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# QT1 - 04 - Probability

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Class notes used in Quantitative Techniques - I course at Praxis Business School, Calcutta

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### QT1 - 04 - Probability

1. 1. Probability Theory Q U A N T T E C H I N T E U Q I A S E V I T 1 0 S
2. 2. Contents <ul><li>Probability </li></ul><ul><ul><li>Classical </li></ul></ul><ul><ul><li>Relative </li></ul></ul><ul><ul><li>Subjective </li></ul></ul><ul><li>Marginal Probability </li></ul><ul><li>Joint Probability </li></ul><ul><ul><li>Exclusive Events </li></ul></ul><ul><ul><li>Independent Events </li></ul></ul><ul><ul><li>Dependent Events </li></ul></ul><ul><li>Bayes Theorem </li></ul>
3. 3. Probability : Basic Terminology <ul><li>Event </li></ul><ul><ul><li>One or more of the possible outcomes of doing something </li></ul></ul><ul><ul><ul><li>Toss a coin </li></ul></ul></ul><ul><ul><ul><ul><li>Getting head </li></ul></ul></ul></ul><ul><ul><ul><ul><li>Getting tail </li></ul></ul></ul></ul><ul><ul><ul><li>Person enters a mall </li></ul></ul></ul><ul><ul><ul><ul><li>Person is male </li></ul></ul></ul></ul><ul><ul><ul><ul><li>Person is a student </li></ul></ul></ul></ul><ul><ul><ul><li>Pick a carton of candies at a store </li></ul></ul></ul><ul><ul><ul><ul><li>Candy is Amul </li></ul></ul></ul></ul><ul><ul><ul><ul><li>Candy is spoilt </li></ul></ul></ul></ul><ul><li>Experiment </li></ul><ul><ul><li>An activity that produces or causes an “event” is referred to as an “experiment” </li></ul></ul><ul><ul><li>Tossing a coin is an experiment. </li></ul></ul><ul><ul><ul><li>Tossing three coins in a row is an experiment </li></ul></ul></ul><ul><ul><li>Choosing a person at the entrance of a mall door and asking questions is an experiment </li></ul></ul><ul><ul><li>Pick a carton of candies at a store </li></ul></ul>
4. 4. Probability : Basic Terminology <ul><li>Mutually Exclusive Events </li></ul><ul><ul><li>Only one of the events can take place at a time </li></ul></ul><ul><ul><ul><li>Either Head or Tail </li></ul></ul></ul><ul><ul><ul><li>Either Man or Woman </li></ul></ul></ul><ul><li>Non Exclusive Events </li></ul><ul><ul><li>Both or more can take place </li></ul></ul><ul><ul><ul><li>Person is male </li></ul></ul></ul><ul><ul><ul><li>Person is student </li></ul></ul></ul><ul><li>Collectively Exhaustive List </li></ul><ul><ul><li>List of events that between them consider all cases </li></ul></ul><ul><ul><ul><li>Toss of coin : Head or Tail – nothing else is possible </li></ul></ul></ul><ul><ul><ul><li>Choice of cold drink </li></ul></ul></ul><ul><ul><ul><ul><li>Coke </li></ul></ul></ul></ul><ul><ul><ul><ul><li>Pepsi </li></ul></ul></ul></ul><ul><ul><ul><ul><li>Anything else </li></ul></ul></ul></ul>
5. 5. Classical Probability <ul><li>Probability of an event is defined as </li></ul><ul><li>[ number of outcomes where event occurs ] </li></ul><ul><li>[ total number of possible outcomes ] </li></ul><ul><li>Experiment of tossing a single coin </li></ul><ul><ul><li>P(Head) = 1 / 2 = 0.5 </li></ul></ul><ul><li>Experiment of rolling a dice </li></ul><ul><ul><li>P(getting 6) = 1 / 6 = 0.167 </li></ul></ul><ul><li>Experiment of choosing a person at the mall </li></ul><ul><ul><li>P(Person is male ) = 0.5 [ unless we know more ] </li></ul></ul><ul><ul><li>P(Person is a student ) = not known as yet </li></ul></ul>Problem with Classical Probability
