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Quantitative Analysis for Management

Quantitative Analysis for Management. Chapter 2 Probability Concepts and Applications. Chapter Outline. 2.1 Introduction 2.2 Fundamental Concepts 2.3 Mutually Exclusive and Collectively Exhaustive Events 2.4 Statistically Independent Events

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Quantitative Analysis for Management

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  1. Quantitative Analysis for Management Chapter 2 Probability Concepts and Applications 2-1

  2. Chapter Outline 2.1 Introduction 2.2 Fundamental Concepts 2.3 Mutually Exclusive and Collectively Exhaustive Events 2.4 Statistically Independent Events 2.5 Statistically Dependent Events 2.6 Revising Probabilities with Bayes’ Theorem 2-2

  3. Chapter Outline - continued 2.7 Further Probability Revisions 2.8 Random Variables 2.9 Probability Distributions 2.10 The Normal Distribution 2.11 The Exponential Distribution 2.12 The Poisson Distribution 2-3

  4. Learning Objectives Students will be able to: • Understand the basic foundations of probability analysis • Understand the difference between mutually exclusive and collectively exhaustive events • Describe statistically dependent and independent events • Use Bayes’ theorem to establish posterior probabilities 2-4

  5. Learning Objectives - continued • Describe and provide examples of both discrete and continuous random variables • Explain the difference between discrete and continuous probability distributions • Calculate expected values and variances • Use the Normal table 2-5

  6. Introduction • Life is uncertain! • We must deal with risk! • A probability is a numerical statement about the likelihood that an event will occur 2-6

  7. Basic Statements About Probability 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: 2-7

  8. Example 2.1 • Demand for white latex paint at Diversey Paint and Supply has always been 0, 1, 2, 3, or 4 gallons per day. (There are no other possible outcomes; when one outcome occurs, no other can.) Over the past 200 days, the frequencies of demand are represented in the following table: 2-8

  9. Quantity Demanded (Gallons) 0 1 2 3 4 Number of Days 40 80 50 20 10 Total 200 Example 2.1 - continuedFrequencies of Demand 2-9

  10. Quantity Frequency Demanded (days) 0 40 1 80 2 50 3 20 4 10 Total days = 200 Probability (40/200) = 0.20 (80/200) = 0.40 (50/200) = 0.25 (20/200) = 0.10 (10/200) = 0.05 Total probability =1.00 Example 2.1 - continuedProbabilities of Demand 2-10

  11. Types of Probability number of times event occurs = P ( event ) total number of trials or outcomes Objective probability: Can be determined by experiment or observation: • Probability of heads on coin flip • Probably of spades on drawing card from deck 2-11

  12. Introduction - continuedTypes of Probability Subjective probability: Based upon judgement Can be determined by: • judgement of expert • opinion polls • Delphi method • etc. 2-12

  13. Mutually Exclusive & Collectively Exhaustive Events • Events are said to be mutually exclusive if only one of the events can occur on any one trial: • Events are said to be collectively exhaustive if the list of outcomes includes every possible outcome: heads and tails as possible outcomes of coin flip 2-13

  14. Outcome of Roll 1 2 3 4 5 6 Probability 1/6 1/6 1/6 1/6 1/6 1/6 Total = 1 Example 2 Rolling a die has six possible outcomes 2-14

  15. Outcome of Roll = 5 Die 1 Die 2 1 4 2 3 3 2 4 1 Probability 1/36 1/36 1/36 1/36 Example 2a Rolling two dice resulting in a total of five spots showing. There are a total of 36 possible outcomes. 2-15

  16. Draw a space and a club Draw a face card and a number card Draw an ace and a 3 Draw a club and a nonclub Draw a 5 and a diamond Draw a red card and a diamond Yes No Yes Yes Yes No Yes Yes No No No No Example 3 Draws Mutually Collectively ExclusiveExhaustive 2-16

  17. Probability of Mutually Exclusive Events P(event A or event B) = P(event A + P(event B) or: P(A or B) = P(A) + P(B) i.e., P(spade or club) = P(spade) + P(club) = 13/52 + 13/52 = 26/52 = 1/2 = 50% 2-17

