Chapter 2 Probability Concepts and Applications

1. Chapter 2 Probability Concepts and Applications 2-1

2. 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 2-2

3. 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 2-3

4. Introduction • Life is uncertain! • We must deal with risk! • What is the chance that Hurricane Frances will hit New Orleans? • Who will win the election? • What is the chance that the oil price will hit $50? • Is it likely that SLU Faculty will get a pay raise since the tuition went up 7%? • A probability is a numerical value that measure the likelihood that an event will occur 2-4

5. How many Christmas trees should be stored? • How long will it take to complete an activity? • What should be the inventory level? 2-5

6. Hurricane Frances 2-6

7. The word "statistics" is used in several different senses. In the broadest sense, "statistics" refers to a range of techniques and procedures for analyzing data, interpreting data, displaying data, and making decisions based on data. This is what courses in "statistics" generally cover. In a second usage, a "statistic" is defined as a numerical quantity (such as the mean) calculated in a sample. Such statistics are used to estimate parameters. • The term "statistics" sometimes refers to calculated quantities regardless of whether or not 2-7

8. they are from a sample. For example, one might ask about a baseball player's statistics and be referring to his or her batting average, runs batted in, number of home runs, etc. Or, "government statistics" can refer to any numerical indexes calculated by a governmental agency. • Although the different meanings of "statistics" has the potential for confusion, a careful consideration of the context in which the word is used should make its intended meaning clear. 2-8

9. 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 2. The sum of the simple probabilities for all possible outcomes of an activity must equal 1. 2-9

10. 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: 2-10

11. Quantity Demanded (Gallons) 0 1 2 3 4 Number of Days 40 80 50 20 10 Total 200 Example 2.1 - continued Frequencies of Demand 2-11

12. Quant. Freq. Demand (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 Prob =1.00 Example 2.1 - continued Probabilities of Demand 2-12

13. Types of Probability Number of times event occurs = P ( event ) Total number of outcomes or occurrences Objective probability: Determined by experiment or observation: • Probability of heads on coin flip • Probably of spades on drawing card from deck 2-13

14. Types of Probability Subjective probability: Based upon judgement Determined by: • judgement of expert • opinion polls • etc. 2-14

15. Mutually Exclusive Events • Events are said to be mutually exclusive if only one of the events can occur on any one trial 2-15

16. Collectively Exhaustive Events • 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 2-16

17. 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 2-17

18. 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 results in a total of five spots showing. There are a total of 36 possible outcomes. 2-18

19. 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 Draw Mutually Collectively Exclusive Exhaustive 2-19

20. Probability : Mutually Exclusive 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% 2-20

21. Probability: Not Mutually Exclusive P(event A or event B) = P(event A) + P(event B) - P(event Aand event B both occurring) or P(A or B) = P(A)+P(B) - P(A and B) 2-21

22. P(A and B)(Venn Diagram) P(A and B) P(A) P(B) 2-22

23. P(A or B) - + P(A) P(B) P(A and B) = P(A or B) 2-23

24. 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) 2-24

25. 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 2-25

26. 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 B has occurred P(A|B) = P(A) 2-26

27. Probability(A|B) Independent Events P(A) P(B) P(B|A) P(A|B) 2-27

28. 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 A bucket contains 3 black balls, and 7 green balls. We draw a ball from the bucket, replace it, and draw a second ball 2-28

29. 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 2-29

30. 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) 2-30

31. Probability(A|B) / P(A) P(AB) P(B) P(A|B) = P(AB)/P(B) 2-31

32. Probability(B|A) / P(B) P(AB) P(A) P(B|A) = P(AB)/P(A) 2-32

33. Statistically Dependent Events • 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) 2-33

34. 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 Events - Continued 2-34

35. 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 2-35

36. Then: P(MT) =P(T|M)P(M) = (0.70)(0.40) = 0.28 Joint Probabilities, Dependent Events - continued Let M represent the event of the stock market reaching the 10,500 point level, and T represent the event that Tubeless goes up. 2-36

37. Bayes’ theorem can be used to calculate revised or posterior probabilities Revising Probabilities: Bayes’ Theorem Prior Probabilities Bayes’ Process Posterior Probabilities New Information 2-37

38. General Form of Bayes’ Theorem 2-38

39. 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 2-39

40. We know that: P(fair) = 0.50 P(loaded) = 0.50 And: P(3|fair) = 0.166P(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 2-40

41. 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 2-41

42. 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 2-42

43. 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 2-43

44. Further Probability Revisions - continued 2-44

45. 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 2-45

46. 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 2-46

47. 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 e 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 2-47

48. 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 2-48

49. Probability Distributions 2-49

50. Expected Value of a Discrete Probability Distribution n = å E ( X ) X P ( X ) i i = 1 i 2-50