6. 6. State Space All Possible Outcomes Event A : Head Event B : Tail Event A : Identical Reading Event B : Different Reading Event A : Three Heads Event B : Two Heads Event C : Anything Else One Coin Tossed Two Coins Tossed Three Coins Tossed H T HH HT TH TT HHH HHT HTT THH HTH THT TTH TTT
7. 7. Probability is NOT certainty ! H T T T T T H T T H P(H) = 0.5 Probability of getting H = 0.5 H H T T T H H H T H T T T T T H H T T T T H H H T H H T H T H H H H H T H H H T 3/10 3/10 6/10 9/20 2/10 11/30 6/10 17/40 8/10 25/50 T T H H T T H T H T 4/10 29/60 H T H H T T H H H T 6/10 499/1000 M a n y m o r e t r I a l s
8. 8. Simple Example <ul><li>Classical Approach </li></ul><ul><ul><li>I have three friends Ram, Gopal, Bharat each of whom is equally likely to visit me for lunch </li></ul></ul><ul><ul><li>What is the probability that today it will be Ram ? </li></ul></ul><ul><li>Total 3 outcomes possible : R G B </li></ul><ul><ul><li>One is the outcome in question : R </li></ul></ul><ul><li>P(R) = 1/3 </li></ul><ul><li>In the past one month </li></ul><ul><ul><li>R has visited 20 times </li></ul></ul><ul><ul><li>G has visited 5 times </li></ul></ul><ul><ul><li>B has visited 5 times </li></ul></ul><ul><li>What is the probability that today it will be Ram ? </li></ul><ul><li>Is it P(R) = 1/3 </li></ul>
9. 9. Relative Frequency <ul><li>Probability calculated from statistical data on the relative frequency of certain occurrences </li></ul><ul><ul><li>Observed relative frequency of an event in a very large number of trials </li></ul></ul><ul><ul><li>Proportion of times that an event occurs in the long run when conditions are stable </li></ul></ul>Carbon monoxide emissions
10. 10. Challenges with relative probability <ul><li>Number of events considered </li></ul><ul><ul><li>Suppose you know of three of your friends who have had jaundice this year </li></ul></ul><ul><ul><li>Can you conclude that the probability of getting jaundice is higher this year than in the last year ? </li></ul></ul><ul><li>Long run stability </li></ul><ul><ul><li>Suppose Tata Motors brings a new model of Indica in the market and in the first three months 10% cars have a clutch failure. </li></ul></ul><ul><ul><li>Can you conclude that the probability of a clutch failure in Indica cars is 0.1 ? </li></ul></ul><ul><ul><li>If you toss a coin </li></ul></ul><ul><ul><ul><li>Two times you may get two heads </li></ul></ul></ul><ul><ul><ul><li>100 times, you would get between 45 – 55 heads </li></ul></ul></ul>
11. 11. Two Questions <ul><li>A yoga club consists of 10 members of whom </li></ul><ul><ul><li>2 Doctors, 3 Engineers, 3 CA and 2 Other </li></ul></ul><ul><ul><li>4 women, 6 men </li></ul></ul><ul><ul><li>7 married, 3 single </li></ul></ul><ul><li>If we were to choose a club secretary by lottery, what is the probability that the secretary is a </li></ul><ul><ul><li>Doctor ? </li></ul></ul><ul><ul><li>Woman ? </li></ul></ul><ul><ul><li>Single person ? </li></ul></ul><ul><li>The following table shows the frequency of marks in a quiz </li></ul><ul><li>If I were to ask a random student what is your score, what is the probability that it is in the range of </li></ul><ul><ul><li>2 – 3 ? </li></ul></ul><ul><ul><li>6 – 7 ? </li></ul></ul>
