noc18-mg33 Lec 07

[Music] [Music] welcome back my dear friends a very good morning good afternoon good evening to all of you and this is the da dm1 lecture and this is the seventh lecture as you know this is the course of 30 hours which will be spread over twelve weeks each week we have five lectures each lecture being off of 30 minutes so if you remember in the last lecture which is the sixth one we were discussing the concept of Bayes theorem and I did mention that how the concept of Bayes theorem could be understood so Bayes theorem is basically coming from the conk on the actual macron win conditional distribution or conditional probability whatever you want to say so with condition and probability we went with the concept of Bayes theorem now rather than basically going through the the equation I basically do a neat diagram explain that how if the overall universal set is broken down by n number of mutually exclusive and exhaustive events being given from by B 1 to B in and then we can actually have an even a which is basically a conglomeration of some intersection of B's such that we can find out the probability of BJ J being from 1 to n so all those ends which were basically mutually named exaust mutually exclusive and exhaustive we can find out the probability of B J's provided a has also already occurred and it can be extended into into the two second stage third stage by the word of second stage third steady means and they are different type of combinations from one stage to the other that means given B you want to find a given a you want to find out C given C you want to find out D and so on and so forth now let us consider a very simple example based theorem and then basically I'll try to build up more such conceptual problems which can be solved using the Bayes theorem so the problem is like this in an examination each question has four alternatives so it can be more than four also less than four also but for our problem you have four alternatives of which only one is correct now the problem again can be extended with more than one as correct multiple you ask questions but let us keep it simple so if a student knows the correct alternative then he or she is definitely able to identify it otherwise he or she picks one of the alternatives at random so we will consider some probability distribution of probability values of cooking picking the random answer and there would be a simple concept a very simple practical concept for that also now given that a student has identified the correct alternative Veon and what what is the conditional probability that here we are required to find out so what is the conditional probability that he or she knew it knew the answer assuming 70% of the students know the correct alternative to the question under consideration so now with this what are the assumptions we will consider and how we'll proceed so let me come to the problem directly and it's very simple one so let us define the following concepts so what are those are the assumptions let us consider a suffix 1 or a 1 be the event that the student identifies the correct alternative B 1 and B 2 which are technically the mutually exclusive and exhaustive set try to remember the diagram which had drawn on the end of the sixth class so so B 1 and B 2 are the mutually exclusive and exhaustive so B 1 denotes the student knows the correct alternative B 2 denotes the student doesn't know the alternative correct alternative so what are the corresponding probabilities data answer it very simply then we know if the student knows the correct alternative it is given by 70% as in the question so it is 0.7 and the complement set of that is the student doesn't know the alternative is 0.3 which is 30% now what are the scenarios given the student knows the correct alternative and if a is the event then the student a 1 is the event the student identifies the correct alternative obviously if he knows it will answer it correctly hence the probability is 1 so probability of P or PA 1 given B 1 the answer is 1 now let us come to the second information now if this student doesn't know the alternative that means he's unknot sure so if he is not sure in front of him he has 4 alternatives so the property of unsealing anyone one of them is basically 1 foot hence the probability that of a that means answering and the correct alternative by the student provided he or she was not aware of the alternatives is given by 1/4 which is 0.25 so if this is the case now we will try to answer the question what is required the answer required is the probability that the student knows the correct alternative provided he has identified the correct alternative we will just formulate it as a simple base theorem given the problem and solve it so the actor so this is what we want to find so this a if you remember and go into this diagram so these are the b1 b2 b3 and I want to basically have a so this is a I want to find out the probability of a so obviously and use the highlighter use the yellow color let me use the green color it will be or the blue it would be easy so I want to find out this probability then this probability I'm going to add them up so what I am marking I want to find out this probability I want to find out this probability I want to find out this probability so if you remember for a let me use the black so this one this probability of a if you basically go through it is basically the sum of this this this this in this so this can be found out put it into the value and the current answer comes out to be zero point nine zero three - I am NOT solving them but I am trying to give you a pictorial Dagenham and you can understand it much better an exercise which I would love the students to do okay remember one thing as we have started now there would be many exercises which come up as we are discussing these exercises are not part and parcel of the assignment not in part and parcel of the enzyme Minister also but please solve them try to understand because as we are going through the concepts you understand that how they would make some sense based on the fact that we had covered one concept will solve one small and then we immediately come up with the small like exercises that you can understand it very clearly the problems and telling you before and are very simple easy understand the concept and solve them so the exercise or the example is like this the marketing manager of a toy manufacturing company is considering the marketing of a new toy in the past 40% of the toys introduced by the company have been successful and 60 have been unsuccessful before the toy is marketed a market research is conducted and a report either favorable and unfavorable is submitted is compiled in the past 80% of the successful toys received a