this will become P e1 Given that you picked bag number one Right and the probability of getting a green ball given you have picked a bag number one will be your five by seven Then this will be P e2 And this will be P getting a green ball given you have selected bag number two Similarly, this is P e3 and getting a green ball given that you have selected bag number three Are you getting my point? So what you wrote without knowing this formula is what this formula actually tells you So there's nothing above logic If you're solving it logically like this, you never need you do not need to remember any kind of formula like this Yes, this is just an eye wash which you have to use in your in your Board exams just to get marks Is this fine? So could you go down one second? Little left there one thing And thank you, sir And sir, yeah, I put something on chat for that and so can you read sir Regarding this question One G three R 39 G 117 R 59 G 177 R can we really go to the first case? Didn't understand anything Yes, sir. So I meant like We eliminated the probability of nine by 18 in the first Method because they weren't equally likely. So in this case we can do you know, sir. Everything is one by four one by four One by four Yeah In each bag in respect of the number of green balls It's equally likely that we pick a green ball. No, sir in a bag. Yes In a bag per se But the event is not just picking from one bag. It is picking up from any one of the three bags Yes, sir, but in this case Tensity is changing. No density is changing. Let's say if I say What is the probability of you finding a tamillion in karnataka? And what you finding tamillion in deli So here, you know that there'll be more chances of getting a tamillion because it is the neighboring state of karnataka So this chance that a tamillion can come in in karnataka, right? But that but deli is like very remote. It's a very less chance that a person will travel from, you know A hub of you know it hub of bangalore and karnataka to a deli to get a job But if you say what is the probability of you getting a tamillion from the entire country Then you'll basically do the total, you know, if if there was a homogeneous distribution Then you'll do total tamillion people divided by total population. Yes, sir But now your chance will change because your distribution of the people are different in different zones No sir, here chance won't change 39. Everything is one by three only. One by four, sorry You're saying that if there's one green and C red? One green three red No, no, no, no a bag one is one green three red back two is 39 117 I chose the wrong numbers very sorry And other bag is 59 and 117 is 7 It's anyway in each of the bag green and green is one Yeah, then it will be same then you can just add all red plus green and then add all green and Then it will be same. Yes. Okay. Got it, sir Next we'll take some questions on this Yeah, let me put the poll Very simple question guys, you should all get it Three groups A, B, C are contesting for position of the board of directors of a company the probability of their winning is 0.5 0.3 0.2 If a wins probability of introducing a new product is 0.7 B wins 0.6 C wins 0.5 respectively find the probability that the new product will be introduced Okay, most of you have gone with option C. Sorry, I'll not say which option We'll stop in 10 seconds So Five Four Three Two One Go All right, most of you have gone with actually option A knew it, sir I was trying to see how many of you marksy when I say that On purpose. Okay. First of all, you have to define it So let Even is a chance that group A wins the election E2 is a chance that group B wins the election And E3 is a chance that your group C wins the election and a is a chance that your product New product whatever Is launched What introduced So what is the chance that the new product will actually come? So probability of occurrence of a will be nothing but summation of probability of e i into p a given e i has occurred So a winning given a winning the product being launched 0.7 correct or B winning Given B winning the product is launched 0.6 C winning given C winning the project is launched is 0.5 So this is 0.35. This is 0.18 and this is 0.10 That's nothing but 0.63 option number a is correct Very easy question Is this fine any questions any concerns? next question Just listen to this question carefully. It's a lengthy question question is Suppose families always have one two or three children Okay, what I'm writing One two or three children Okay One child I should write not children Two children Three children Okay, so probability of the family having one child is one child is one child So what is the probability of the family having one child is one child is one child probability of the family having one child is one fourth probability having two children is half and probability having three children is again one fourth So there's more chance of having two children other than one and three Okay, suppose everyone eventually gets married and has children Okay Suppose everyone eventually marries and has children Find the probability that Find the probability that probability that A couple has A couple has Exactly four grandchildren Exactly four grandchildren Think carefully and then answer Question is understood, right? So the chance that a family has one children is one fourth two children is half three children is one fourth Okay So let's say there was a couple So I don't know how many children they had And those children had children So basically grandchildren So how many way, what is the probability that that couple will have exactly four grandchildren Good try Amla, but that is not correct Good try Ravi But not correct That's correct Patam, very good Anybody else who would like to contribute? Yes sir, one second No Trippan, no Shisti Okay, let's see this Let's say there's a couple Okay If this couple has one child If this couple has one child Then there is no way that he can actually give four Or they can not have four grandchildren Okay Correct If this family has Let's say I take case one If this family has two children Let's say this couple has two children Let's say Chintu, Mintu Then either Chintu has one, Mintu has three Or Chintu has three, Mintu has one Or both have two And then only this couple can have