12. 12. Points to note <ul><li>Numerical value of a probability of any event is between 0 and 1 </li></ul><ul><ul><li>0 : if it is impossible, e.g. The probability of teacher being an alien from outer space </li></ul></ul><ul><ul><li>1 : if is totally certain, e.g. The probability of teacher being a human being </li></ul></ul><ul><li>Sum of the probabilities of events that must be 1, provided the events are </li></ul><ul><ul><li>Mutually exclusive </li></ul></ul><ul><ul><li>Part of a collectively exhaustive list </li></ul></ul><ul><li>Something must happen !! </li></ul>
13. 13. Subjective Probability <ul><li>Based on beliefs, not hard facts .. </li></ul><ul><ul><li>Because hard facts are simply not available and you cannot sit and do nothing because you do not have facts </li></ul></ul><ul><li>Widely used in cases where the event in question occurs very rarely </li></ul><ul><li>To be replaced with statistical models if possible </li></ul><ul><li>Consider </li></ul><ul><ul><li>Selection of a candidate for a job </li></ul></ul><ul><ul><li>Sanction of a loan </li></ul></ul><ul><li>Earlier these were all subjective but now with more data being collected </li></ul><ul><li>Strategic Decisions are still taken on the basis of subjective probability !! </li></ul><ul><li>Classical Probability </li></ul><ul><li>Relative Probability </li></ul><ul><li>Subjective Probability </li></ul>
14. 14. Marginal Probability [unconditional probability] <ul><li>Probability of an event is written as </li></ul><ul><ul><li>P(A) = n where </li></ul></ul><ul><ul><li>A is the event that secretary is </li></ul></ul><ul><ul><ul><li>Doctor </li></ul></ul></ul><ul><ul><ul><li>Engineer </li></ul></ul></ul><ul><ul><ul><li>CA </li></ul></ul></ul><ul><ul><ul><li>Other </li></ul></ul></ul><ul><ul><li>n is the value of the probability 0 <= n <= 1 </li></ul></ul><ul><li>Marginal or Unconditional Probabilities are as follows : </li></ul><ul><ul><li>P(Doctor) = 0.2 </li></ul></ul><ul><ul><li>P(Engineer) = 0.3 </li></ul></ul><ul><ul><li>P(CA) = 0.3 </li></ul></ul><ul><ul><li>P(Other) = 0.2 </li></ul></ul><ul><ul><li>yoga club of 10 </li></ul></ul><ul><ul><ul><li>2 Doctors, 3 Engineers, 3 CA and 2 Other </li></ul></ul></ul><ul><ul><ul><li>4 women, 6 men </li></ul></ul></ul><ul><ul><ul><li>7 married, 3 single </li></ul></ul></ul>
15. 15. Venn Diagram / Exclusive Events <ul><li>P(Doctor) = 0.2 </li></ul><ul><ul><li>2 </li></ul></ul><ul><ul><li>2 + 3 + 3 + 2 </li></ul></ul><ul><li>P(Engineer) = 0.3 </li></ul><ul><ul><li>3 </li></ul></ul><ul><ul><li>2 + 3 + 3 + 2 </li></ul></ul><ul><li>P(Doctor OR Engineer) </li></ul><ul><ul><li>2+3 </li></ul></ul><ul><ul><li>2 + 3 + 3 + 2 </li></ul></ul><ul><li>P(A or B) = P(A) + P(B) </li></ul><ul><li>Provided the events are exclusive </li></ul>Doctor 2 CA 3 Engineer 3 Other 2
16. 16. Venn Diagram / Non-Exclusive Events <ul><li>P(Engineer) = 0.3 </li></ul><ul><ul><li>3 </li></ul></ul><ul><ul><li>2 + 3 + 3 + 2 </li></ul></ul><ul><li>P(Married) = 0.7 </li></ul><ul><ul><li>7 </li></ul></ul><ul><ul><li>7+3 </li></ul></ul><ul><li>P(Engineer OR Married) </li></ul><ul><ul><li>= P(Eng) + P(Mar) – P(Eng AND Mar) </li></ul></ul><ul><li>P(A or B) = P(A) + P(B) - P( AB ) </li></ul><ul><li>Provided the events are non-exclusive </li></ul>Single 3 Married 7 Doctor 2 CA 3 Engineer 3 Other 2
17. 17. Double Counting of Non Exclusive Events <ul><li>P(Engineer OR Married ) </li></ul><ul><ul><li>= P (Single Engineer) + P(Married Engineer) + P(Married NonEngineer) </li></ul></ul><ul><ul><li>= (3 – x)/10 + x/10 + (7 -x)/10 </li></ul></ul><ul><ul><li>= 3/10 + 7/10 – x/10 </li></ul></ul><ul><ul><li>= P(Engineer) + P(Married) – P(Married AND Engineer) </li></ul></ul><ul><li>Since Profession and Marital Status are NOT EXCLUSIVE </li></ul>All Engineer = 3 All Married = 7 Single Engineer = 3 - x Married, non Engg = 7 - x Married, Engg = x Engineers, married = x
