Lec 7, Introduction to Probability-II

[Music] [Music] [Applause] [Music] [Applause] [Music] dear students will continue that the concept of probability in lecture number seven will take one sample then we will try to understand the concept of marginal probability joint probability conditional probability in this problem the problem is a company data reveal that 155 employees work at any of four types of positions shown here again is the raw values matrix also called a contingency table with the frequency count for each category and subtotals and totals containing a break down of these employees by topic type of position and the section look at this contingency table in the row it is given what kind of position they are holding whether managerial professional technical clerical in column what is their sex suppose if an employee of the company is selected randomly what is the probability that the employee is female the contingency table row represents the type of position column represent 6 the type of position they may hold whether they can have them they can work as a managerial position professional position technical clerical we have asked the other six also this data says the question is if an employee of the company is selected randomly what is the probability that the employee is female or professional worker that is what is a P of F Union P F is the female P is the professional so as per the law of addition of probability P of F Union P equal to P of AF plus P of P minus P of F intersection P P FA F we can find out and going to the previous slide so the P of F is there are 55 females total there are 155 so when you divide 55 greater by 155 you will get 0. 355 then P of P going to previous light P of P is professionals there are 44 professionals 44 / 155 that will give you 0. 28 for then - P of F intersection P that is good female is a tape same time they're working in a professional type of position that is 13 so 13 Tudor by going back 13 - 150 I will give you 0.

08 for that is equal to zero point five five five shown here are the raw values matrix and corresponding probability matrix of the result of a National Survey of 200 execute EES who are asked to identify the geographical location of their company and their company's industry type so they are asking to question what is their company's geographic location and what kind of industry they are working the executi we're only allowed to select one locale and one industry because they can work only one industry in one location okay this table shows you see row there is a industry type may be finance manufacturing communications they're calling it ABC the geographic location may be north east south east Midwest and west so for example in the finance yay there are 56 people are working manufacturing be there are 70 people are working the communications 74 in the Northeast location there are 82 people in southeast location 34 people and Midwest location 42 and West 42 the question is what is the probability that the responded days from the Midwest directly we can read this answer from the table second question is what is the probability that the respondent is from the communications industry are from the Northeast so here the addition of the two probability that is a P of C and P of D the third question is what is the probability that the respondent is from the southeast are from the finance industry okay so from the given table we find out the probability that means we have divided each element in the cell donor by the the grass total after dividing that we got 0. 12 point zero five point zero four point zero four now this is a matrix a conditional probability from this table we can pick up whatever answer we wanted to get for answering that question for example okay and going back suppose what is the probability that the respondent is from the Midwest so I'm going to next light Midwest is EF so it is a point two one going back what is the probability that the responder DS from the communication industry are not this so you are to see P of P of C Union D equal to P of C plus P of D minus P of C intersection D you can find out then what is the probability that respondent is from south east are the same thing P of e Union J equal to P of E Plus P of a minus P of e intersection yay these values you can directly pick up from the previous table from this table now we'll go for mutually exclusive events suppose okay suppose if you want to find out P of T Union C here T is the those who are technical those who are technical professional P of T Union C is P of T plus P of C because generally it will be minus P of T intersection C but that is not possible because a person cannot work in two industry at a time so it is mutually exclusive event in the mutually exclusive event in the the intersection will become zero that is a P of T Union C equal to P of T plus P FC that another term that is minus P of T intersection C will become zero so P of T is a 69 by 125 plus P FC is 31 out of a one-foot when you simplify it is zero point six four five this example of mutually exclusive event there is another P of P you you see P of P plus P FC there won't be any intersections even that formulas which I told the two slides before also that intersection component will become zero because the person cannot work in two industries okay there is an example of mutually exclusive event in the law of multiplication P of X intersection y equal to P of X multiplied by B of Y given X equal to P of y equal to P of X given Y this is a law of multiplication what will happen if event x and y are independent event simply we can multiply P of X into P of Y then we will see later you'll see another problem your company has so 140 employees of which 30 are supervisors 80 of the employees are married and 20 percentage of the married employees are supervisors if your company employees randomly selected what is the probability that the employee is married and is a supervisor wherever this kind of problem comes if you're able to construct a contingency table whatever questions being asked you can pick up from there directly so from the given data first we will construct a contingency table in the contingency table you see there are 140 employees out of which 30 or supervisors out of 140 80 people are married and for example those who are married at the same time supervisors what you have to do that is you have to multiply that how will you that if you multiply 80 into 32 by 140 you will get this answer so for example you see that how we are getting that we have the previously we get point 1 1 4 3 we will see how we are getting so we know the probability of married people 1800 by 140 this is given those who are supervisor at the same time married 0. 2 there is a 20% is given if you want to know those who are married at the same time their supervisors so we would use conditional probability P of M multiplied by P of yes by M yes given M so P of M is known P of point was given so 0.