four grandchildren, right So within this we have these cases Okay Let me write it as small one, small two, small three Correct The third case is let's say this couple itself has three children Let's say Chintu, Mintu And Pintu Okay So what are the chance that they will have four children together So this has one, one, two Or various combinations of it So you can have one, two, one Or one, one, two, whatever Or Or, or, or, or Any other way? No One, one, one also cannot happen Two, two, zero cannot happen Okay That all combinations I will take care of Don't worry So these are the cases in which The couple will have four grandchildren So let's say if the couple had A number of ways in which they have four Grandchildren Four grandchildren Would be Either they have two children Mintu and Pintu whose chance Itself is half because Any parent having two children Or having set of parents Having two children, chance is half Okay And then Chintu has one, Mintu has three The chance will be Into one fourth And there are two such cases like this Where Chintu has three, Mintu has one Okay And if they have two, two, the chance is half Square So this is the probability that The couple will have four grandchildren Given that they have two children So this is like your P, E1 And this is like your P Let's say this event was A So this is A by E1 Okay Three children Chintu, Mintu and Pintu Then the chance that two of them have Two children Two of them have one child Okay And the other one has got See have I repeated anything I think I have repeated this again So two having One one is one Fourth Square and other having Two is half And there are three such cases Correct One fourth, one fourth, half One fourth, half, one fourth One, sorry, half, one fourth, one fourth So they have written this three times So this will be your total probability So let's check how much does it come out to be So this is half I think this will come out to be One One sixteenth Plus one eight So Oh sorry, I have written Let me just write it properly Yeah So this is Half of one eighth, one fourth And this is one 32 Okay So this is half of 12 by 32 Have I missed out Anything, anything, anything One 28 Wait, wait, wait This is one fourth Thank you 64 into 22 This also Okay So this will be Six, that is 24 by 128 plus 3 by 128 Answer is 27 by 128, this is the answer Is this fine Any questions All right, so there are enough questions we can take On law of total probability But I would like to introduce to you very quickly Base theorem because time is running out Base theorem Base theorem is also called reverse probability Because when you Read the question of a base theorem You would get a feeling that you are going In a backward direction To calculate something Okay, so let's go back to our Old question where we had those Bags and balls inside it So let me just give you the same scenario So there are three bags Of equal sizes Bag 1, bag 2, bag 3 Again 5 red, 6 green Some random numbers I am taking 3 red, 4 green 2 red, 1 green Okay Now again you are going And picking up a ball from this bag And again you are let's say Blindfolded Okay You pick up a ball, you found that ball is Green What is the probability it came from Bag number 2 So what I am trying to find out Given your ball is green What is the probability it came from Bag number 2 As you can see the look and feel Of the wording of the question will suggest That you know the final result And then you are going to check Which of the MECE event Was associated with that event Which of those MECE events Was associated with that event So this is what we call As the base theorem Now ultimately what is this This is a conditional probability only So you will say As per the conditional probability Formula or as per the Multiplication theorem formula This is going to be this Your numerator is just going to be Probability of picking bag 2 And probability of getting a green ball Given you have picked bag 2 And your denominator will be The total number of phase in which You are picking up the green ball So bi into pg Given bi has occurred I will go from 1 to 3 Isn't it So what is happening see in a nutshell You are using conditional Probability So base theorem is nothing but conditional probability Nothing else But what is happening in addition To conditional probability in one of the Unknowns that you are trying to find out Denominator you are using LTP Law of total probability And numerator is nothing but the use Of multiplication theorem Of course the whole thing is coming from Multiplication theorem only So as you can see this concept Incorporates all the sub topics That you have already studied so far Multiplication theorem, conditional probability LTP and of course we are studying it Under base theorem So when you are trying to Find out any such questions You can relate the law of total probability first For example if I have to solve this question I will first calculate In how many ways can I get a green ball So one third Six by eleven One third Four by seven One third one by three Correct So this will sit in my denominator And what will sit on the numerator Is where I have got my green ball From the bag number two This will sit on the numerator So your answer for this question will be One third into four by seven Upon Upon This whole expression that you have written That is one third six by eleven One third four by seven One third one third Are you getting my point I have just explained it from a simple example If you remember this example All base theorem question you will find it Very very easy Understood Okay Now there are certain terms that we normally use For this Your e1, e2 etc Which is basically let me just give you the theorem first And then I will name it So is this example making sense to you Does this example make sense to you Yes sir Yes So when you talk about things in general Again let's say You have You have e1, e2 etc As your Mutually exclusive and collectively Exhaustive events And if you are trying to find out Probability of occurrence of Okay let me just write a also And a occurs with With your eis Okay And if you know that a has occurred What are the probability that it came Because of some ek event Then base theorem says that P ek into P a by ek By Summation of Summation of P i into P a by e i i from 1 to n In short your denominator