  18. Probability of Events Which Are Not Mutually Exclusive P(event A or event B) = P(event A) + P(event B) - P(event A and event B both occurring) or P(A or B) = P(A) +P(B) - P(A and B) 2-18

  19. Probability(A and B)(Venn Diagram) P(A) P(A and B) P(B) 2-19

  20. Probability(A or B) - + P(A) P(B) P(A and B) = P(A or B) 2-20

  21. Statistical Dependence • Events are either • statistically independent (the occurrence of one event has no effect on the probability of occurrence of the other) or • statistically dependent (the occurrence of one event gives information about the occurrenceof the other) 2-21

  22. Which Are Independent? • (a) Your education (b) Your income level • (a) Draw a Jack of Hearts from a full 52 card deck (b) Draw a Jack of clubs from a full 52 card deck • (a) Chicago Cubs win the National League pennant (b) Chicago Cubs win the World Series 2-22

  23. Probabilities - Independent Events • Marginal probability: the probability of an event occurring [P(A)] • Joint probability: the probability of multiple, independent events, occurring at the same time [P(AB) = P(A)*P(B)] • Conditional probability (for independent events): • the probability of event B given that event A has occurred [P(B|A) = P(B)] • or the probability of event A given that event Bhas occurred [P(A|B) = P(A)] 2-23

  24. Probability(A|B) Independent Events P(A) P(B) P(A|B) P(B|A) 2-24

  25. A bucket contains 3 black balls, and 7 green balls. We draw a ball from the bucket, replace it, and draw a second ball. 1. P(black ball drawn on first draw) P(B) = 0.30 (marginal probability) 2. P(two green balls drawn) P(GG) = P(G)*P(G) = 0.70*0.70 = 0.49 (joint probability for two independent events) Statistically Independent Events 2-25

  26. 1. P(black ball drawn on second draw, first draw was green) P(B|G) = P(B) = 0.30 (conditional probability) 2. P(green ball drawn on second draw, first draw was green) P(G|G) = 0.70 (conditional probability) Statistically Independent Events - continued 2-26

  27. Probabilities - Dependent Events • Marginal probability: probability of an event occurring [P(A)] • Conditional probability (for dependent events): • the probability of event B given that event A has occurred [P(B|A) = P(AB)/P(B)] • the probability of event A given that event B has occurred [P(A|B) = P(AB)/P(A)] 2-27

  28. Probability(A|B) / P(A) P(AB) P(B) = P(A|B) = P(AB)/P(B) 2-28

  29. Probability(B|A) / P(AB) P(B) P(A) = P(B|A) = P(AB)/P(A) 2-29

  30. 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) Then: P(WL) = 4/10 = 0.40 P(WN) = 2/10 = 0.20 P(W) = 6/10 = 0.60 P(YL) = 3/10 = 0.3 P(YN) = 1/10 = 0.1 P(Y) = 4/10 = 0.4 Statistically Dependent Events 2-30

  31. Then: P(L|Y) = P(YL)/P(Y) = 0.3/0.4 = 0.75 P(Y|L) = P(YL)/P(L) = 0.3/0.7 = 0.43 P(W|L) = P(WL)/P(L) = 0.4/0.7 = 0.57 Statistically Dependent EventsContinued 2-31

  32. Your stockbroker informs you that if the stock market reaches the 10,500 point level by January, there is a 70% probability the Tubeless Electronics will go up in value. Your own feeling is that there is only a 40% chance of the market reaching 10,500 by January. What is the probability that both the stock market will reach 10,500 points, and the price of Tubeless will go up in value? Joint Probabilities, Dependent Events 2-32

  33. Let M represent the event of the stock market reaching the 10,500 point level, and T represent the event that Tubeless goes up. Then: P(MT) = P(T|M)P(M) = (0.70)(0.40) = 0.28 Joint Probabilities, Dependent Events - continued 2-33

  34. Bayes’ theorem can be used to calculate revised or posterior probabilities Revising Probabilities With Bayes’ Theorem Prior Probabilities Bayes’ Process Posterior Probabilities New Information 2-34

  35. A cup contains two dice identical in appearance. One, however, 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 have no idea which die is which, but we select one by chance, and toss it. The result is a 3. What is the probability that the die rolled was fair? Posterior Probabilities 2-35

