1. Basic Probability

2. Uncertainties Managers often base their decisions on an analysis of uncertainties such as the following: What are the chances that sales will decrease if we increase prices? What is the likelihood a new assembly method method will increase productivity? What are the odds that a new investment will be profitable?

3. Application in Business Probability in Manufacturing Manufacturing businesses can use probability to determine the cost-benefit ratio or the transfer of a new manufacturing technology process by addressing the likelihood of improved profits. In other instances, manufacturing firms use probability to determine the possibility of financial success of a new product when considering competition from other manufacturers, market demand, market value and manufacturing costs. Other instances of probability in manufacturing include determining the likelihood of producing defective products, and regional need and capacity for certain fields of manufacturing.

4. Application in Business Scenario Analysis Probability distributions can be used to create scenario analyses. For example, a business might create three scenarios: worst-case, likely and best-case. The worst-case scenario would contain some value from the lower end of the probability distribution; the likely scenario would contain a value towards the middle of the distribution; and the best-case scenario would contain a value in the upper end of the scenario. Risk Evaluation In addition to predicting future sales levels, probability distribution can be a useful tool for evaluating risk. Consider, for example, a company considering entering a new business line. If the company needs to generate $500,000 in revenue in order to break even and their probability distribution tells them that there is a 10 percent chance that revenues will be less than $500,000, the company knows roughly what level of risk it is facing if it decides to pursue that new business line.

5. Application in Business Sales Forecasting One practical use for probability distributions and scenario analysis in business is to predict future levels of sales. It is essentially impossible to predict the precise value of a future sales level; however, businesses still need to be able to plan for future events. Using a scenario analysis based on a probability distribution can help a company frame its possible future values in terms of a likely sales level and a worst-case and best-case scenario. By doing so, the company can base its business plans on the likely scenario but still be aware of the alternative possibilities.

6. Important Terms Probability – the chance that an uncertain event will occur (always between 0 and 1) Event – Each possible outcome of a variable Simple Event – an event that can be described by a single characteristic e.g. Tossing a coin: Getting head is one event Getting tail is another event Here, “Tossing a coin” is referred as ‘EXPERIMENT’ If we collect all the outcomes or events of an experiment that is called sample space. Sample Space – the collection of all possible events e.g. Sample space can be written as : S = {Head tail}

7. Sample Space The Sample Space is the collection of all possible events e.g. All 6 faces of a die: e.g. All 52 cards of a bridge deck:

8. Events Simple event An outcome from a sample space with one characteristic e.g., A red card from a deck of cards Complement of an event A (denoted A’) All outcomes that are not part of event A e.g., All cards that are not diamonds Joint event Involves two or more characteristics simultaneously e.g., An ace that is also red from a deck of cards Favorable event Events in which we are interested e.g. Interest is have even number when rolling a die Equally likely events Non preference of occurring of any event e.g. tossing a coin, event can be head or tail

9. Visualizing Events Contingency Tables Purchasing a big screen TV No 100 650 750 Yes 200 50 250 Total 300 700 1000 Yes No Total Sample Space Actually purchased Planned to purchase

10. Bradley has invested in two stocks, Markley Oil and Collins Mining. Bradley has determined that the possible outcomes of these investments three months from now are as follows. Investment Gain or Loss in 3 Months (in $000) Markley Oil Collins Mining 10 5 0 -20 8 -2 Example: Bradley Investments An Experiment and Its Sample Space

11. Subjective Method An analyst made the following probability estimates. Exper. Outcome Net Gain or LossProbability (10, 8) (10, -2) (5, 8) (5, -2) (0, 8) (0, -2) (-20, 8) (-20, -2) $18,000 Gain $8,000 Gain $13,000 Gain $3,000 Gain $8,000 Gain $2,000 Loss $12,000 Loss $22,000 Loss.20.08.16.26.10.12.02.06 Example: Bradley Investments

12. Probability as a Numerical Measure of the Likelihood of Occurrence 0 1.5 Increasing Likelihood of Occurrence Probability: The event is very unlikely to occur. The occurrence of the event is just as likely as it is unlikely. The event is almost certain to occur.