is coming from Your law of total probability Okay And whole thing is coming from conditional probability Are you getting my point By the way there are certain terms That we normally use P a, e1, e2 etc You can say P eis This is actually called as Priori Priori probabilities Nothing in the name just you should know Okay And P E i given a has occurred Is actually called Prosperity Probabilities Today my spelling is Probabilities Okay Just a name that is given to it So this is one of your Posteriori probability Okay And this probability of E1, E2 etc They are called Priori probabilities Now before we take a break I would definitely like to take a question on this Without a question Things will not be Let's take a question Question is I will just make a question over here Let's say There is a bag Bag 1 and there is another bag Bag 2 This bag contains 3 red And 5 green balls Okay And this bag contains 5 red And 3 green balls Fine Okay Two balls are transferred From bag number 1 to bag number 2 Without noticing their color Okay Two balls are transferred from bag number 1 To bag number 2 without noticing their color And again Two balls were drawn from bag number 2 Okay And they were found to be A red and a Green Okay What is the probability that The two balls transferred from Bag number 1 to bag number 2 Was both red Find the probability that The balls transferred from Bag number 1 To bag number 2 Were both red Were both red Given that You got a red and a green From bag number 2 After this transfer was done This is what I am asking you Sir here red green green Same thing over here sir Sorry Again when you are drawing two balls One of them is red and one of them is green That's it Anyway it doesn't matter Okay let's discuss this See here What are your MEC events Your MEC event could be You could transfer both red balls From bag number 1 to bag number 2 Correct E2 could be you can transfer One red and one green From bag number 1 to bag number 2 E3 could be you can transfer Both green balls from Bag number 1 to bag number 2 So these are your MEC events Correct Now MEC events it is very easy to identify them By the use of the fact that The sum of their probability should come out to be One always please remember this This is very important If your MEC events are such that They are coming out to be one That means something is wrong in your working And what is the event A Event A is you get both Red from bag 2 Correct Now what did I tell you When I was discussing base theorem So we are basically trying to find out Given A has occurred what is the probability That both balls Transferred were red in color Right this is what we are finding out So as per our base theorem This is Pe1 into PeA by E1 By summation of PeI into PeA by EI Okay But I normally do not proceed with a formula In mind this is just for your school exam What I do is plain and simple My approach is very plain and simple I first sit and find out in how many ways Can I get two red balls from bag 2 Right Indirectly I am trying to find the denominator First So I can get I can get a red and a green ball from bag 2 Oh my bad My event was not both red balls My event was I am so sorry My event was getting a red and a green from bag 2 Okay sorry So what do I do first I will first see how many ways can I get a red and a green from bag 2 So if two reds were transferred What is the probability first That two reds will be transferred So that will be 3C2 by 8C2 So two red out of three can be chosen in 3C2 Right and total 8 balls were there So 8 by 8C2 And now that Two reds are transferred into this The composition of red ball here will now become 5 Sorry 7 And 3 Now from 7 and 3 You are picking up one red and one green So 7 And 3 If you are picking one red and one green The probability for that will be 7C1 Into 3C1 by 10C2 Am I right Okay Or Let us say I had transferred one red And one green to the other bag So one red and one green I can transfer it in 3C1 5C1 Upon 8C2 And given that I have transferred One red and one green The composition will change where it will become 6 red And 4 green From 6 red and 4 green The chance of me getting one red and one green Will be 6C1 4C1 by 10C2 Okay Or I could end up transferring Two green balls from bag number 1 to bag number 2 Whose chance itself will be 5C2 Upon 8C2 And given that I have transferred two green balls The composition will become 5C5 each And then you are getting One green ball and one red ball Is 5C1 into 5C1 By 10C2 Okay This will become your The denominator term So this will become your denominator term So this guy is going to be nothing But your denominator term Out of this one is going to sit on the numerator Which of them is going to sit on the numerator Where you had transferred both red balls So the first guy Is going to sit on the numerator So the answer for the question that you are looking Out will be Will be 3C2 by 8C2 Into 7C1 3C1 by 10C2 Whole divided by Whole divided by this whole lot 3C2 8C2 7C1 3C1 10C2 Is there any questions Anybody based on this No I am not calculating it For the interest of time So please do so And you can proceed from here This is your answer Any questions Any questions Okay Now you can take a break You can take a break A letter is known to have come Either from Krishnagiri or Dharmapuri On the postmark Only the two consecutive letters R and I are visible What is the probability that It came from Krishnagiri Let me put the poll on for this Yeah I know many people will mark that option Where somebody has already marked Remember the distribution of RI is not uniform So it is not about seeing How many RIs are there And how many of them are present in Which one of them So it is slightly more complicated There is one similar question that says London and Clifton So they will say The letter is supposed to either come from London or Clifton And if ON is visible on the Two consecutive letters What is the chance that it came from London or Clifton This is just an Indian version of the same thing Another question was I think Tata Nagar and Kolkata If TA is visible What is the chance that it came from Tata Nagar Or what is the chance that it came from Kolkata So all of these are basically talking about A similar type of concept We will close this in another 30 seconds So if you want to vote you can do so Only one person have