18. 18. Non Exclusivity : Question <ul><li>Probability of Either of Two Events A, B = </li></ul><ul><li>P( A OR B ) </li></ul><ul><li>= P(A) + P(B) </li></ul><ul><ul><li>If A and B are exclusive events </li></ul></ul><ul><li>= P(A) + P(B) – P(A AND B) </li></ul><ul><ul><li>If A and B are not exclusive </li></ul></ul><ul><li>= P(A) + P(B) - P(AB) </li></ul><ul><li>A Quality Control inspector checks cartons of breakfast cereals for </li></ul><ul><ul><li>Correctness of weight </li></ul></ul><ul><ul><li>Damage to cartons </li></ul></ul><ul><li>Historical data shows that </li></ul><ul><ul><li>1% cartons are underweight </li></ul></ul><ul><ul><li>0.5% cartons are damaged </li></ul></ul><ul><ul><li>0.5% cartons have both flaws </li></ul></ul><ul><li>If he inspects 1000 cartons how many would he expected to reject </li></ul>
19. 19. Where are we ? 1 <ul><li>Classical Probability </li></ul><ul><li>Relative Probability </li></ul><ul><li>Subjective Probability </li></ul><ul><li>Marginal Probability P(A) </li></ul><ul><li>Probability of A or B </li></ul><ul><li>Exclusive Events P(AorB) = P(A) + P(B) </li></ul><ul><li>Non Exclusive Events P(AorB) = P(A)+P(B)-P(AB) </li></ul>
20. 20. Joint Probability of Two Events <ul><li>What are event pairs ? </li></ul><ul><ul><li>Person is DOCTOR, Person is MARRIED </li></ul></ul><ul><ul><li>Carton is DAMAGED, Carton is UNDERWEIGHT </li></ul></ul><ul><li>Major Question </li></ul><ul><ul><li>Are the two events in the pair dependent or independent </li></ul></ul><ul><li>Consider the following </li></ul><ul><ul><li>Person is DOCTOR, Person is MARRIED </li></ul></ul><ul><ul><ul><li>Two events seem to be independent </li></ul></ul></ul><ul><ul><li>Person is DOCTOR, Person has PAN number </li></ul></ul><ul><ul><ul><li>Two events seem to be dependent </li></ul></ul></ul><ul><li>Whether Independent or Not Independent is to be determined OUTSIDE Probability theory </li></ul>
21. 21. Back to the Yoga Club <ul><li>A yoga club consists of 10 members of whom </li></ul><ul><ul><li>2 Doctors, 3 Engineers, 3 CA and 2 Other </li></ul></ul><ul><ul><li>4 women, 6 men </li></ul></ul><ul><ul><li>7 married, 3 single </li></ul></ul><ul><li>If we were to choose a club secretary by lottery, what is the probability that the secretary is a </li></ul><ul><ul><li>Doctor AND Woman </li></ul></ul><ul><ul><li>Man AND Single </li></ul></ul><ul><ul><li>Engineer AND Married </li></ul></ul><ul><li>Events are deemed to be independent </li></ul><ul><ul><li>Profession, Gender and Marital Status are independent of each other </li></ul></ul><ul><li>We are interested in </li></ul><ul><ul><li>P(Doc Wom) </li></ul></ul><ul><ul><li>P(Man Sing) </li></ul></ul><ul><ul><li>P(Engg Marr) </li></ul></ul>
22. 22. Joint Probability under Statistical Independence <ul><li>Probability of two or more independent events happening together or in succession is the product of their individual marginal probabilities </li></ul><ul><li>P(AB) = P(A) x P(B) </li></ul><ul><li>Consider the Yoga Club </li></ul><ul><ul><li>A = person is DOCTOR, B = person is WOMAN </li></ul></ul><ul><ul><li>P(A) = 0.2 P(W) = 0.4 </li></ul></ul><ul><ul><li>P(DW) = 0.2 x 0.4 = 0.08 </li></ul></ul><ul><ul><li>=> that out of hundred possible outcomes, there will be 8 where the secretary will be a WOMAN DOCTOR </li></ul></ul><ul><ul><li>Is this correct ? Let us check </li></ul></ul>