  36. We know that: P(fair) = 0.50 P(loaded) = 0.50 And: P(3|fair) = 0.166 P(3|loaded) = 0.60 Then: P(3 and fair) = P(3|fair)P(fair) = (0.166)(0.50) = 0.083 P(3 and loaded) = P(3|loaded)P(loaded) = (0.60)(0.50) = 0.300 Posterior Probabilities Continued 2-36

  37. A 3 can occur in combination with the state “fair die” or in combination with the state ”loaded die.” The sum of their probabilities gives the unconditional or marginal probability of a 3 on a toss: P(3) = 0.083 + 0.0300 = 0.383. Then, the probability that the die rolled was the fair one is given by: Posterior Probabilities Continued P(fair and 3) 0.083 = = = 0 22 P(fair | 3) . 3 P ( ) 0.383 2-37

  38. General Form of Bayes’ Theorem P(AB) = P ( A | B) or P(B) P ( B | A ) P ( A ) = P ( A | B ) _ _ + P ( B | A ) P ( A ) P ( B | A ) P ( A ) _ = where A the complement of the event A _ For example : if A is the event " fair die, " then A is the event " unfair die" or " loaded die." 2-38

  39. To obtain further information as to whether the die just rolled is fair or loaded, let’s roll it again. Again we get a 3. Given that we have now rolled two 3s, what is the probability that the die rolled is fair? Further Probability Revisions 2-39

  40. P(fair) = 0.50, P(loaded) = 0.50 as before P(3,3|fair) = (0.166)(0.166) = 0.027 P(3,3|loaded) = (0.60)(0.60) = 0.36 P(3,3 and fair) = P(3,3|fair)P(fair) = (0.027)(0.05) = 0.013 P(3,3 and loaded) = P(3,3|loaded)P(loaded) = (0.36)(0.5) = 0.18 P(3,3) = 0.013 + 0.18 = 0.193 Further Probability Revisions - continued 2-40

  41. Further Probability Revisions - continued P(3,3 and fair) = P(fair | 3,3) P(3,3) 0.013 = = 0.067 0.193 P(3,3 and loaded) = P(loaded | 3,3) P(3,3) 0.18 = = 0.933 0.193 2-41

  42. Further Probability Revisions - continued To give the final comparison: P(fair|3) = 0.22 P(loaded|3) = 0.78 P(fair|3,3) = 0.067 P(loaded|3,3) = 0.933 2-42

  43. Random Variables • Discrete random variable - can assume only a finite or limited set of values- I.e., the number of automobiles sold in a year • Continuous random variable - can assume any one of an infinite set of values - I.e., temperature, product lifetime 2-43

  44. Variables (Numeric) 2-44

  45. Variables (Non-numeric) 2-45

  46. Probability Distributions Figure 2.5 Probability Function 2-46

  47. Expected Value of a Discrete Probability Distribution n = å E ( X ) X P ( X ) i i = 1 i 5 = å E(X) X P(X ) i i = 1 i = + + + + X P(X ) X P(X ) X P(X ) X P(X ) X P(X ) 1 1 2 2 3 3 4 4 5 5 = 5 0 1 + 4 0 2 + 3 0 3 + 2 0 3 + 1 0 1 ( )( . ) ( )( . ) ( )( . ) ( )( . ) ( )( . ) = 2 9 . 2-47

  48. Variance of a Discrete Probability Distribution n 2 2 s = - å [ X E ( X )] P ( X ) i i = 1 i 2 2 2 2 s = 5 - 2 9 0 1 + 4 - 2 9 0 2 + 3 - 2 9 0 3 ( . ) ( . ) ( . ) ( . ) ( . ) ( . ) 2 2 + 2 - 2 9 0 3 + 1 - 2 9 0 1 ( . ) ( . ) ( . ) ( . ) = 0 441 + 0 242 + 0 003 + 0 243 + 0 361 . . . . . = 1 29 . 2-48

  49. Probability Distribution Continuous Random Variable 2 é ù ( ) - 1 2 - m / X ê ú ê ú 2 1 ê ú s ë û = f(X) e s p 2 • Probability density function - f(X) Normal Distribution: 2-49

  50. Normal Distribution for Different Values of  - Fig. 2.7 =50 =40 =60 2-50

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