html5-img
1 / 78

Chapter 2 Probability Concepts and Applications

Chapter 2 Probability Concepts and Applications. Prepared by Lee Revere and John Large. Learning Objectives. Students will be able to: Understand the basic foundations of probability analysis. Describe statistically dependent and independent events.

dareh
Télécharger la présentation

Chapter 2 Probability Concepts and Applications

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Chapter 2 Probability Concepts and Applications Prepared by Lee Revere and John Large 2-1

  2. Learning Objectives Students will be able to: • Understand the basic foundations of probability analysis. • Describe statistically dependent and independent events. • Use Bayes’ theorem to establish posterior probabilities. • 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 and use the Normal table. 2-2

  3. 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 2.7 Further Probability Revisions 2-3

  4. Chapter Outline continued 2.8 Random Variables 2.9 Probability Distributions 2.10 The Binomial Distribution 2.11 The Normal Distribution 2.12 The Exponential Distribution 2.13 The Poisson Distribution 2-4

  5. Introduction • Life is uncertain! • We must deal with risk! • A probability is a numerical statement about the likelihood that an event will occur. 2-5

  6. Basic Statements about Probability • 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 • The sum of the simple probabilities for all possible outcomes of an activity must equal 1. 2-6

  7. 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 frequencies of demand are represented in the following table: 2-7

  8. Quantity Freq. Demand (days) 0 40 1 80 2 50 3 20 4 10 Total days = 200 Probability (Relative Freq) (40/200) = 0.20 (80/200) = 0.40 (50/200) = 0.25 (20/200) = 0.10 (10/200) = 0.05 Total Prob =1.00 Diversey Paint Example (continued) Probabilities of Demand Note: 0  P(event)  1  P(event) = 1 and 2-8

  9. Types of Probability Number of times event occurs = P ( event ) Total number of outcomes or occurrences Objective probability is based on logical observations: Determined by: • Relative frequency – Obtained using historical data (Diversey Paint) • Classical method – Known probability for each outcome (tossing a coin) 2-9

  10. Types of Probability Subjective probability is based on personal experiences. Determined by: • Judgment of experts • Opinion polls • Delphi method • Others 2-10

  11. Mutually Exclusive Events • Events are said to bemutually exclusiveif only one of the events can occur on any one trial. Example: a fair coin toss results ineithera head or a tail. 2-11

  12. Collectively Exhaustive Events • Events are said to becollectivelyexhaustiveif the list of outcomes includes every possible outcome. • Heads and tails as possible outcomes of coin flip. Example: a collectively exhaustive list of possible outcomes for a fair coin toss includes heads and tails. 2-12

  13. Outcome of Roll 1 2 3 4 5 6 Probability 1/6 1/6 1/6 1/6 1/6 1/6 Total = 1 Die Roll Example This is a collectively exhaustive list of potential outcomes for a single die roll. The outcome is a mutually exclusive event because only one event can occur (a 1, 2, 3, 4, 5, or 6) on any single roll. 2-13

  14. Twin Birth Example A woman is pregnant with non- identical twins. Following is a list of collectively exhaustive, mutually exclusive possible outcomes: Outcome Probability of Birth Boy/Boy ¼ Boy/Girl ¼ Girl/Girl ¼ Girl/Boy ¼ What is the probability that both babies will be girls? / boys? 2-14

  15. Draw a spade 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 In-Class Practice Assuming a traditional 52-card deck, can you identify if these outcomes are mutually exclusive and/or collectively exhaustive ?? 2-15

  16. Law of Addition: Mutually Exclusive P (event A or event B) = P (event A) + P (event B) or: P (A or B) = P (A) + P (B) Example: P (spade or club) = P (spade) + P (club) = 13/52 + 13/52 = 26/52 = 1/2 = 50% 2-16

  17. Law of Addition: not Mutually Exclusive P(event A or event B) = P(event A) + P(event B) - P(event Aandevent Bbothoccurring) or P(A or B) = P(A)+P(B) - P(A and B) 2-17

  18. Venn Diagram P(A and B) P(A) P(B) 2-18

  19. Venn DiagramP(A or B) - + P(A) P(B) P(A and B) = P(A or B) 2-19

  20. In-Class Example: Specialized University Specialized University offers four different graduate degrees: business, education, accounting, and science. Enrollment figures show 25% of their graduate students are in each specialty. Although 50% of the students are female, only 15% are female business majors. If a student is randomly selected from the University’s registration database: • What is the probability the student is a business or education major? • What is the probability the student is a female or a business major? 2-20

  21. Specialized University Solution The probability that the student is a business or education major is mutually exclusive event. Thus: P(Bus or Edu) = P(Bus) + P(Edu) = .25 + .25 = .50 or 50% The probability that the student is a female or a business major is not mutually exclusive because the student could be a female business major. Thus: P(Fem or Bus) = P(Fem) + P(Bus) – P(Fem and Bus) = .50 + .25 - .15 = .60 or 60% 2-21

  22. 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 occurrence of the other). 2-22