13. Visualizing Events Venn Diagrams Let A = Planned to purchase Let B = Actually purchased A B A ∩ B = Planned and actually purchased A U B = Planned or actually purchased A B Planned Actually Purchased

14. Fundamental Concepts 1.The probability, P, of any event or state of nature occurring is greater than or equal to 0 and less than or equal to 1. That is: 0  P (event)  1 2.The sum of the simple probabilities for all possible outcomes of an activity must equal 1

15. Diversey Paint Example Demand for white latex paint at Diversey Paint and Supply has always been either 0, 1, 2, 3, or 4 gallons per day Over the past 200 days, the owner has observed the following frequencies of demand QUANTITY DEMANDED NUMBER OF DAYSPROBABILITY 0400.20 (= 40/200) 1800.40 (= 80/200) 2500.25 (= 50/200) 3200.10 (= 20/200) 4100.05 (= 10/200) Total200Total1.00 (= 200/200)

16. Diversey Paint Example Demand for white latex paint at Diversey Paint and Supply has always been either 0, 1, 2, 3, or 4 gallons per day Over the past 200 days, the owner has observed the following frequencies of demand QUANTITY DEMANDED NUMBER OF DAYSPROBABILITY 0400.20 (= 40/200) 1800.40 (= 80/200) 2500.25 (= 50/200) 3200.10 (= 20/200) 4100.05 (= 10/200) Total200Total1.00 (= 200/200) Notice the individual probabilities are all between 0 and 1 0 ≤ P (event) ≤ 1 And the total of all event probabilities equals 1 ∑ P (event) = 1.00

17. objective probability Determining objective probability Relative frequency Typically based on historical data Types of Probability P (event) = Number of occurrences of the event Total number of trials or outcomes Classical or logical method Logically determine probabilities without trials P (head) = 1212 Number of ways of getting a head Number of possible outcomes (head or tail)

18. Types of Probability Subjective probability Subjective probability is based on the experience and judgment of the person making the estimate Opinion polls Judgment of experts Delphi method Other methods

19. Example Taking Stats Not Taking Stats Total Male 84145229 Female 76134210 Total160279439 Find the probability of selecting a male taking statistics from the population described in the following table:

20. Mutually Exclusive Events mutually exclusive Events are said to be mutually exclusive if only one of the events can occur on any one trial Mutually exclusive events Events that cannot occur together example: Tossing a coin will result in either a head or a tail Events A and B are mutually exclusive A = defective; B = non-defective

21. Collectively Exhaustive Events collectively exhaustive Events are said to be collectively exhaustive if the list of outcomes includes every possible outcome One of the events must occur The set of events covers the entire sample space example: Sample : 400 units are there in one lot Experiment : drawing an unit from the lot Simple event: A = Defective; B = Non defective Collectively exhaustive events : 350 are non defective and 50 are defective units.

22. Joint Probability P (Planned and purchased) = P (Planned and did not purchase) = P (not Planned and purchased) = Similarly others… Marginal Probability P (Planned to purchase) = Actually purchased Planned to purchase Computing Joint and Marginal Probabilities

23. The probability of a joint event, A and B: Computing a marginal (or simple) probability: Where B 1, B 2, …, B k are k mutually exclusive and collectively exhaustive events

24. Bradley has invested in two stocks, Markley Oil and Collins Mining. Bradley has determined that the possible outcomes of these investments three months from now are as follows. Investment Gain or Loss in 3 Months (in $000) Markley Oil Collins Mining 10 5 0 -20 8 -2 Example: Bradley Investments An Experiment and Its Sample Space