voted so far If 4 3 2 1 Go There is a confused response B, C and D Have almost got equal number of votes But more votes has gone to B Let's check If you pick up Krishna Giri What is the chance that If you pick up any two consecutive It would be RI So how many consecutive alphabets Can be picked up 1, 2, 3, 4, 5, 6, 7, 8, 9, 10 So out of 10 There are two cases where you end up Getting RI RI So you can say 2 by 10 is the chance that Given you have picked up Krishna Giri Let's say I call that as even even You will get an RI Okay Now Dharmapuri Dharmapuri How many first of all Two consecutive alphabets Can be picked up 1, 2, 3, 4, 5, 6, 7, 8, 9 It's actually one less than The total number of letters over here 1, 2, 3, 4, 5, 6, 7, 8, 9, 10 10 minus 1 is 9 Out of which only one case Is where you end up getting RI So probability that you will see RI Given Dharmapuri has been picked up Will be 1 by 9 Now My question was what is the probability that If you see RI it has come from It has come from Krishna Giri So this is what we are looking for As per Our base theorem formula This is PI into PA by EI By PEI PA by EI Plus PE2 Into PA by E2 So PE1, PA by E1 By Summation of PEI Into PA by EI So first of all I always Write the denominator term Half the chance it comes from Sorry Krishna Giri And given that it comes from Krishna Giri The chance of you having Two consecutive letters As RI is 2 by 5 Or it comes from Dharmapuri And the chance that you get RI as the two consecutive letters Is 1 by 9 Out of this term Will be there on the numerator So your answer will be 1 by 10 divided by 1 by 10 1 by 10 Plus 1 by 18 So that will give you 18 Upon 28 18 upon 28 is 9 upon 14 Option C is correct Got the point So those who are just counting The RI There are two RI's here And that one RI here So 2 by 3 is the chance That is not the right way to proceed Let's take another one J2005 question Okay let me put the poll on Is this up this up in one more minute time Yes Venkat do you have any questions Yes sir I am typing Because while talking I will make it very muddled up Okay 5 4 3 2 1 Go Okay most of you have gone with C Okay the late problem parts are adding up to 1 So that So it doesn't mean Take that out okay So here is a very good question If you say The probability of him Let me just do this part first See what are the questions saying The question is saying that a person goes to office Either by car Scooter, bus or train The probability of which being one seventh So Car is one seventh Scooter is three seventh Bus is two seventh and train is one seventh Probability that he reaches office late If he takes Car is two by nine If he takes Scooter one by nine If he takes bus four by nine If he takes train again one by nine Given that he reached office in time What is the probability That he traveled By car Okay so first of all I first write down the total probability That he is not late That means he is on time Correct When will he not be late If he had taken a car And Given that he has taken a car He is not late which is seven by nine Or he had taken A bus Sorry scooter Which is three by seven And given that he has taken a scooter He is not late that will be Eight by nine Or he had taken a bus Two by seven and given that he had Taken a bus he is not late five by nine Or he had taken This And that given that he has taken A train he is not late Is eight by nine Out of this where he had Taken car That is your first Expression this will be on the numerator So given that he is Not late what is the probability That he had taken a car it will be One by seven into seven by nine By the whole thing Which is one by seven into seven by nine Three by seven into eight by nine Two by seven into five by nine One by seven into eight by nine Is that clear Is that clear Yes sir This answer You can easily calculate by dropping 63 from the denominator so it is Seven by seven plus 24 Plus plus plus plus plus Ten plus eight This will come out to be seven by 49 Which is one by seven Okay Now the question is What is the probability that He used car given he was Late Now Is it one minus this That is what I think is your question Correct Yes sir So if you calculate it you will see What is the relationship Between these two Let's try to figure out So if I write this term P C by L Bar it is P C intersection L bar By P L bar Correct Now PC intersection L bar Is what PC Minus P C intersection L Correct And L bar is one minus P L Do you think this will simplify To give you this term No sir And we will also take it through this example Now given that he is late What is the probability that he was a car So now this answer will be Given that he is late The probability that he will use a car will be Correct me if I am wrong I am just quickly writing down the expression That means it will be 2 upon 2 plus 3 plus 8 plus 1 What is this This is 2 upon 15 I guess No sorry 14 This is also 1 by 7 So they need not add to give you 1 They did not add to give you a 1 That is very much evident from the General formula also This is not same as 1 minus PC by L So the other part Of the probability comes like he took bus And late bus and early and stuff like that Yes So many people who have done this Given that he is late What is the probability that he used A car to come And then you did 1 minus that you will not get the answer You will not get the answer Okay Let's take this one So what is the bottom What is the moral of the story What is the value Sir Yes Sir if we have to calculate The total probability that he came late Or should we just add Those 4 probabilities No you will have to do Taking a car and being late Plus taking a scooter and being late Plus taking a train and being late Taking a bus and being late Yes Now this is a passage based question Which came in the J advance exam 2006 Read this question I first invite you to solve the 12th one Yes Do you want me to run a poll One minute sir Now you can see that they have Attached Or they have made it directly proportional To I that means The chance of picking the first turn Is not the same as the chance of picking the second one Okay so now There is no equal chance of picking up And like we used to do one third one third One third or half half That is ruled out now Okay guys let's do it Because we have to start with binomial Probability distribution also So it will