23. 23. Classical Definition of Probability When we choose a secretary, there are 100 possible outcomes .. But only 8 of these outcomes will give us a WOMAN DOCTOR
24. 24. Points to note <ul><li>Can be extended to more than two events </li></ul><ul><ul><li>P(ABC) = P(A) x P(B) x P(C) </li></ul></ul><ul><ul><li>P(Married Woman Doctor) </li></ul></ul><ul><ul><li>= P(Married) x P(Woman) x P(Doctor) </li></ul></ul><ul><ul><li>= 0.7 x 0.4 x 0.2 = 0.056 </li></ul></ul><ul><ul><li>= 56 out of 1000 possible outcomes </li></ul></ul><ul><li>If we consider all possible outcomes ... </li></ul><ul><ul><li>P(M-W-D) + P(S-W-D) + P(M-M-D) + P(S-M-D) + </li></ul></ul><ul><ul><li>P(M-W-E) + P(S-W-E) + P(M-M-E) + P(S-M-E) + </li></ul></ul><ul><ul><li>P(M-W-C) + P(S-W-C) + P(M-M-C) + P(S-M-C) + </li></ul></ul><ul><ul><li>P(M-W-O) + P(S-W-O) + P(M-M-O) + P(S-M-O) </li></ul></ul><ul><ul><li>= 1.000 ! </li></ul></ul>
25. 25. Points to note <ul><li>This result holds only if the outcomes are independent of each other .. </li></ul><ul><ul><li>Will not hold if we have a situation where </li></ul></ul><ul><ul><ul><li>ALL (or MOST) of the DOCTORS are WOMAN </li></ul></ul></ul><ul><ul><ul><li>ALL (or MOST) of the ENGINEERS are MEN </li></ul></ul></ul><ul><ul><li>We do not know what fraction of any Profession is </li></ul></ul><ul><ul><ul><li>Man or Woman </li></ul></ul></ul><ul><ul><ul><li>Married or Single </li></ul></ul></ul><ul><ul><li>Which means that even if we are told that a DOCTOR has been selected as secretary our estimate of the probability of P(W) is still 0.4 </li></ul></ul>
26. 26. Conditional Probability under Statistical Independence <ul><li>The probability of Event B given that Event A has happened = P(B/A) </li></ul><ul><ul><li>Given that the secretary is DOCTOR, what is the probability that secretary is WOMAN </li></ul></ul><ul><ul><li>P(WOMAN / DOCTOR) </li></ul></ul><ul><li>In the case of INDEPENDENT events </li></ul><ul><ul><li>P(B/A) = P(B) </li></ul></ul><ul><li>In the case of the Yoga Club </li></ul><ul><ul><li>P(WOMAN / DOCTOR ) = P(WOMAN) = 0.4 </li></ul></ul>Mathematical Definition of Independence
27. 27. Where are we ? 2 <ul><li>Classical Probability </li></ul><ul><li>Relative Probability </li></ul><ul><li>Subjective Probability </li></ul><ul><li>Marginal Probability P(A) </li></ul><ul><li>Probability of A or B </li></ul><ul><li>Exclusive Events P(AorB) = P(A) + P(B) </li></ul><ul><li>Non Exclusive Events P(AorB) = P(A)+P(B)-P(AB) </li></ul><ul><li>Two Events under Statistical Independence </li></ul><ul><li>Joint probability P(AB) = P(A) x P(B) </li></ul><ul><li>Conditional Probability P(B/A) = P(B) </li></ul>
28. 28. When are things independent ? <ul><li>Is gender independent of profession ? </li></ul><ul><ul><li>Most police / fire brigade personnel are men </li></ul></ul><ul><ul><li>Most kindergarten teachers are women </li></ul></ul><ul><li>Is the probability of having a PAN card independent of profession ? </li></ul><ul><ul><li>Are rickshaw pullers likely to have PAN cards </li></ul></ul><ul><li>Is the probability of defective products independent of factory ( or machine) where it is produced ? </li></ul><ul><ul><li>Old plants, worn out machinery likely to cause more defects </li></ul></ul><ul><li>Is the probability of loan default independent of gender ? </li></ul><ul><ul><li>Ask Mohd Younus !! </li></ul></ul>
