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Hello students, my name is Dr. Gajendra Purohit and you are watching our YouTube channel. If you are doing BSc engineering or preparing for any competitive exam, then our YouTube channel will be very helpful for you. So today, We will be discussing a concept in probability that is Bayes theorem. Let us say that we are given any conditional probability as a by b. So, in that case we need to find its inverse that means probability of b by a.
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or we need to find the total probability. So, we will learn its concept today. So, what is Bayes theorem and how to solve questions based on it? Let's start. MUSIC First, we will discuss that what is Bayes theorem. if an event A occurs with any of the events b one b two b three b n then we can find the probability of A by the following rule. Suppose B-one, B-two, B-N are mutually exclusive events.
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whose union is sample space. That means if you take union of B-one, B-two, B-three, B-N, then the sample space is created. and probability of each one is not equal to zero. Okay, then for any event A related to B one, B two, B N and then P A is equal to P B I plus P A. conditional probability. B.I. you might wonder what exactly this is. So, let us learn something. I am going to try to explain this concept to you with an example ok? Let us say that suppose we have a company.
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and from three different vendors A, B and C. It gets bolts.
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From vendor A, it orders thirty percent. Then from vendor B, it orders forty percent.
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And from vendor C, the company orders, thirty percent. Students, this means that the products are ordered in this ratio. But, what is wrong here? There is a problem with A. So, what does A do? It sends two percent defective with this thirty percent. and b. sends three percent defective. and the vendor C sends five percent of the products defective. This is their track record that they will send defective products inside the product lot right. It is obvious.
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So, students what happened here all these products are collected together. All the products have been collected. Now the bolts are all the same. We can't tell which one is from A, which one is from B and which one is from C. So, what did we do? We just randomly picked a product from here. and that came out defective.
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So, whether that is defective from A. or defective from B. or defective from C. We will be calculating this using Bayes' theorem, right? But, if we take out a product from here. what is the probability of it being defective? So, using the formula I just explained. We will find it. So, that is the concept of total probability. Let me explain it to you with an example. So, students here we are given. that A here is supplying, thirty percent goods, that is point three.
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and B is supplying forty percent goods that will be point four.
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and then C gives point three that is, thirty percent. Now, we are given that vendor A gets a lot of defects in its production. So, how much is that? that comes out to be zero point. zero two After that we are given something about production of B. So, according to the given data the defective comes out to be zero point zero three. and the defective from C. that comes out to be zero point zero five. We know this, right? First question being asked is
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that let us say we have randomly picked a product from here. Tell me what is the probability of that being defective? So, if we need to Find the probability of defective ones. We will just multiply these together. that means, Pa then P. D by A and multiply these together, then Multiply. this by this okay. so P D by B. plus PC. P. D by C, right? This is the probability of it being defective. We calculate this in total probability. How much will it be if I calculate this?
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When we multiply this together, it will be point three into this. So, this will be point three into. zero point zero two, plus. This will be valued at point four into. point zero, point zero three. and this will be point three. into point zero five right and then if we solve this. So, from here we will get six and twelve. eighteen and fifteen. So, fifteen and fifteen, twenty-three and thirty-three. So, from here we will get its value as zero point.
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zero three three. So, what will we call this? it will be the value, right? Now, if you are asked this question. that it's great that you figured out that this is the defective value. Tell me. whatever defective we have calculated from here. so is it the defect from A. or is it the defect from B or defectives from C. What is the probability of this? that means it is okay that you have identified the defective ones. So, what have we done here? We have found the defective.
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So, tell me what is the probability? of this being defective from A. and probability. of this being defective from B. What is the probability of this being defective from C? In this way, we can find the concept of Bayes theorem. So, students what does this mean? We are given D divided by A. Now we have to find out what is A by D. It means we have to find the inverse probability of this. And when we have to do this, then we use.
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Bayes theorem. A question might arise in your mind that how do we calculate the value of this? What we need to do is Since it's A by D, so we need to multiply the previous one by this over here. We have the formula, so we don't need to memorize anything. Whatever question we are given, we will directly put it over here and this is the total probability so this entire value will come down right here. which will be PA into this will be p. d by a. plus. P B.
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plus this will come as P, B P. D by B. and we will get this as P C. and P, D by C, right? Do you understand it? We need to calculate for this. So, after dividing we will get the total probability ok?. We will multiply PA. by D to A. As soon as we multiply this by this, we will get zero point zero. zero six, divided by, we have already calculated the total probability that is zero point. zero three three. So, we will get its value from there. Six by thirty three right?