take a bit of time So Anybody who wants to vote please do so In the next 10 seconds For the 12th part 5 Okay not many people Only 5 of you have voted Most of you have voted for B option B for Bengaluru Now let's discuss this Let's discuss this See first of all The probability of you picking A earn Is directly proportional To I so let's say I'll call it as Ki But since your UI Is a MEC event We know that summation of PUI Should give you a 1 Right that means summation of Ki Okay I will go from 1 to N should give you a 1 That means K times N into N plus 1 by 2 is equal to 1 So your proportionality constant Will be 2 by N N plus 1 Okay Now as per our question We are asked What is the probability of PW and after that Will take N tending to infinity So first let us take that Can I say this is a question of LTP So the total way of getting A white ball is the product Of the probability of picking the Ith earn And then given that you have picked up an Ith earn Getting a white ball from it Okay so first of all Ith earn itself will be Ki That is 2I by N into N plus 1 Right And The chance of you getting Sorry And the chance of you getting A white ball in the Ith earn Is how much I by N plus 1 I by N plus 1 so basically you have to do this Okay So in short we have to find summation Of Summation of 2I square Upon N N plus 1 Square So this is a constant The I square will undergo Summation process So I square summation will be N N plus 1 Into 2N plus 1 by 6 N N gone This will be 2 third One of this will also go So it will be N plus 2N plus 1 By 3 times N plus 1 This is your PW Now they are asking you what is the limit of this As N tends to infinity So limit of this will be 2 third Something like this So answer is 2 third Option number B is correct for the first one Second 12th one Next If P UI is a constant Then What is the probability that you picked up Nth earn Given you have found a white ball Given that you have found a white ball What is the probability that you have got it From the Nth earn Whole Is on for 13th one Okay Tripan So 13th one is a question of base theorem Okay we will stop In another 15 seconds Okay 5 4 3 2 1 Most of you have gone with option A Let's check So if you want to know What is the probability that N was chosen given you got a white ball So it is nothing but probability Of choosing earn number N And getting a white ball From it By total number of ways Of getting a white ball Okay Now U N Choosing U N itself is a constant See the chance that you get a white ball From earn number N is N by N plus 1 Because any i tharn will contain i white balls Out of total N plus 1 balls And this will be C into 1 by N plus 1 C into 2 by N plus 1 C into 3 by N plus 1 Da da da da da Till C into N by N plus 1 Am I right? Okay let's do one thing Let's directly Remove this C Let's directly remove N plus 1, N plus 1, N plus 1, N plus 1 So it is nothing but N by 1 plus 2 Plus 3 up till N Which is nothing but N into N plus 1 by 2 N N gone, answer is 2 by N plus 1 Okay So option number A is correct Janta was absolutely right Next 14th question If N is even and E denotes the Event of choosing even numbered earn What is this I Okay it is given that P U i is 1 by N Then the value of P Getting a white ball Given that you have chosen an even number Okay This is a conditional probability question I'll put the poll on for this This time the chance of Picking any earn is fixed Which is 1 by N This is the question 14th question That question Oh Passage Let's finish this We have less half an hour to discuss Probability distribution function Let's do another 15 seconds So wrap this up Okay 5 4 3 1 So we have to find out what is the probability Of getting a white ball from an even numbered Earn From an even numbered earn Most of you have gone with B option Chela will check See Where should I write So probability Of you getting White ball from an even numbered Earn is nothing but probability Of white ball From an even numbered earn By probability of choosing an even numbered earn Right So you getting a white ball from an even numbered earn You have to start with The second earn The chances of you getting a white ball from there Is this And you choosing fourth earn And the chances of you getting a white ball from there Is 4 by n plus 1 Correct And this will go till 1 by n Into n by n plus 1 Okay Am I right n is even sir Yes n is even Okay Divided by the total ways of getting even Which is 1 by n 1 by n 1 by n This will be written n by 2 times Correct No So what you get from here you get We can take I think 2 by n n plus 1 Common so 2 by n n plus 1 Common And on the numerator you will get 1 plus 2 Plus 3 all the way till n by 2 And denominator Is n by 2 times 1 by n Okay n n gone this will go up So it will become 4 by n n plus 1 And this will become n by 2 n by 2 plus 1 by 2 How much is it simplified So first of all n n 4 4 gone this will become n plus 2 By 2 n plus 1 Is there any option which says like this n plus 2 by 2 n plus 1 n plus 2 by 2 n plus 1 Yes option number b says so Is this fine Any questions any concerns Okay Okay So now we are going to talk about pdfs Probability distribution Functions Probability Distribution Functions Now before I start talking about it I would like to talk about something called Random discrete variable Random Discrete variable Okay There is something called random continuous variable But that will be studied under the concept of z distribution Which is also called normal distribution That is another type of function Which is called normal distribution function We will talk about that in your engineering Not It's not that we will talk about it Your professor will talk about it Sir you will be only taking our engineering classes also By the way I can teach till first year But not beyond that First year math is all about You know abstract thing linear algebra And all will come into picture Anyways In order to explain this I will give you a simple example Let us say you have 3 coins You have 3 coins Okay Let's say you have 1 coin You are tossing it 3 times That's another way to look at it So you have a coin You have a coin which you are tossing Which you are tossing 3 times Okay If I ask you what is the probability of Getting heads On these 3 tosses Mind you my question is just saying What is the probability of getting heads When you are tossing 