25. Subjective Method An analyst made the following probability estimates. Exper. Outcome Net Gain or LossProbability (10, 8) (10, -2) (5, 8) (5, -2) (0, 8) (0, -2) (-20, 8) (-20, -2) $18,000 Gain $8,000 Gain $13,000 Gain $3,000 Gain $8,000 Gain $2,000 Loss $12,000 Loss $22,000 Loss.20.08.16.26.10.12.02.06 Example: Bradley Investments

26. The union of events A and B is denoted by A  B  The union of events A and B is the event containing all sample points that are in A or B or both. Union of Two Events Sample Space S Sample Space S Event A Event B

27. Union of Two Events Event M = Markley Oil Profitable Event C = Collins Mining Profitable M  C = Markley Oil Profitable or Collins Mining Profitable (or both) M  C = {(10, 8), (10,  2), (5, 8), (5,  2), (0, 8), (  20, 8)} P ( M  C) = P (10, 8) + P (10,  2) + P (5, 8) + P (5,  2) + P (0, 8) + P (  20, 8) =.20 +.08 +.16 +.26 +.10 +.02 =.82 Example: Bradley Investments

28. The intersection of events A and B is denoted by A  The intersection of events A and B is the set of all sample points that are in both A and B. Sample Space S Sample Space S Event A Event B Intersection of Two Events Intersection of A and B

29. Intersection of Two Events Event M = Markley Oil Profitable Event C = Collins Mining Profitable M  C = Markley Oil Profitable and Collins Mining Profitable M  C = {(10, 8), (5, 8)} P ( M  C) = P (10, 8) + P (5, 8) =.20 +.16 =.36 Example: Bradley Investments

30. The addition law provides a way to compute the probability of event A, or B, or both A and B occurring. Addition Law The law is written as: P ( A  B ) = P ( A ) + P ( B )  P ( A  B 

31. Event M = Markley Oil Profitable Event C = Collins Mining Profitable M  C = Markley Oil Profitable or Collins Mining Profitable We know: P ( M ) =.70, P ( C ) =.48, P ( M  C ) =.36 Thus: P ( M  C) = P ( M ) + P( C )  P ( M  C ) =.70 +.48 .36 =.82 Addition Law (This result is the same as that obtained earlier using the definition of the probability of an event.) Example: Bradley Investments

32. Mutually Exclusive Events Two events are said to be mutually exclusive if the events have no sample points in common. Two events are mutually exclusive if, when one event occurs, the other cannot occur. Sample Space S Event A Event B

33. Mutually Exclusive Events If events A and B are mutually exclusive, P ( A  B  = 0. The addition law for mutually exclusive events is: P ( A  B ) = P ( A ) + P ( B ) There is no need to include “  P ( A  B  ”

34. Adding Not Mutually Exclusive Events P (event A or event B ) = P (event A ) + P (event B ) – P (event A and event B both occurring) P ( A or B ) = P ( A ) + P ( B ) – P ( A and B ) P ( five or diamond )= P ( five ) + P ( diamond ) – P ( five and diamond ) = 4 / 52 + 13 / 52 – 1 / 52 = 16 / 52 = 4 / 13 The equation must be modified to account for double counting The probability is reduced by subtracting the chance of both events occurring together

35. Adding Mutually Exclusive Events We often want to know whether one or a second event will occur When two events are mutually exclusive, the law of addition is – P (event A or event B ) = P (event A ) + P (event B ) P (spade or club)= P (spade) + P (club) = 13 / 52 + 13 / 52 = 26 / 52 = 1 / 2 = 0.50 = 50%

36. Venn Diagrams P ( A ) P ( B ) Events that are mutually exclusive P ( A or B ) = P ( A ) + P ( B ) Events that are not mutually exclusive P ( A or B )= P ( A ) + P ( B ) – P ( A and B ) P ( A ) P ( B ) P ( A and B )

37. Example P(Red or Ace) = P(Red) +P(Ace) - P(Red and Ace) = 26/52 + 4/52 - 2/52 = 28/52 Don’t count the two red aces twice! Black Color Type Red Total Ace 224 Non-Ace 24 48 Total 26 52