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In the same way, if you are asked, that what is the probability of this, you do not need to worry. So, we will be multiplying these two values together for this. Am I right? We will put it above. What did we do in previous one? We have to calculate its value so we multiply these in the above. We have to calculate the probability of the defective in B. What will we do? We will multiply this. We will multiply these two and this will come as P B P D by B. divided by
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I am directly writing this entire value as P D because writing it repeatedly is difficult. This means it is the total probability. when we multiply both of these. Right? When we multiply both of these equations or you can say when we calculate this four with this, then this will become zero point.
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zero one two, divided by, zero point zero. three three. So, we will get this as twelve by thirty three. Right? If you are asked that what is the probability that the calculated defective is of C, then no need to worry. So, we will multiply this equation with this okay. This will come as PC into P. We will get D by C from here and what will we get below? We will get P D here meaning the total value. When we multiply this by this, it will be zero. Point three.
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Into zero point zero five. So, this will be zero point. zero, one five, and this will be divided by. zero point. Three three. the value that we will get from here will be this right? we can calculate this. in this way if you add all these. So, we will get one again. Is this clear? Students, if you are asked in the exam that what would the probability be? If it is only defective, then we will use this concept. if you are told that suppose a product is taken out from here.
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and it turned out to be defective. If it is defective from A, Is it defective from B? Is it defective from C? In that case, we will use Bayes theorem, this is total concept of. this class, right? I will explain with more questions which will help you understand concept. clearly. I will take one question and through this I will try to. explain it, in two thousand thirteen. There will be three candidates for the post of Professor X, Y and Z.
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whose chances of getting the promotion are in the proportion of four is to two is to three. So, as we all can see here that the chances of X becoming a professor. Chances of y becoming a professor. and well as the chances of z becoming a professor. Am I right? So, the ratio is provided here. So, if we look at the ratio of x. How much is it? Four upon, add all these in denominator, four plus two plus three, it's nine. So, students this will come as four by nine ok.
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Here, we have the chance of X becoming a professor. If you're asked the chance of y becoming a professor, then it's two by nine. The total denominator is nine and two is the chance. of Y becoming professor in the same way the chance of Z becoming a professor will be. three by nine. The probability that X if selected would introduce a new subject for today's requirement in the department is point three five. meaning if x becomes a professor.
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then he'll introduce new subject and its value is given here. Similarly, if y or z would introduce new subjects then its probability is given as point five two and point eight zero. What is the probability that there will be a new subject in the department in the year two thousand fourteen? You are being asked that what is the probability that a new subject will be introduced, here it is not being asked that what is the probability of a new subject being introduced by Professor X?
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that what is probability of Y introducing new subject or Z introducing new subject. It simply asked that what is probability of new subject being introduced. We will calculate the total probability in this case. If x becomes a professor here, then a new subject is introduced. I am denoting it with n. students the probability of that is given to us is zero point. Three-five. Similarly, if y would be selected here and introduces a new subject.
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The probability of this is given to us as zero, point five two, as you can see here on the screen. And in such case where Z is selected, and if Z introduces a new subject then its probability is given as And if you are being asked that suppose someone is selected Then, what is the probability of him introducing a new subject? So, we will calculate P n here right. What will we do? We will multiply this by this. That is, p x into p. n by x. then we will multiply this with this, that is PY. P
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P N by Y. and If we do this with this, then this will come as P Z. p and we will get N here. So, this is what we get as the value and if we put it here then we will get the answer. We will multiply this with each other from here and this comes as. four by nine. into point three five. plus two by nine into point five two. we will get plus three by nine into point. eight zero. We will take out one by nine as common right. This three point five is given here.
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This will come as one point four here. Am I right? And when we will multiply this point three five by four. we will get one point four. Similarly, when you multiply this with this. this will come as one point zero four as the value for this it is point five two into two because what have we done with this nine here It's common, right? And when we solve this in the same way then this will come as, two point four that is three into eight zero.
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and if we simplify this then this will become two point four while this will become three point four four. And we will get this as four point eight four, am I right students? This will be four point. eight four, divided by nine okay. Here we get its probability. If you want You can try this with three here. According to me, it is not divisible by three here. So the value of this will be if you want this is double. And if you remove this point here, then it will be four eight four divided by
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nine hundred. What we will get here is it's probability. Next question, listen carefully. It is very interesting. There are three boxes containing Respectively, one white, two red and three black balls. Three different boxes are given here. It's an interesting question. Students, look here, you are provided with three boxes. This is first box, this is second box and this is third box. Three boxes are given, right?