3 times I never said how many heads So your number of heads here Will be your Random discrete variable Because it is coming from a Random experiment Discrete because it can take Discrete values like you can get 0 head 1 head, 2 head, 3 head Variable because it is variable Because it can get either 0, 1, 2 or 3 Okay So that would be called a random discrete variable So let's say if I say What is the probability of Of getting Getting heads Okay When you toss a coin So how many heads That is A random discrete variable Because it can be 1, 2 or 3 in number Okay So if you write down a table If you write down a table Where you are Listing the probability of Getting 0 head 1 head, 2 head Or 3 head Such a table will be called As a probability distribution table Okay Correct So we all know that if you are tossing a coin 3 times the probability of getting 0 head Is 1 by 8 because out of 8 Outcomes 1 of them will be 0 head at all 1 head will be 3 by 8 2 head will also be 3 by 8 And 3 head will be 1 by 8 So when you write such a table You are basically writing the Probability distribution table Now you will ask Where is function here Okay So if you write the same as a form of a formula Which I am going to write down in front Of you right now So the probability of you getting x heads Is 3 C X Half to the power of X Half to the power of 3 minus x So if you put x as 0 you will end up getting 1 by 8, you can check it out Now from where do I get this formula That is something which I will tell you later on So as of now this is a Probability distribution function So this will be your PDF Okay As of now just understand what is the Random discrete variable And when we write such a table We call it as a probability distribution table Okay Now few things that You must know about this table So why do we need this table What are the inferences We can draw from this table I will tell you in some time So let us try to understand Before that something called Bernoulli's Trial Bernoulli's Trial If you are performing an experiment If you are performing an experiment Whose outcome Can be written as a Success or a failure So any trial Whose outcome Can be Considered to be a success Or a failure Now when I say failure and success Do not take it in a literal sense It just means Something and something Complimentary of it For example You tossing a coin is a Bernoulli Trial Because either you get a head Or you don't get a head So if you consider head to be a success So either you get a success or you don't Correct Let's say A ship is trying to reach From the source to destination So ship reaching is a success If you are calling it by that And not reaching it is a failure There is nothing positive or negative Attached to it Don't think like overreached it is success Good thing no success Bad thing means no For an enemy the ship not reaching is a success So it depends on the question What are you calling as a success What are you calling as a failure Do not attach any literal Emotion feeling to that Particular section and the failure So let's say it raining today Is a failure For us And a success for the farmers So it depends upon With respect to what you are solving the question That is success or a failure In short basically I want to say Anything which results into The event happening or the event Not happening is a Bernoulli Trial Ok Now if a Bernoulli Trial Is performed independently End times Then we basically use Binomial probability distribution For such cases So let me give you an example You tossing a coin Three times You tossing a coin Three times This is actually a Bernoulli Trial Because a coin can give you Either head or no head So if let's say I say Write down the probability distribution Of the number of heads That you get on tossing of this coin Which is basically the same example Which I gave you Now this is a question where You getting a head is a let's say success Which I call as P And you getting a tail Is a failure Which I call as Q normally So failure is Now you want to write down The probability distribution Of the number of heads that you get So what do we do in such cases We use Binomial Theorem And that is why that is called A binomial distribution function Now how do we use binomial theorem All of you please understand here So what is the chance that You will get You will get What is the chance that you will get No success at all What is the chance that you will get No success at all So when you don't get a success means Every time you are tossing You are getting tail, tail, tail Isn't it Yes or no So read this as if you have written Q Right Correct If I ask you what is the chance you get This one head can come in three ways Either the first row gives you head And the other throws do not give you head Or The first gives you tail, second gives you head Third doesn't give you Or First doesn't give you, second doesn't give you Third gives you, right So if you see here You are basically writing One by two whole cube three times Indirectly you are writing Three C one One success you want Or you can say Two failures and one success So this term that you have written Is actually read as this This term that you have written Is actually read as this Okay, so one success two failures And that success can be Positioned in three ways In those independent trials So three C one Q square P to the power one Similarly if you want two successes Again It can be done in three C two Q into P square If you want three successes It is three C three Q to the power zero P to the power three And just to make things rhyme up Over here this will be three C one P to the power zero Three C one Q to the power three P to the power zero If you see these three terms It would remind you As if you are writing Your writing probability You are writing binomial theorem For this expansion isn't it So if you expand it Don't you get three C zero This which is your first term over here You get three C one Q square P which is your second term here You get three C two Q P Square which is your third term here And you get three C three Q zero P Q which is your fourth Okay so since this probability Distribution can be written By the use of binomial theorem We call them as Binomial probability distribution So when Bernoulli trials are performed Independently Independently Okay We use binomial Probability