38. Computing Conditional Probabilities A conditional probability is the probability of one event, given that another event has occurred: Where P(A and B) = joint probability of A and B P(A) = marginal probability of A P(B) = marginal probability of B The conditional probability of A given that B has occurred The conditional probability of B given that A has occurred

39. Event M = Markley Oil Profitable Event C = Collins Mining Profitable We know: P ( M  C ) =.36, P ( M ) =.70 Thus: Conditional Probability = Collins Mining Profitable given Markley Oil Profitable Example: Bradley Investments

40. What is the probability that a car has a CD player, given that it has AC ? i.e., we want to find P(CD | AC) Conditional Probability Example Of the cars on a used car lot, 70% have air conditioning (AC) and 40% have a CD player (CD). 20% of the cars have both.

41. Conditional Probability Example No CDCDTotal AC 0.20.50.7 No AC 0.20.1 0.3 Total 0.40.6 1.0 Of the cars on a used car lot, 70% have air conditioning (AC) and 40% have a CD player (CD). 20% of the cars have both. (continued)

42. Conditional Probability Example No CDCDTotal AC 0.20.50.7 No AC 0.20.1 0.3 Total 0.40.6 1.0 Given AC, we only consider the top row (70% of the cars). Of these, 20% have a CD player. 20% of 70% is about 28.57%. (continued)

43. Multiplication Law The multiplication law provides a way to compute the probability of the intersection of two events. The law is written as: P ( A  B ) = P ( B ) P ( A | B )

44. Event M = Markley Oil Profitable Event C = Collins Mining Profitable We know: P ( M ) =.70, P ( C | M ) =.5143 Multiplication Law M  C = Markley Oil Profitable and Collins Mining Profitable Thus: P ( M  C) = P ( M ) P ( C|M ) = (.70)(.5143) =.36 (This result is the same as that obtained earlier using the definition of the probability of an event.) Example: Bradley Investments

45. Independent Events If the probability of event A is not changed by the existence of event B, we would say that events A and B are independent. Two events A and B are independent if: P ( A | B ) = P ( A ) P ( B | A ) = P ( B ) or

46. The multiplication law also can be used as a test to see if two events are independent. The law is written as: P ( A  B ) = P ( A ) P ( B ) Multiplication Law for Independent Events

47. Event M = Markley Oil Profitable Event C = Collins Mining Profitable We know: P ( M  C ) =.36, P ( M ) =.70, P ( C ) =.48 But: P ( M)P(C) = (.70)(.48) =.34, not.36 Are events M and C independent? Does  P ( M  C ) = P ( M)P(C) ? Hence: M and C are not independent. Example: Bradley Investments Multiplication Law for Independent Events

48. Independent Events 1. A black ball drawn on first draw P ( B ) = 0.30 (a marginal probability) 2. Two green balls drawn P ( GG ) = P ( G ) x P ( G ) = 0.7 x 0.7 = 0.49 (a joint probability for two independent events) A bucket contains 3 black balls and 7 green balls We draw a ball from the bucket, replace it, and draw a second ball

49. Independent Events 3. A black ball drawn on second draw if the first draw is green P ( B | G ) = P ( B ) = 0.30 (a conditional probability but equal to the marginal because the two draws are independent events) 4. A green ball is drawn on the second if the first draw was green P ( G | G ) = P ( G ) = 0.70 (a conditional probability as in event 3) A bucket contains 3 black balls and 7 green balls We draw a ball from the bucket, replace it, and draw a second ball

50. When Events Are Dependent Assume that we have an urn containing 10 balls of the following descriptions 4 are white ( W ) and lettered ( L ) 2 are white ( W ) and numbered ( N ) 3 are yellow ( Y ) and lettered ( L ) 1 is yellow ( Y ) and numbered ( N ) P ( WL ) = 4 / 10 = 0.4 P ( YL ) = 3 / 10 = 0.3 P ( WN ) = 2 / 10 = 0.2 P ( YN ) = 1 / 10 = 0.1 P ( W ) = 6 / 10 = 0.6 P ( L ) = 7 / 10 = 0.7 P ( Y ) = 4 / 10 = 0.4 P ( N ) = 3 / 10 = 0.3