favorable market research report and 30% of the unsuccessful if we see a favorable market research so it may be possible that you do a marketing survey so it is not hundred-person that you are able to predict everything so even if the marketing survey survey the data may give you that the market is actually bad but still it is given a good set of information to you or it may be possible the market is good but still the marketing surveys give you giving your bad survey report so any any combinations are possible and obviously other two combinations are the marketing survey does really analyze the market and it is good and it gives you a good a report and it unless the market the market is bad and it he gives you a bad report so those are actually what it is and you are getting the information but the other two instances where it is good but you get a bad report or it is bad you get a good report are the problematic area but and obviously we will consider such problems later in the hypothesis testing so the marketing manager wants to know the probability that the toy will be successful if it receives a favorable report so basically you analyze the prom market which one should be b1 which one should be 2 which one should be b3 such that B 1 B 2 B 3 or whatever how they however different values of n you have you basically combine them and try to find out that they are mutually exclusive and exhaustive and solve the problems accordingly now what will mean by independence of events now here be careful what I mentioned and note down it very carefully so we will say two events are independent if the prop the probability of the intersection of a and B a and B are in in events which are and part and parcel of the universal set so we'll see a and B are mutually independent of each other taking two at a time if the probability of a intersection B is given by the multiplicative value of probability a into probability B now if I am saying that if there are three events if they are independent how would we change analyze it will be given the prop that three events a1 a2 a3 are independent of each other if probability of a 1 intersection a 2 intersection a3 is given by probability of a 1 multiplied by probability a 2 multiplied by the probability of a 3 so now further on now what will now consider is basically a little bit different we'll consider the different above distributions and we'll come to the uses of distribution in different general examples now depending on what is the outcome of an experiment you will basically have a random variable to denote that what is the actual thing we are going to study what we are going to learn and you will consider that is to be random because the fact that we do not know what should the outcome be of the experiment which you are trying to study so this random variable which is RV would be used to denote the outcome of an experiment and will try to basically denote the random variable by either the capital symbol of the variable X or capital Y or capital Z now the reason why we use capital 1 is basically denotes the random variable which is unknown before we do the experiment or as we are doing the experiment the moment we complete the experiment the value which is in front of us is known it is known as the reales value and the random variable may take those reales value depending on what is the actual probability function based on which you are trying to map the random variable and the different type of Els value so this basic a mapping where you take a random variable X put it into the function that function what do you want to find out is the probability distribution function and the output which will get would base be based on the fact that what capital X takes that's small realize values small X based on when you put the small DL as well as small X into the function you get an actual function and value which is known as the probability distribution function so let us consider a very simple example consider a coin and it is an unbiased one so there's a head there is a table consider that rather than head and 10 the coins in one face you have red color another case-based face you're basically yellow color not so now before you toss or as the coin is tossed in the air we are interested to find out that what face will come out so we'll denote the random variable capital X by the face which which will mean that we want to study the face which will be coming out on the top and we know that for that face which is a random variable which we do not know as the experiment is being done but what we actually know is that once the face actually comes out they are basically two realist values one is the red one one is the yellow one now you may say that why you are basically trying to depict them as colors but then how does the value of numbers come in to as the probability sense now consider the example that those red and yellow are removed but you put that in the case of a coin where the faces are now black and white or consider you consider the coin is in such a way where the faces in one face there is one not and another face is on these two dots so your actual aim is to find out the outcome but what you are trying to do is that that outcome is mapped onto a function which is basically a mathematical functional formulation which gives us that what is the outcome and what is the corresponding probability based on which we are trying to study that outcome now consider the second example again the probability example not of the coin but of a dice consider the faces are marked 1 to 6 so obviously this why I'm saying 1 to 6 or why I told 1 dot for the coin or 2 dots for the coin is basically to give you an example such that the underlying factor of the experiment which you are trying to do need not have any quantitative values assigned to them but the actual output based on which we are trying to understand the concept of probability are all quantitative in nature because there are some functional form such that once the capital X whether qualitative or quantitative whatever it is it sticks is various value once you put the realest value into the functional form you get an output which is in number terms that is the probability now consider the dice example if you roll the dice the probability of getting any face that is 1 to 6 is given by 1/6 now replace that experiment with the same site of sort of dice say exact same density same weight same shape whatever it is same material but now the faces are marked by the letters ABCD so a for 1 B for 2 and so on and so forth now you again do the experiment in the initial case the experiment outcomes were dots 1 to 6 any one of them now but now in this second experiment that the they are not the dot but they are the faces are the alpha v alpha