distribution Are you getting my point Let's take a simple example of this Of course this is one of the examples Which I have taken But before that I would try to Generalize this So binomial probability distribution function Is basically nothing but If you are performing a Bernoulli trial Where chances of success is Q Chances of failure Sorry chances of success is P And chances of Failure is Q Then let's say The number of successes you get Then according to Bernoulli trial It is basically the expansion Of this term so it will be NC zero Q to the power n P to the power zero NC one Q to the power n minus one P to the power one Next term also I will write NC two Q to the power n minus two P square Till NC n Q to the power Zero P to the power n Is this fine By the way whenever you are using Binomial theorem for this Which of course you have to use always Write the failure first and then the success So this is failure success And this is the number of trials you are taking Is it fine Let's take a small question on the same Let's take this one The probability that A student is not a swimmer is one fifth Find the probability That out of four students Out of five students four are swimmers Out of Out of Five students four are swimmers Okay now let's say I do this problem in a more elaborative way Let's say if I say What is the probability of You choosing a person which is swimmer Okay now I did not mention How many swimmer So out of five people What is the probability of you Choosing swimmers So if you write a table out of it This is how you will write So first of all The X values can range From one to five because There are five people Either you can get no swimmer One swimmer Two swimmer Three swimmer Four swimmer or five swimmer Now in order to fill this I will make use of binomial theorem So you getting a swimmer is a success Over here And you not getting a swimmer is a failure So failure you write first Okay So failure is one-fifth That is Q You getting a swimmer is this Okay And how many trials are you taking There are five people out of it So five trials you are taking Okay so when you write this In a binomial form You will end up getting the first term as Five C zero Over of five Into four by five to the power zero You don't have to write five C zero And four by five to the power zero It is just one-fifth to the power five That means you end up getting No swimmers in that Okay Now one means one success So you want one swimmer to be selected So that will be this If you want two swimmers to be selected Five C two This Four swimmers This And five swimmers, five C five One by five to the power zero Four by five to the power five Now what is the question asking you The question is asking you What is the probability that out of five students Four are swimmers So basically they are asking what is P four Okay So right now you wrote for everything It is like you picked up the entire Himalaya as a Hanuman And we are not able to find out that Sanjeevani Jalli Voti But now they are asking you just one of the features So this is the one they are asking you So it answers five C four One-fifth into four-fifth to the power four Clearly option number A is correct Are you getting my point So several types of questions can be framed For example I may ask you What is the probability that you get At least four swimmers Then you have to add zero one two three four Oh sorry At least four swimmers At most four swimmers Then you have to add zero one two three four Correct A few things that you would all notice over here Is that these all terms if you add You are actually writing The sum of all these terms is actually This term and this term is clearly a one So few observations That you can have I will write here itself So important points to be noted The summation of Your PXI will always be one Always So in any probability distribution Whether it is binomial or nonbinomial The sum of probability should always be one Right Second thing that you need to know is Something called the expected value of X Many people call it As mean value of X That means If at all you are picking up Five students What is the chance that In those What is the chance of getting a swimmer In those five students Or what is the average number of swimmers You will end up picking if you pick up five students Are you getting my point For example let's say I am tossing a coin Three times Three times I am tossing a coin What is the average number of heads I will get Common sense is three by two Heads right Fifty-fifty chance is head and tail Out of three times One point five heads I will get Even though theoretically it is not possible Practically it is not possible But this is the theoretical way of looking at it That is called the expected value That value This formula is actually given by The same formula that you have learned In your mean statistics That is summation XI Or you can say summation F I into XI By summation FI Right Here the name is slightly getting changed Here you can write F1 F1 X1 By Total number F2 X2 By total number and so on This is Fn Xn By total number This sum is actually the probability Of occurrence of X1 So you write it as PX1 Into X1 This is PX2 PX2 Into X2 and so on So in short the formula For getting the expected number Of successes When you perform the experiment End times Expected number of successes When you perform the experiment End times is summation PXI Into XI I from 1 to N Okay So let's Apply this to our coin Question So let's say I am tossing a coin 3 times What is the expected number of heads I will get So I will use the data Which I had already written In the first slide So your expected number of heads Will be 0 into 1 by 8 1 into 3 by 8 2 into 3 by 8 Plus again 3 into 1 by 8 If you sum this up You will end up getting 3 plus 6 How much is This 3 plus 6 9 9 6 is 12 12 by 8 12 by 8 is 1.5 Didn't we guess that Out of 3 times 50 percent Or 1.5 heads you will get Even though practically it is not possible This gives you a rough idea about how many successes you need And the third thing that you need to know Is the how Scattered is your successes Or what is the variance of the success So this is given by Summation of PXI Into X I minus X Bar which is your expected value whole Square Let me take this in the other page Anybody who wants to note this on Please note it Yes