51. When Events Are Dependent 4 balls White ( W ) and Lettered ( L ) 2 balls White ( W ) and Numbered ( N ) 3 balls Yellow ( Y ) and Lettered ( L ) 1 ball Yellow ( Y ) and Numbered ( N ) Probability ( WL ) = 4 10 Probability ( YN ) = 1 10 Probability ( YL ) = 3 10 Probability ( WN ) = 2 10 Urn contains 10 balls

52. When Events Are Dependent The conditional probability that the ball drawn is lettered, given that it is yellow, is P ( L | Y ) = = = 0.75 P ( YL ) P ( Y ) 0.3 0.4 Verify P ( YL ) using the joint probability formula P ( YL ) = P ( L | Y ) x P ( Y ) = (0.75)(0.4) = 0.3

53. Joint Probabilities for Dependent Events P ( MT ) = P ( T | M ) x P ( M ) = (0.70)(0.40) = 0.28 If the stock market reaches 12,500 point by January, there is a 70% probability that Tubeless Electronics will go up There is a 40% chance the stock market will reach 12,500 Let M represent the event of the stock market reaching 12,500 and let T be the event that Tubeless goes up in value

54. Marginal Probability Marginal probability for event A: Where B 1, B 2, …, B k are k mutually exclusive and collectively exhaustive events

55. Visualizing Events Contingency Tables Purchasing a big screen TV No 100 650 750 Yes 200 50 250 Total 300 700 1000 Yes No Total Sample Space Actually purchased Planned to purchase

56. Posterior Probabilities Bayes’ Process Revising Probabilities with Bayes’ Theorem posterior probabilities Bayes’ theorem is used to incorporate additional information and help create posterior probabilities Prior Probabilities New Information

57. Example Let's say we want to know how a change in interest rates would affect the Value of a stock market index. All major stock market indexes have a plethora of historical data available so you should have no problem finding the outcomes for these events with a little bit of research. For this example we will use the data below to find out how a stock market index will react to a rise in interest rates. P(SI) = the probability of the stock index increasing P(SD) = the probability of the stock index decreasing P(ID) = the probability of interest rates decreasing P(II) = the probability of interest rates increasing Interest Rates Stock Price DeclineIncreaseUnit Frequency Decline2009501150 Increase80050850 Unit Frequency1000 2000 P(SD|II) =

58. Posterior Probabilities A cup contains two dice identical in appearance but one is fair (unbiased), the other is loaded (biased) The probability of rolling a 3 on the fair die is 1 / 6 or 0.166 The probability of tossing the same number on the loaded die is 0.60 We select one by chance, toss it, and get a result of a 3 What is the probability that the die rolled was fair? What is the probability that the loaded die was rolled?

59. Posterior Probabilities We know the probability of the die being fair or loaded is P (fair) = 0.50 P (loaded) = 0.50 And that P (3 | fair) = 0.166 P (3 | loaded) = 0.60 We compute the probabilities of P (3 and fair) and P (3 and loaded) P (3 and fair)= P (3 | fair) x P (fair) = (0.166)(0.50) = 0.083 P (3 and loaded)= P (3 | loaded) x P (loaded) = (0.60)(0.50) = 0.300

60. Posterior Probabilities We know the probability of the die being fair or loaded is P (fair) = 0.50 P (loaded) = 0.50 And that P (3 | fair) = 0.166 P (3 | loaded) = 0.60 We compute the probabilities of P (3 and fair) and P (3 and loaded) P (3 and fair)= P (3 | fair) x P (fair) = (0.166)(0.50) = 0.083 P (3 and loaded)= P (3 | loaded) x P (loaded) = (0.60)(0.50) = 0.300 The sum of these probabilities gives us the unconditional probability of tossing a 3 P (3) = 0.083 + 0.300 = 0.383