29. 29. Conditional Probability under Statistical Dependence <ul><li>Another yoga club has the following kinds of members </li></ul><ul><ul><li>3 employed men </li></ul></ul><ul><ul><li>1 employed women </li></ul></ul><ul><ul><li>2 student men </li></ul></ul><ul><ul><li>4 student women </li></ul></ul><ul><li>Once again we select one person at random to work as the secretary of the club </li></ul><ul><li>Suppose that we know that the secretary is an employed person ( not a student) </li></ul><ul><li>What is the probability that the secretary is a </li></ul><ul><ul><li>Man ? </li></ul></ul><ul><ul><li>Woman ? </li></ul></ul><ul><li>We want to know </li></ul><ul><ul><li>P( M / E ) or </li></ul></ul><ul><ul><li>P( W / E ) </li></ul></ul>Conditional Probability that secretary is Man given that he is employed
30. 30. Conditional Probability under Statistical Dependence <ul><li>Person is named 1 to 10 </li></ul><ul><li>Selection of person n is event n </li></ul><ul><li>Since all persons are equally likely </li></ul><ul><ul><li>P(any specific individual) = 1/10 </li></ul></ul><ul><ul><li>= 0.10 </li></ul></ul>We first use Classical Probability to list out all possibilities
31. 31. Conditional Probability under Statistical Dependence <ul><li>P( Man / Employed ) = 3 / 4 = 0.75 </li></ul><ul><ul><li>¾ the of all employed members are men </li></ul></ul><ul><li>P( Woman / Employed ) = 1 / 4 = 0.25 </li></ul><ul><li>P( M / E ) + P( W / E ) = 1.00 </li></ul>Employed Student 3 Employed man 1 Employed woman 2 Student man 4 Student Woman Employed 3 Employed man 1 Employed woman
32. 32. Rule for Conditional Probability under Statistical Dependence <ul><li>P(Man/Employed) </li></ul><ul><li>P(Man AND Employed) </li></ul><ul><li>P(Employed) </li></ul><ul><li>3/10 </li></ul><ul><li>4/10 </li></ul><ul><li>0.75 </li></ul><ul><li>similarly </li></ul><ul><li>P(Woman/Employed) = 0.25 </li></ul>= = =
33. 33. Joint & Conditional Probability under Statistical Dependence <ul><li>P( Man / Employed) </li></ul><ul><li>P ( Man AND Employed) </li></ul><ul><li>P ( Employed) </li></ul><ul><li>P( A / B) </li></ul><ul><li>P ( A B ) </li></ul><ul><li>P ( B ) </li></ul>= = Conditional Probability Joint Probability Marginal or Unconditional Probability
34. 34. Conditional Probabilities from Joint Probabilities <ul><li>P(E/M) = P(E M) / P(M) = [3/10] / [5/10] = 3/5 </li></ul><ul><ul><li>P(S/M) = P(S M) / P(M) = [2/10] / [5/10] = 2/5 </li></ul></ul><ul><li>P(E/W) = P(E W) / P(W) = [1 / 10] / [ 5 / 10] = 1 / 5 </li></ul><ul><ul><li>P(S/W) = P(S W) / P(W) = [4/10] / [5/10] = 4/5 </li></ul></ul><ul><li>P(M/E) = P(E M) / P(E) = [3 / 10] / [ 4 / 10 ] = 3/4 </li></ul><ul><ul><li>P(W/E) = P(W E) / P(E) = [1/10] / [4/10] = 1/4 </li></ul></ul><ul><li>P(M/S) = P(M S) / P(S) = [2 / 10 ] / [ 6 / 10 ] = 2 / 6 </li></ul><ul><ul><li>P(W/S) = P( W S) / P(S) = [4/10] / [6/10] = 4/6 </li></ul></ul>
35. 35. Conditional => Joint Probability under Statistical Dependence <ul><li>P(Man/Employed) </li></ul><ul><li>P(Man AND Employed) </li></ul><ul><li>P(Employed) </li></ul><ul><li>P(A / B) </li></ul><ul><li>P(A B) </li></ul><ul><li>P(B) </li></ul><ul><li>P(Man and Employed) </li></ul><ul><li>= P(Man / Emp) x P(Emp) </li></ul><ul><li>P (A B ) = P (A / B ) x P(B) </li></ul>= =
36. 36. Joint Probabilities from Conditional Probabilities <ul><li>P(E M) = P(E/M) x P(M) = 3/5 x 5/10 = 3/10 </li></ul><ul><ul><li>= P(M/E) x P(E) = 3 / 4 x 4/10 = 3/10 </li></ul></ul><ul><li>P(E W) = P(E/W) x P(W) = 1/5 x 5/10 = 1/10 </li></ul><ul><li>P(S M) = P(S/M) x P(M) = 2/5 x 5/10 = 2/10 </li></ul><ul><li>P(S W) = P(S/W) x P(W) = 4/5 x 5/10 = 4/10 </li></ul>