sir one second Sir Yes sir Sir in this previous question that you did The swimmer question So there the swimmers are like Distinctness They are not identical swimmers Distinctness yes So if we have like 5 identical Balls And then we say that If we have identical things Identical things You will not be able to apply this 5C1, 5C2 etc Okay In that case the process will change This will not be the process Okay sir And sir if we have more than Two outcomes possible Then can we use multinomial theorem also That's a very good Question to ask You can yes You can use multinomial theorem Yeah So variance I will talk about variance of X in So variance of X is defined As summation of PXI into X minus X Bar the whole square Remember X bar Is your expected value of X Okay See this is very similar to the definition That you followed for variance In a statistics also Do you remember you used to write FI into XI minus X bar Whole square By summation FI The same thing I have written over here In the language of probability So if you expand this I will just show you something very interesting If you expand it you get X bar Minus to X bar X Plus X bar square So this will give you X square Or you can say Summation PXI Into XI square Minus to X bar Summation PXI into XI And here you get X bar square times summation PXI This term let it be as it is This term that you see is back to X bar Isn't it So it is X bar bar And this term Is 1 We have already seen that the sum of the probabilities will be 1 So if you simplify it One of them will get cancelled over here With this So it will give you summation of PXI Into XI square Minus expected value Square So you can also write it as Summation PXI XI square Minus expected value Square Now for special case Of binomial distribution So for our Binomial distribution The expected value of the success Is always NP Variance is NPQ Now how? I'll just take one minute for you to prove Expected value is NP Arrest you can prove it as a homework So when you write a binomial distribution When you write a binomial distribution This is NC0 Q to the power N P to the power 0 NC1 Q to the power N minus 1 P to the power 1 NC2 Q to the power N minus 2 P square NCN Q to the power 0 P to the power N So as for the definition of Expected value The definition is summation If you do 0 into 0 into this that will be 0 So you can say Just to write less I'll say 0 into C0 And I'm not writing it Q to the power N P to the power 0 Then you have 1 into C1 Q to the power N minus 1P Then 2 into C2 Q to the power N minus 2 P square and so on till N into CN P to the power N Now if you please pay attention I will take Q to the power N Common so it will be 0 C0 P by Q to the power of 0 This will be 1C1 P by Q to the power of 1 Then 2C2 P by Q square Till N into NCN P by Q to the power N Ok so we want to find this Now just look at this This will remind you of C0 C1X 2C2X2 Dot dot dot till NCN X to the power N Where X is your P by Q correct Now what is the way to find this Now go back to your childhood days When we did the binomial theorem In binomial theorem we learned that In order to find this We can use our calculus method So let me just write it Just allow me 5 more minutes Time You people say You give me 5 minutes extra Ok Right so first multiply It with a X By the way you don't want C0 To occur right so it is 0 C0 So don't worry so just differentiate it With respect to X N1 plus X to the power N minus 1 C0 will go for a toss You will get C1 You will get 2C2X You will get 3C3X Square All the way till NCN X to the power N minus 1 But we need an X X square here So we will multiply again with an X So just multiply an X So furnish one X to it Ok That means this term that you have written Can be written as So back to this term again Q to the power N Now N X 1 plus X To the power N minus 1 Am I right Ok let's try to simplify this So here if you take an LCM Inside the bracket it will become P plus Q To the power N minus 1 By Q N minus 1 By Q this is N And this is Q to the power N This will get cancelled P plus Q is known to be 1 So the answer will be NP Into 1 to the power N minus 1 Which is NP So you don't have to directly apply this formula Even this is a universal formula This is used for any distribution Universal formula But if you are using a binomial distribution You can directly use this formula To save your time Ok And variance also Please try to prove this Variance please try to prove That this expression This expression Gives you Under NPQ Please prove this as a homework Same approach Not different Ok I would like to take One more question on this Before we wrap today's session Pozons I will be discussing when I am doing Crash course don't worry That is not that important That will take another half an hour easily Ok let's take this question X is a random variable With distribution given below Ok What is the value of K And what are the variance of this experiment Or what are the variance of the Random discrete variable Ok done This is K is what K can be obtained by using the fact That this will be 1 So it is very obvious That 8K is 1 so K is 1 by 8 Ok So this is 1 by 8 3 by 8 Very much like that coin thing question So what are the variance of the data Variance of the data as we had already discussed It is summation of PXI Into XI square Minus expected value Ok Now first of all we need to find out The expected value Expected value is summation PXI Into XI Ok You can directly find it by the original formula But original formula is slightly difficult to apply This is more convenient to apply So this is 0 into 1 8 1 into 3 8 2 into 3 8 Plus 3 into 1 8 Which we had already figured out was coming out to be 3 by 2 So now This will become 0 into K square Which is 0 1 into Sorry 1 square into 3 by 8 So I am applying this thing first Then 2 square into 3 by 8 And then 3 square into Again 1 by 8 Minus 3 by 2 square Let's see how much it comes out to be So 3 by 8 12 by 8 Again 3 Sorry 9 by 8 Minus 18 by 8 So how much is this How much does it come out to be 3 by 2 sir 3 by 2 4 Option number D is correct Is the idea clear how it works Okay So with this we close our Session here Thank you so much It was a pleasure teaching you all We will again meet tomorrow With a crash course on Tignometry It will run for 2 days Monday and Tuesday Take care bye bye Have a good day Yes sir