  23. 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 2-23

  24. 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 eventBgiven that eventAhas occurred: P(B|A) = P(B) • or, the probability of eventAgiven that eventB has occurred:P(A|B) = P(A) 2-24

  25. Venn Diagram: P(A|B) P(A) P(B) P(B|A) P(A|B) 2-25

  26. 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) Independent EventsExample A bucket contains 3 black balls and 7 green balls. We draw a ball from the bucket, replace it, and draw a second ball. 2-26

  27. 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) Independent Events Example continued 2-27

  28. 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(A) • The probability of event A given that event B has occurred:P(A|B) = P(AB)/P(B) • Joint probability: The probability of multiple events occurring at the same time: P(AB) = P(B|A)*P(A) 2-28

  29. Venn Diagram: P(B|A) P(A) P(B) P(A and B) / P(B) P(AB) P(A) P(B|A) = P(AB)/P(A) 2-29

  30. Venn Diagram: P(A|B) P(A) P(B) P(A and B) / P(A) P(AB) P(B) P(A|B) = P(AB)/P(B) 2-30

  31. Dependent Events Example • 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 • 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) 2-31

  32. Dependent Events Example Frequencies in table format 2-32

  33. Dependent Events Example Probabilities in table format 2-33

  34. Dependent Events Example 2-34

  35. Then: P(Y) = .4 - marginal probability P(L|Y) = P(YL)/P(Y) = 0.3/0.4 = 0.75 - conditional probability P(W|L) = P(WL)/P(L) = 0.4/0.7 = 0.57 - conditional probability Dependent Events Example (continued) 2-35

  36. Your stockbroker informs you that if the stock market reaches the 10,500 point level by January, there is a 70% probability that 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? Dependent Events: Joint Probability Example 2-36

  37. Then: P(MT) =P(T|M)P(M) = (0.70)(0.40) = 0.28 Dependent Events: Joint Probabilities Solution LetMrepresent the event of the stock market reaching the 10,500 point level, andTrepresent the event that Tubeless goes up. 2-37

  38. Bayes’ theorem can be used to calculate revised or posterior probabilities. Revising Probabilities: Bayes’ Theorem Prior Probabilities Bayes’ Process Posterior Probabilities New Information 2-38

  39. General Form of Bayes’ Theorem 2-39

  40. 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 Example 2-40

  41. We know that: P(fair) = 0.50 P(loaded) = 0.50 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 Example(continued) - marginal probability - joint probability 2-41

  42. 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 marginal probability of a 3 on a toss: P(3) = 0.083 + 0.030 = 0.383 Then, the probability that the die rolled was the fair one is given by: Posterior Probabilities Example continued - marginal probability - conditional probability 2-42

  43. 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 2-43

  44. We know from before that: P(fair) = 0.50, P(loaded) = 0.50 Then: P(3,3|fair) = P(3|fair)*P(3|fair) = (0.166)(0.166) = 0.027 P(3,3|loaded) = P(3|loaded)*P(3|loaded) = (0.60)(0.60) = 0.360 So: P(3,3 and fair) = P(3,3|fair)*P(fair) = (0.027)(0. 50) = 0.013 P(3,3 and loaded) = P(3,3|loaded)P(loaded) = (0.36)(0.50) = 0.180 Thus, the probability of getting two 3s is a marginal probability obtained from the sum of the probability of two joint probabilities: P(3,3) = 0.013 + 0.180 = 0.193 Further Probability Revisions continued 2-44

  45. Further Probability Revisions continued • Using the probabilities from the previous slide: 2-45

  46. 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 2-46

  47. 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. 2-47

  48. Random Variables (Numeric) Experiment Outcome Random Variable Range of Random Variable Stock 50 Number of X = number of 0,1,2,, 50 Xmas trees trees sold trees sold Inspect 600 Number Y = number 0,1,2,…, items acceptable acceptable 600 Send out Number of Z = number of 0,1,2,…, 5,000 sales people people responding 5,000 letters responding £ £ Build an % completed R = % completed 0 R 100 apartment after 4 after 4 months building months £ £ Test the Time bulb S = time bulb burns 0 S 80,000 lifetime of a lasts - up to light bulb 80,000 (minutes) minutes Discrete Continuous 2-48

  49. RandomVariables (Non-numeric) Experiment Outcome Random Range of Variable Random Variable Students Strongly agree (SA) X = 5 if SA 1,2,3,4,5 respond to a 4 if A Agree (A) questionnaire Neutral (N) 3 if N Disagree (D) 2 if D Strongly Disagree (SD) 1 if SD One machine is Defective Y = 0 if defective 0,1 inspected Not defective 1 if not defective Consumers Good Z = 3 if good 1,2,3 respond to how Average 2 if average they like a Poor 1 if poor product 2-49

  50. Probability Distributions • Probability distribution – the set of all possible values of a random variable and their associated probabilities. • In a discrete probability distribution a probability between 0 and 1 is assigned to each discrete variable.The sum of the probabilities sum to 1. 2-50

More Related