1 143 okay otherwise if you go back in this light if you multiply 80 with 30 and divided by 140 you will get the same answer you see that once you know that one cell in the contingency table filling the remaining cell is so easy okay and whatever value you want to become we can pick up for example I have filled the first 0. 1 143 I know what is the row total and column total from that I can substrate it I can get the remaining on the side so that is an application of contingency table suppose P of s equal to 1 1 minus P of s that we know that P of s bar equal to 1 that is a 0. 5 87 I am saying this one this location this location 0.

7 and 5/7 if you want to know P of M bar intersection s but that means those were not M those are not married at the same time or not Supervisors so that that is nothing but this location 0. 326 those were not married at the same time they are not supervisor at these locations P of M intersection s but those are married but not the supervisors right so that is a P of M minus P of M - yes so this value is 0. 5714 - this one so we will get this point nothing but in the contingency table if you know one cell the remaining things can be found out the special law of multiplication for independent event generalize if P of X intersection y equal to P of ax material B P of Y given X equal to P of Y multiplied by P of X given Y special law that is if x and y are independent so P of X equal to P of X given Y because when the event x and y are independent the outcome of X is not depending on outcome of Y so P of x given Y will become P of P of X itself so similarly P of Y and a exofit non-independent P of Y given X will become P of Y itself and you substitute there so P of X intersection Y will be P of X into P of Y only when they both even sir independent this also law of conditional probability this also we have seen previously also the conditional probability of x given Y is joint probability of x and y divided by marginal probability of Y so joint probabilities intersection the P of X intersection Y tournament P of Y so this can be just by readjusting B of Y given X multiplied by P of X by P of Y little detailed explanation on conditional probability a conditional probability is the probability of one event given that another event has occurred suppose if I say P of a given B so first you have to find the intersection of P of a and B then divide by P of B so the conditional probability of a given that B has occurred this is an explanation for this suppose if you want to know P of B given a has occurred so P of a intersection B divided by P of a where P of a and B equal to joint probability of a and B so P of a is marginal probability of event E a pure B is marginal probability of event B we'll take an example how to find out the conditional probability of the car sunny a used-car lot 70 percentage how we are conditioning and forty percent I have CD player twenty percent of the cars have both so what is the probability that car has a CD player given that it has AZ that means we want to find out P of CD given that yes is there as I told you just to you draw the contingency table because all the values are given so what is the value I am going back see for example 70 percentage of the cars having a AC so this value and going back 40 percentage of the cars having CD player so this value and you substitute minus 1 you will get that one another information is given 20% of the cars have both go back see that point to this value so once you know these values other cells can be easily find out so if you want to know PFC D given AC as per the definition PFC D and EAC VP of AC so this is a point - this is by P of a AC is 0.

7 0. 2 8 5 7 so given AC we only consider the top row 70 percents the cars of these and 20 personal CD player so 20 percentage of 70 percentages 28. 5 son person age okay there so we are getting the conditional probability so this conditional probability can be explained with the help of a tree diagram because the tree diagram is easy to visualize so having AC having not AC having CD having not CD having CD having not CD so 0.

7 0. 2 0. 5 0.

2 0. 1 so if you want to know having CD so you have to divide point to dude by 0. 7 okay point two for example if you want to know this this arc so if this is a point five by 0.