37. 37. Probabilities under Statistical Dependence Conditional Probability P(B/A) = P(B A) / P(A) P(A/B) = P(B A) / P(B) Joint Probability P(B A) = P(B / A) x P(A) = P(A / B) X P(B)
38. 38. Where are we ? 3 <ul><li>Classical Probability </li></ul><ul><li>Relative Probability </li></ul><ul><li>Subjective Probability </li></ul><ul><li>Marginal Probability P(A) </li></ul><ul><li>Probability of A or B </li></ul><ul><li>Exclusive Events P(AorB) = P(A) + P(B) </li></ul><ul><li>Non Exclusive Events P(AorB) = P(A)+P(B)-P(AB) </li></ul><ul><li>Two Events under Statistical Independence </li></ul><ul><li>Joint probability P(AB) = P(A) x P(B) </li></ul><ul><li>Conditional Probability P(B/A) = P(B) </li></ul><ul><li>Two Events under Statistical Dependence </li></ul><ul><li>Conditional Probability P(B/A) = P(AB) / P(A) </li></ul><ul><li>Joint Probability P(AB) = P(B/A) x P(A) </li></ul>
39. 39. Marginal Probabilities under Statistical Dependence <ul><li>Marginal Properties under statistical dependence are computed by simply summing up the probabilities of all the joint events where the simple event occurs </li></ul><ul><ul><li>P(A) = P(AB) + P(AC) </li></ul></ul><ul><ul><li>P(Man) = P(Man Employed) + P(Man Student) </li></ul></ul><ul><ul><li>P(Student) = P(Man Student) + P(Woman Student) </li></ul></ul>
40. 40. Marginal Probabilities under Statistical Dependence <ul><li>P(M) = P( E M) + P(S M) = 0.3 + 0.2 = 0.5 </li></ul><ul><li>P(W) = P(E W) + P(S W) = 0.1 + 0.4 = 0.5 </li></ul><ul><li>P(E) = P(E M) + P(E W) = 0.3 + 0.1 = 0.4 </li></ul><ul><li>P(S) = P(S M) + P(S W) = 0.2 + 0.4 = 0.6 </li></ul>
41. 41. Where are we ? 4 <ul><li>Classical Probability </li></ul><ul><li>Relative Probability </li></ul><ul><li>Subjective Probability </li></ul><ul><li>Marginal Probability P(A) </li></ul><ul><li>Probability of A or B </li></ul><ul><li>Exclusive Events P(AorB) = P(A) + P(B) </li></ul><ul><li>Non Exclusive Events P(AorB) = P(A)+P(B)-P(AB) </li></ul><ul><li>Two Events under Statistical Independence </li></ul><ul><li>Joint probability P(AB) = P(A) x P(B) </li></ul><ul><li>Conditional Probability P(B/A) = P(B) </li></ul><ul><li>Two Events under Statistical Dependence </li></ul><ul><li>Conditional Probability P(B/A) = P(AB) / P(A) </li></ul><ul><li>Joint Probability P(AB) = P(B/A) x P(A) </li></ul><ul><li>Multiple Events under Statistical Dependence </li></ul><ul><li>Marginal Probality P(A) = P(AB) + P(AC) + P(AD) + ..... </li></ul>
42. 42. Bayes Theorem revising prior estimates of probability <ul><li>Central Theorem that connects </li></ul><ul><ul><li>Joint Probability </li></ul></ul><ul><ul><li>Conditional Probability </li></ul></ul><ul><ul><li>Marginal Probability </li></ul></ul><ul><li>Using the basic formula </li></ul><ul><ul><li>P(B A) = P(B/A) x P(A) </li></ul></ul><ul><ul><li>P(B A) = P(A/B) x P(B) </li></ul></ul><ul><ul><li>P(B/A) x P(A) = P(A/B) x P(B) </li></ul></ul><ul><li>Its value lies in being able to calculate P(B/A) from the value of P(A/B) </li></ul>