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Let's read the question once. It reads, there are three boxes containing respectively one white. two red and three black balls. So, what has been given? in the first box it says. that the box contains one white ball while We are provided with two red balls as well as we are also given three black balls and in the second box we are provided with two white balls and after that we have three red balls. and one black ball. This is the second box which we are currently discussing.
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In the third box, we are provided with three white, one red and two black, right? So, you are given three white balls. and one red ball. and two black balls. Students, all of this is provided to you. A box is chosen at random from it. and two balls are drawn meaning we have chosen one box out of the three boxes and then we drew. two balls from it. The two balls are one red and one white, meaning one of them is red and the other is white. White.
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find the probability that these came from first box you are being asked what is the probability of These coming from first box, this is a very easy question. What do you need to do first? The first box, pay attention. The second box. and the third box and if we choose any one box out of these three boxes. the probability of choosing one box out of these three boxes. is about how much students one by. Three, we will choose one out of the three.
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What have we done here? We have chosen one box. Am I right? What did we do? We chose the first box. We are taking two balls out from this one is red. and another is white. We are taking out one red and one white ball from first box. How many total balls do we have from the first box? Six. We take out two balls meaning it's six C two, the total cases. Out of which one is red and other is white, we get one C-one. and Two C one. If you look then the value that we get from here is two.
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and the six c two we have is six into five upon one into two. If we solve this equation then we get its value as two by fifteen. This is the probability. we are getting for this. then I am asked to choose the second box. What is the probability of getting red and white from it? This comes to us as six C two again here. We are getting one red ball and one white ball. So, this will become two C one. And three C one, If you look here, then the numerator is six and denominator is fifteen.
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Six C-two's value is fifteen. In the same way, you are being told that. If third box is chosen, which has one red and one white. So, we will get six C two here. and now you are being told that one is white. and other is red. Then, as you can see the value that we get here is three by fifteen. And now coming back. to the question you are being asked. to find the probability that this came from the first box. It means that what is the probability that these one red and one white
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that we got is from the first box itself. I mean that you are asked. that what is the probability that this one red and one white ball that we get is from the first box. meaning you are asked the inverse probability. of this. We will multiply these two and put it on top. So if you look here, This will become first. into This will come to us as red and white. First, right, we multiply these two and put them on top. If you were told to do second, we would multiply second and put it on top.
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divided by what will we have here we will get the total probability. What will be the total probability? We will multiply these two, so first. into this p. red white. first. The next will be second. then this will come as P. red white. second. plus this will become Third, and P. this will be red white. Third right? Now put the value of all of these. As soon as you put all of its values, then what will we have here? Multiply these two and put it on top.
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We will multiply one third by two by fifteen because these two will be multiplied. then this will come as two upon forty five. Right? divided by what will we do here, we will multiply this. this will come to us as two by forty five. plus if we multiply these two, we get six by forty five. and when we multiply these two, we get this as plus three by As you can see, forty five is common, hence it will be cancelled. Two will remain as numerator and will calculate the LCM as forty five. we will add
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six plus three nine, nine plus two eleven. So, this will come as two by forty five upon eleven by forty five. So, two by eleven. will be the answer and forty five will be cancelled by forty five. We can solve it like this. Right students? Look at the question it reads, a newly constructed bridge may fall down either due to wrong designing or by inferior material used in construction. Chance that the designing is faulty is ten percent meaning if the design is wrong, then it is ten percent right?
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Now, this question says. and the probability of its collapse if the designing is faulty, we have conditional probability here. but the probability that the designing is wrong and it will collapse is ninety five percent. and due to the bad material it is forty five percent so if it collapses. Due to the bad material, its probability is given as forty five percent. if the bridge collapses. Find the chance that it was due to wrong designing. You are asked, that what is the probability
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that it collapses due to wrong. Designing, What are we given here? We are told here look carefully the first given probability is ten percent. which is for the wrong designing. Is this much clear? if I explain this. We will take this as WD here. WD stands for wrong designing and the given probability for this is ten percent. That is one by. ten. And the other probability for the collapsing is the inferior material, either the bridge will collapse due to the wrong designing.
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or it could be due to inferior material meaning bad material right. I will write this used inferior material as the bad material here. Let me do one thing, let me write the word inferior material here right. if this is one by ten. The probability of this will be nine by ten. Either the bridge will collapse due to this or due to this. because the sum of these two will always be one. As you can see, the sum of all the questions I explained earlier is always same. Even in the earlier questions,
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Sum will be same. If we are given two values, we can subtract third from one. We can easily calculate it in this way. Now we are given. The probability for the bridge collapsing, but it is due to the faulty designing this time. So what is the probability? the probability for this is given as ninety five percent right. We are given ninety five percent if you want then you can write as ninety five by one hundred. and then in such case where the inferior material is used.