betting values ABCD whatever so when you're doing the experiment in both the cases the capital X the random variable where in one case the random variable denotes the the outcome which is unknown in the second case also the outcome is unknown but when actually X takes place and you know that in the first case small X which is the realest value gives us third point which is either one or two or three or four or five or six in the second case when the realist values comes they are the value either A or B or C or D or E or F but the corresponding functional mapping when you put where is 1 to 6 or a 2f into the functional function f of X the outcome is the probability which is 1 by 6 in both the cases provided they are unbiased quaint now we will consider that the the outcomes which you are doing for the experiment can be either continuous or they can be random discrete so now you'll be asking what do you mean by continuous and discrete is the X values which are the outcomes can be either be the discrete or continuous based on the fact hence we'll have the probability mass function for the case when it's discrete and probability distribution function for the case when they're continuous so these would basically given by the functional form of small f of X in both the cases and we'll analyze them in details later on so let me continue reading it so we will usually denote the random variable using X Y our set and the corresponding probability will given as I just mentioned that by small f of X or small F of Phi or small F of Z now this F of small X my F of small wise F of small Z they are not small F of capital X or not small F of Capital y because those capital values denote the random variable once you plug in the values which is the realest one that will give you what is actually the the PDF or the PMF or the distribution we are trying to consider now if it is a discrete distribution we will have basically the probability mass function that means the probabilities are all concentrated on masses and if it is continuous it is known as a probability density function where the random variable if you are trying to understand or trying to study are continuous in nature now let us come back to the example of a probability mass function this example which because in these two examples of a tossing a coin are rolling a dying now when you basically me roll the 9 the probabilities are concentrated at the values of realize values which are either any one of them either one so hence the corresponding probability which is being assigned or the masses which are being assigned to the value 1 and the corresponding probability is 1 by 6 if you come to the example of say for example when you are trying to analyze the probability for the real as well as 2 again it is 1 by 6 where it is 3 it is again 1 by 6 4 it is again 1 by 6 5 onion 1 by 6 6 again 1 by 6 which means that the probabilities are concentrated at the varies 1 to 6 but any probability values at the point other than 1 to 6 are 0 now you may be asking that what does it mean it doesn't show such then when you are trying to form a plot the value of the CDF which is the cumulative distribution function I will come to that later on but in the case when you are trying to analyze the property density function we do not consider the densities are are concentrated on only points and alack basically analyze that in later not mean more details so now we will consider the discrete distributions and I'll and and whatever I've talked about showing that once life for about 10 minutes I will go into the details about that discussion later on also so the dismiss distribution which will consider for our example in this da dm1 course would be the uniform discrete distribution the binomial distribution the negative binomial distribution the geometric distribution the hyper geometric distribution the Poisson distribution the log distributions and there are other distribution also which will not consider in detail these would be more than sufficient to basically at least cover the probability mass function now before we discuss the probability and discrete probability distribution or the PMF probably mass functions we like to basically go into the concept very simple concept of burning your trials now in the burning or trial we have basically two outcomes it can be more than two outcomes also not for the binary travels but for the experiment but we are not going to go into the details of that so there are I'm an experiment our trial which you do or an X or some random experiment it you do and there are two possible outcomes say there's a success and a failure so that very good example would be a head and a tail but if you are considering rolling the die it is not if and only if we are able to divide the rolling other die into real not by into even numbers and odd numbers obviously there are two categories then it can be said depending on how you have been able to formulate the experiment as Bernell a trial but let us not go into such details but let us come to the simple example so each trial or the binary tree I have put has two possible outcomes which is a success or a failure or a head or a tail or a yes or no whatever it is these trials are independent that means when we do the experiment the the outcome of the head or outcome the yes an outcome of the red we are outcome of of white whatever it is doesn't affect the complimentary event of that that is for the yes it is a no for a head is a tail for a white is a black whatever it doesn't affect that and vice versa also that means the trials are independent also which basically means that if I do the experiment repeatedly like I roll the die and I want to find out the probability of getting a even number or what number or the experiment which I do is basically I toss the coin I want to find out the head or the tail if I keep doing this tosses these are each trials these trials are independent of each other that means one trial does not affect the next or the next trial does not affect the form one so they are independent of each other now the point third point I have already mentioned it means the probability of a success or a failure remains the same and and it is not doesn't change from experiment to experiment now there are other important point which I also mentioned consider that you have a coin and you keep tossing it so the trials do not affect each other they are independent that was point number two as mentioned in the slide point number three that the outcome of ahead or outcome of a tale whatever the outcomes are we have denoted in toss in the quiet are independent of each other do not affect that means it is not that if the outcome of the head and the tail for an unbiased coin is half and half the probability