43. 43. Brand Conscious Customer <ul><li>Suppose there are two types of customers </li></ul><ul><ul><ul><li>Type A : Brand Conscious; Type B : Brand Indifferent </li></ul></ul></ul><ul><ul><li>Demographic information tells us that 20% customers are Type A and rest 80% is Type B </li></ul></ul><ul><ul><ul><li>P(A) = 0.2 ; P(B) = 0.8 </li></ul></ul></ul><ul><ul><ul><li>P(A) + P(B) = 1 since events are mutually exclusive </li></ul></ul></ul><ul><li>Probability of sale of Designer Shirt is </li></ul><ul><ul><ul><li>60% if the customer is Brand Conscious </li></ul></ul></ul><ul><ul><ul><li>10% if the customer is Brand Indifferent </li></ul></ul></ul><ul><ul><ul><li>P(S/A) = 0.6, P(S/B) = 0.1 </li></ul></ul></ul><ul><li>A person walks in and purchases a Designer Shirt </li></ul><ul><ul><ul><li>What is the probability that he is Brand Conscious </li></ul></ul></ul><ul><ul><ul><li>Calculate P(A/S) </li></ul></ul></ul>
44. 44. Brand Conscious Customer : 1 Demographics tells us that P(A) = 0.2; P(B) = 0.8 We know the conditional probabilities P(S/A) = 0.6, P(S/B) = 0.1 Joint Probability of Sale and given type of customer hence P(SA) = P(S/A)xP(A) Marginal Probability of Sale is sum of the two joint probabilities P(S) = P(SA) + P(SB)
45. 45. Brand Conscious Customer : 2 <ul><li>After purchase of ONE branded shirt </li></ul><ul><ul><li>Probability that customer is brand conscious has now increased from 20% to 60% </li></ul></ul>Conditional Probability P(A/S) = P(SA) / P(S) P(B/S) = P(SB) / P(S)
46. 46. Brand Conscious Customer : 3 a second sale <ul><li>After purchase of TWO branded shirts </li></ul><ul><ul><li>Probability that customer is brand conscious has now increased from 20% to 90% </li></ul></ul><ul><li>Consider Loyalty Program / Free Gifts !! </li></ul>The TWO sales are INDEPENDENT of each other hence P(2 Sales) = P(1 Sale) x P(1 Sale)
47. 47. Error in Machine Setup <ul><li>A carton sealing machine has to be aligned correctly before it starts packaging </li></ul><ul><ul><li>10% of the time it is aligned badly : P(B) = 0.1 </li></ul></ul><ul><ul><li>90% of the time it is aligned correctly : P(C) = 0.9 </li></ul></ul><ul><li>If the machine is aligned is aligned badly then there is a 50% chance that the carton will be found defective by the QC staff </li></ul><ul><ul><li>P(D / B ) = 0.5 </li></ul></ul><ul><li>If the machine is aligned correctly, the probability of a defect is is 5% </li></ul><ul><ul><li>P(D/C) = 0.05 </li></ul></ul><ul><li>Given that the first carton is rejected, what is the probability that alignment is bad ? What is P(B/D) </li></ul>
48. 48. Error in Machine Setup : 1 <ul><li>After the detection of ONE defect on the first carton </li></ul><ul><ul><li>Probability that alignment is erroneous has now increased from 10% to 53% </li></ul></ul><ul><li>Let us rework the numbers such that </li></ul><ul><ul><li>Probability of BAD alignment = 2% </li></ul></ul><ul><ul><li>Probability of Defective carton with </li></ul></ul><ul><ul><ul><li>Bad alignment is 5% </li></ul></ul></ul><ul><ul><ul><li>Good alignment is 1% </li></ul></ul></ul><ul><li>Is one defective carton important enough ? </li></ul>
49. 49. Error in Machine Setup : 2 <ul><li>After the detection of ONE defect on the first carton </li></ul><ul><ul><li>Probability that alignment is erroneous has now increased from 2% to 9% </li></ul></ul><ul><li>Is it worth stopping production and fixing alignment ? </li></ul><ul><li>Let us consider 2 defects in the first 5 cartons </li></ul>
50. 50. Error in Machine Setup : 3 inconsistent behaviour <ul><li>After TWO defects in FIVE cartons </li></ul><ul><ul><li>Probability that machine is badly aligned has now increased from 2% to 31% </li></ul></ul><ul><li>Consider Stopping Production and fixing the machine </li></ul>P(2 D in 5) = 0.05 x 0.05 X 0.95 X 0.95 X 0.95 = 0.00214 P(2D in 5) = 0.01 x 0.01 x 0.99 x 0.99 x 0.99 = 0.00010