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and the bridge collapses because of it. So, the given probability for this is forty five percent. Forty-five by one hundred these are what we are given here. You are told that if the bridge collapses Find the chance that it was due to the wrong designing. You are being told that what is the probability. that the bridge has collapsed and it has happened because of the wrong designing? Meaning the inverse probability is asked, that is Bayes theorem. What will we do here? We will multiply.
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the given probability meaning whose inverse probability needs to be calculated. and we will put it on top in the numerator. the rest of total probability which we will have will be written down in the denominator, this has to be kept in mind. Is this clear? This is the inferior material and. Here is the collapse. because of the inferior material. but we will keep this multiplied. this will come as ninety five divided by one thousand okay. When we multiply this by this, then divided by we get
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ninety-five upon one thousand. And we will multiply this by this. This will come as forty five multiplied by nine. it is better to do nine into forty five right. divided by how much will we get from here. It is thousand. Take thousand as common and cancel it. We will get ninety five here. we will get ninety five in the numerator. What will we do with these two? We will add it, ninety five. plus nine into forty five. So, when we multiply and add these.
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then the value we will get for this will be five hundred right. because as we multiply this. we will get four hundred five and ninety five the value we will get as denominator will be five hundred right? if we cancel this by five. nineteen will be the numerator and one hundred will be the denominator that is we will get zero point one nine as the probability in this way we can easily solve. this question. It's an interesting question and Such questions are often asked in exams.
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There are three true coins and one false coin. This means that we have three true coins and we have one false coin. With. head on both sides. What do we have on both sides? We have head A coin is chosen at random and tosses four times, meaning we have chosen a coin. and we have tossed. it four times, if head occurs all the four times. Then what is the probability that the false coin has been chosen and used? You are being asked that
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What is the probability that a coin has been randomly chosen here and? Tossed about four times and head came on all the four times from the false coin? It's an interesting question. First of all tell me that how many coins do we have here in the question? We are given three true coins here and what one coin are we given here. We have a false coin here, if you look. So the probability of selecting a true coin from these four coins is three out of four. And...
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What is the probability of selecting a false coin? It is one out of four. This should be clear to you first. What did we do now? Pay attention students, what did we do now? We have selected the true coin. and we tossed it. Heads came up on all four times. What is its? Probability. So, as we already know that if we toss any coin for about four times. The sample points that we have are for two to the power four.
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which has sixteen points in total and there is only one case where we get heads four times in a row. So, the probability that we will have in such case will be one by. Sixteen. The second coin that we have has heads on both sides. Pay attention to what I am saying. with head on both sides meaning this coin has heads on both of its sides. This coin has head and tail on its sides. The same goes for other coin. head on one side and tail on the other. You need to pay attention.
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It has heads on both sides. Toss it as many times as you want. But if we select the false coin, then we will always get heads. The probability for this is one hundred percent. right? clear? The question that you are asked now is that there is a coin and the coin has been randomly selected. Am I right? and the coin was tossed four times and after tossing we got heads in all those four times. What is the probability that the coin was a false coin?
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What will we do since we are asked its inverse probability? We will multiply this. and then keep it on top as the numerator, then this is what we will get. right. divided by, what will we get as denominator below? P T into P. Here, this will come as true plus. and then we will write P false here. Head then head, again head and head. and then false. We will calculate its value, we will put it in the numerator after multiplying. This will become one by four into one, right?
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divided by multiply this with this. this will come as three by sixty four.
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plus we will multiply this, so this will be one by four okay. Let us take LCM of this one by four. and the LCM of this sixteen and three comes as nineteen. It will be nineteen by sixty four. if you reverse this equation, that is, one by four. into sixty four by nineteen, here it will be. When we cancel this, what will we get? It will be sixteen by nineteen and it is the probability. So, in this way you can very easily solve this.
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Students, this question is for the comment box and I request that you solve it. You can also tell me through the comments that how much time it took for solving the question in the comment box. If you want to prepare for CSIR, NET GATE, CUET PG or IIT JAM exams. This is our MathScare app which is available on Android and iOS. You can download this app and students this is our telegram. You can join this. Any latest update or any information.
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is posted here on this platform. So you can obtain those. information here. You can watch the video on how to download the Mathcare app here. You can also watch how to take the subscription. The playlist on our old videos on stats and probability is also available for you to watch here. You can subscribe the channel here and you can also download the app from here. Thank you very much to all of you. Bye Bye.
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