doesn't change from trial to trial it remains half and head doesn't affect the tail in the next outcome or head doesn't affect the head from the last outcome and it or tails tail doesn't affect the tail or tail doesn't affect the head and it continues in that way in case if you have a trial whether three outcomes the same concept can be you place where the trials will be independent of each other there will be three tails whatever the probabilities are they can be P 1 P 2 P 3 and the probabilities of getting the first one which is P 1 that would not affect p2 or p3 and vice-versa in the sense P 2 de would not affect P 1 and P 3 similarly P 3 would not affect P 1 and P 2 and it will continue in this way so let us consider the first discrete distribution which is the uniform discrete case so X is the random variable and we denote as I'm just hovering my pen here it means UD which is the uniform destitute distribution and this a and B are the parameters based on which we are trying to basically find out the uniform discrete distribution case so U is uniform d is discrete a and B other parameters now the f of X which is the PMF is denoted by 1 by n n is basically the number of such trials which we have and the realest values off of X which is the random variables are in the in the least case you can have basically the minimum value which is a and the corresponding masses are concentrated at points which are a plus K that means the next jump is basically from a to a plus K the next jump further on would be a plus k plus K which is a plus 2k and it will go on till the last value which is B so that means the masses are concentrated at the reales values which are a a plus K a plus 2k so on and so it'll be and they are such n number of such masses where the probabilities can be found out here a and B are the parameters which you're on the airline the expected value I'll come to the expected value and and the variances concept later on when we solve the problems so they expand and they will be utilized later rather also the expected value is given by a plus K by 2 into n minus 1 and the variance is given by K square into n square minus 1 divided by 12 so what can be the example example can be generating the random numbers 1 2 3 4 till 10 here the random distribution of the probability mass function will be given by X is renewed from distorted between 1 and 10 so a is 1 B is 10 where the values of K provided they're only discrete number like 1 2 3 4 in that case a is 1 then a plus K is 2 so hence K would be 1 and you can find out that what are the different numbers of ends which you have based on which you can formulate the ppm ppm F which is the probability mass function we can find out the expected value can find out the variance also now remember one thing for the uniform distribution case if you are able to draw it how it will look like I'll come to that later on and as also signify the main concept of the expected value there's a reason for that and I'll come to the variances expected value for the other distributions later on also so this is a uniform discrete distribution I have do it in very simple X n so the the values are of the XS are from 1 to 20 the corresponding probabilities are given by 0.1 so this is a uniform discrete so the heights are all equal so f of X is plotted along the y axis and X values are plotted along the x axis and if you find out the expected value would be center of gravity of this figure so obviously the expected value would be denoted by somewhere in between so is this I'm just using a very simple concept center gravity which we know in physics are very simple plus 10 not to a twelve concept of simple sciences so babe so center gravity is basically where you can basically balance it and the weight would basically be kept balance like if you have a scale scale of 1 feet which is 12 inch or 30 centimeters if you balance it at the point which is 15 centimeters it will be keep well in the pose because the weights on the left-hand side and the right-hand side are equal so it is the same way what what we denote so we will consider now the binomial distribution so binomial distribution is given by if you remember the binomial expansion which we are done in hinting in class so it is given by basically a plus X to the power n so you want to basically expand it so the first term would basically be NC 0 a to the power n X to the power 0 and the last term would basically be NC n a to the power 0 X to the power n so basically you find out all the powers of X and a as they keep changing such that X increases from 0 to n and simultaneously a basically changes from n to 0 such that the power values when they added up for a a and X are always n now here n NP are the parameters and the random variable is X B is the binomial distribution the parameters are P and n now P is the probability of the of the good thing or the bad thing whatever it is however you do not or yes or no so if it is P is for yes then probability of no would be 1 minus P if P is no then the probability of es would be 1 minus P whatever way you denote now the f of X is given by NC xB to the power X Q to the power n minus X and the X values are from 0 to n so a X being 0 means the corresponding probability of getting the good item is 0 there is no good items and prob and and the values of the all the bad atoms are all then and if basically X is n it means all the good items are are there while the number of bad items are 0 so N and P are the parameters so P is between 0 & 1 and is integers positive expected value is given by n into P varies is given into NP Q so consider an example you are checking the quality of the product coming out from the shop floor a product can either pass with the probability appointed and fail with probability 0.2 it can also be the other warehouse like me I want to find out the probability of all the bad items always mean that is P is 1 2 and Q is point a now you are checking 50 such items which n is 50 if X is the random variable which denotes the number of success then for the inspection of such 50 items it will be 50 CX which is n is 50 P to the bar X which is 0.8 to the bar X and Q to the power n minus X which is 50 minus X would be given by 0.2 to the power 50 minus X now remember if you want to find out the bad items again the formula remains the same initial part 50 CX would remain the same it will be point 2 to the power X and point 8 to the power 50 minus X but in that case X is a number of bad items while in this case which is shown on the slide is the number of good items so with this I will end this seventh lecture and continue the discussion further on with examples for the binomial distribution so on and so on and so forth have a nice day thank you very much [Music] [Music] [Music]