7 and so on because the tree diagram is very easy to understand then we will see the definition of independent event if an x and y are independent events the occurrence of Y does not affect the probability of X occurring so x and y are not connected similarly if X in X and y as independent even the occurrence of Y does not affect the probability of Y occurring you see that P off if x and y are independent P of x given y equal to P of X P of Y given X equal to P of Y this we have seen the please also okay this is an another example two events are independent this is the condition P of a by B equal to P of a given B equal to P of a so this condition is for testing independent even a and B are independent when the probability of one event is not affected by other event so we will take an example we'll check the the practical application of this concept of independent events this also this data previously given so we have asked which what kind of industry you are working whether you finance manufacturing communications then we ask you the juggling locations now you see that the question is tested the matrix for the 200 executed respondents to determine whether the industry type is independent of geographical location that means we were to find out is there any dependency between the jariabek location and what kind of Industry okay for example in India if you look at their most of the software companies are in South so is there any connection between the geographic location and kind of Industry which are located ok we'll take this example finance and the best reason so when you go this if you want to know P of Y a given G that is P of a intersection G will be P of Z P of a intersection G we can directly read from the table Oh point zero seven this one this value then P of G point to one directly we can read from the table so what do you do that value is 0. 33 but P of a when you look at P of a so P of a is 0. 28 so now what is happening the P of a given G is not equal to P of a if it is equal both are independent since it is not equal there is a kind of your dependency between the geographical location and the type of Industry which are located there so this is a one way to test the in dependency for example you take or another example because J given D this is an another example soapy of here a intersection D is 8 here the actual count is given okay anyway you can get to both ways PRV a intersection D is y 8 little bit P of D P of D is 34 so we are getting this value but you see the P of a is 20 divided by 85 22 by 85 so both P of a by D and P of a are same so these are independent events there's an example if it is same P of a given D equal to P of a both the events are independent this is the way to test the independent next we are going to an important application that is the Bayes rule or Bayes theorem it is used to revise the probabilities it has lot of applications in higher level of probability theory an extension to the conditional law of probabilities enables revision of original probability with the new informations so P of x given y equal to P of Y given X into P of X divided by the summation of this one so I'll tell you then let the next slides for example suppose see that see P of X Y P of Oh Y X so this can be written as P of X intersection Y triggered by P of Y this can be written as P of Y X intersection Y little by P of X you look at this here also P of X intersection Y this also P X intersection Y so this can be written is P of X Y multiplied by P of y equal to P of Y by Y X multiplied by gfx you see that if I look at this suppose I know P of X given Y suppose if we want to know the reverse of this that is P of y by X you see that I know P of X by Y I'm getting the reverse of that that is P of Y by X so from this you can write it P of Y by X is nothing but P of X by Y tutor by P of Y divided by P of X here the P of X is only because the only two outcomes there are if there are more outcomes here the Sigma of P X will come the Sigma of P X is nothing but different combination of joint probabilities that we will see with the help of an example this is a very typical example machine a and B and C all produce the same two parts x and y all of the parts produced machine a produces 60 percentage machine B produces 30 percentage and mission C produces 10% A's in addition 40 percentage of the parts made by a mission a are part X 50 percentage of the parts made by mission BR part X 70 percent of the part made by machine see his part X a part produced by this company is randomly sampled and determined to be an expert with the knowledge that it is an expert he revised the probabilities that the part came from machine a B and C first we will solve this one with the help of a tabular format for example there are three mission as their mission a and B see that 60 percentage of that part was produced by Machine a 30 percentage was produced by machine B 10 percent is by C previously we have seen how that formula for conditional probability has come now I will tell you an application of Bayes theorem suppose there are two say there are two supplier supplier ei supplier B I know that say the 40 percentage of the product supplied by supplier B remaining 60% ages supplied by supplier B from my past experience I know that the two percentage out of 40 person days the 2 percentage will be defective product which are supplied by supplier a from supplier B I know from my past experience he used to supply 3 percentage of defective production out of 16 by using their products that I have assembled a a new machine now the machine is not working the machine is not working but I want to know what is the probability that product was supplied by supplier a the machine is not working what is the probability that the product was supplied by supplier B ok so this is the application of your BS theorem ok we will see with the help of an example the different options there it the problem is a particular type of printer ribbon is produced by only two companies that company name is our Alamo ribbon company and South Jersey products suppose allama produces 60 percentage of the ribbons and South Jersey produces 35 percentage of the ribbon from over experience look at this 8 percentage of the ribbons produced by Aloma are defective and 12 percentage of the South Jersey ribbons are defective from our past experience a customer purchases a new ribbon what is the probability that allama produces the ribbon otherwise what is the probability that South Jersey produced the ribbon like in the previous example the machine is not working what is the probability that product was supplied by supplier eh what is the probability that product was supplied by supplier me the same example ok now first we will find out the marginal probability and conditional probability so P of Aloma that is 6 if a percentage of the product was supplied by Aloma south jersey 35 percentage from the past experience I know the defective parts which was supplied by Aloma supplier is point zero eight similarly the defective products which were supplied by South Jersey's point one two you see that this sum won't be 100 but this sum will be hundred because this is a total they supply here eight percent days is in the sixty five you out of sixty five eight percent days it defective products are supplied by a llama person now as per the formula now we look at this V now it is a reverse now we know it is it defective then we want to find out what is the probability that was supplied by Aloma so we look at this P of D given by Aloma multiplied by P of Aloma so look at this this component this is the sum of all possibilities P of D given a llama multiplied by P of P Aloma plus P of D given by e soldier C multiplied by P of soldier C so this point zero it is given 0.

65 is given so this was combination of 0. 08 and point six five this and this this was combination of 0. 35 and point one two so when you but here because we have to add these two so when you divide by this is 0.

55 three that is if the prod the ribbon is defective then there is a 50% chance it was supplied by supplier Aloha similarly the product is defective what is the probability that it was supplied by soldier see this MT 0. 12 multiple by 0. 35 because P of D allama is given then point zero eight this is the all combination this denominator is same so 0.