61. Posterior Probabilities P (loaded | 3) = = = 0.78 P (loaded and 3) P (3) 0.300 0.383 The probability that the die was loaded is P (fair | 3) = = = 0.22 P (fair and 3) P (3) 0.083 0.383 If a 3 does occur, the probability that the die rolled was the fair one is revisedposteriorprobabilities These are the revised or posterior probabilities for the next roll of the die prior probability We use these to revise our prior probability estimates

62. Bayes Calculations Given event B has occurred STATE OF NATURE P ( B | STATE OF NATURE) PRIOR PROBABILITY JOINT PROBABILITY POSTERIOR PROBABILITY A P ( B | A )x P ( A )= P ( B and A ) P ( B and A )/ P ( B ) = P ( A | B ) A’A’ P ( B | A’ )x P ( A’ )= P ( B and A’ ) P ( B and A’ )/ P ( B ) = P ( A’ | B ) P(B)P(B) Given a 3 was rolled STATE OF NATURE P ( B | STATE OF NATURE) PRIOR PROBABILITY JOINT PROBABILITY POSTERIOR PROBABILITY Fair die0.166x 0.5= 0.0830.083 / 0.383 = 0.22 Loaded die0.600x 0.5= 0.3000.300 / 0.383 = 0.78 P (3) = 0.383

63. General Form of Bayes’ Theorem We can compute revised probabilities more directly by using where the complement of the event ; for example, if is the event “fair die”, then is “loaded die”  A Bayes’ theorem is the extension of conditional probability

64. General Form of Bayes’ Theorem This is basically what we did in the previous example If we replace with “fair die” Replace with “loaded die Replace with “3 rolled” We get

65. Example A manufacturer claims that its drug test will detect steroid use (that is, show positive for an athlete who uses steroids) 95% of the time. Further, 15% of all steroid-free individuals also test positive. 10% of the rugby team members use steroids. Your friend on the rugby team has just tested positive. The probability that he uses steroids is? Consider: E = the event that a rugby team member tests positive F = the event that a rugby team member uses steroids P(E|F) = 0.95; P(E|F') = 0.15; P(F) = 0.1; P(F') = 0.9

66. Example A drilling company has estimated a 40% chance of striking oil for their new well. A detailed test has been scheduled for more information. Historically, 60% of successful wells have had detailed tests, and 20% of unsuccessful wells have had detailed tests. Given that this well has been scheduled for a detailed test, what is the probability that the well will be successful?

67. Let S = successful well U = unsuccessful well P(S) = 0.4, P(U) = 0.6 (prior probabilities) Define the detailed test event as D Conditional probabilities: P(D|S) = 0.6 P(D|U) = 0.2 Goal is to find P(S|D) Example (continued)

68. Example (continued) Apply Bayes’ Theorem: So the revised probability of success, given that this well has been scheduled for a detailed test, is 0.667

69. Given the detailed test, the revised probability of a successful well has risen to 0.667 from the original estimate of 0.4 Example Event Prior Prob. Conditional Prob. Joint Prob. Revised Prob. S (successful)0.40.6(0.4)(0.6) = 0.240.24/0.36 = 0.667 U (unsuccessful) 0.60.2(0.6)(0.2) = 0.120.12/0.36 = 0.333 Sum = 0.36 (continued)

70. Application Questions An automobile dealer has kept records on the customers who visited his showroom. Forty percent of the people who visited his dealership were women. Furthermore, his records show that 37% of the women who visited his dealership purchased an automobile, while 21% of the men who visited his dealership purchased an automobile. a. What is the probability that a customer entering the showroom will buy an automobile? b. Suppose a customer visited the showroom and purchased a car. What is the probability that the customer was a woman? c. Suppose a customer visited the showroom but did not purchase a car. What is the probability that the customer was a man?