1. 1 MAT137 Business Statistics Week 2 Probability DePaul University Bill Qualls

2. 2 Objectives At the end of this section you should be able to answer questions concerning elementary probability. Specifically, you should understand: sample space, event vs. simple event union and intersection complementary events additive rule of probability conditional probability independent events mutually exclusive events permutations vs. combinations

3. 3 Descriptive Statistics (continued)

4. 4 But first… But first we have a few more descriptive statistics topics… –Chebyshev's Theorem –Empirical Rule –Quantiles and percentiles

5. 5 Chebyshev's Theorem Chebyshev's theorem: Regardless of the shape of the distribution… –At least 75% of the observations will fall within the interval [x – 2s, x + 2s] –At least 89% of the observations will fall within the interval [x – 3s, x + 3s]

6. 6 Chebyshev's Theorem Recall the invoice data from last week: 82 77 97 100 99 105 112 68 93 72 126 71 97 84 98 76 67 109 83 100 86 94 77 121 115

7. 7 Chebyshev's Theorem Given x=92.4, s=16.7, and the ordered stem and leaf: 6 78 7 12677 8 2346 9 347789 10 0059 11 25 12 16 ±2s: 92.4 ± 2*16.7 = [59, 125.6] 24 observations, or 96% 96% > 75%, as expected ±3s: 92.4 ± 3*16.7 = [42.3, 142.5] 25 observations, or 100% 100% > 89%, as expected

8. 8 Empirical Rule Empirical rule: Given a "mound" shaped distribution… –Approximately 68% of the observations will fill within the interval [x – 1s, x + 1s] –Approximately 95% of the observations will fill within the interval [x – 2s, x + 2s] –Approximately 99.7% of the observations will fill within the interval [x – 3s, x + 3s] Why is this important? Because observations which fall outside the interval [x – 2s, x + 2s] are often considered "unusual" or "significant".

9. 9 Empirical Rule 2.5%

10. 10 Empirical Rule Given x=92.4, s=16.7, and the ordered stem and leaf: 6 78 7 12677 8 2346 9 347789 10 0059 11 25 12 16 ±1s: 92.4 ± 16.7 = [75.7, 109.1] 17 observations, or 68% ±2s: 92.4 ± 2*16.7 = [59, 125.6] 24 observations, or 96% ±3s: 92.4 ± 3*16.7 = [42.3, 142.5] 25 observations, or 100%

11. 11 Percentiles and Quantiles The n th percentile (P n ) is that point below which n % of the observations fall. Recall the median is the halfway point of interval data. The median is also known as Q 2 or P 50. The lower quartile (Q L or Q 1 or P 25 ) is the median of the lower half. The upper quartile (Q U or Q 3 or P 75 ) is the median of the upper half.

12. 12 Percentiles and Quantiles Given: 12 5 22 30 7 36 14 42 15 53 25 Ordered: 5 7 12 14 15 22 25 30 36 42 53 Q 1 = 13 Q 2 = 22 Q 3 = 33

13. 13 Percentiles and Quantiles Given: 12 5 22 30 7 36 14 42 15 53 25 65 Ordered: 5 7 12 14 15 22 25 30 36 42 53 65 Q 1 = 13 Q 2 = 23.5 Q 3 = 39

14. 14 Percentiles and Quantiles But consider this simple example: 1 2 3 4 5 6 7 Is Q 1 = 2 or 2.5? Is Q 3 = 5.5 or 6? SAS software has five different ways of determining percentiles and quartiles! You will NOT be expected to find percentiles or quantiles, but should know what they are.

15. 15 Example: Credit Report

16. 16 Probability

17. 17 Probability Probability is man's attempt to quantify risk or uncertainty. Probability is a measure of the likelihood of an event occurring.

18. 18 Experiment Definition: An experiment is an act or process that leads to a single outcome that cannot be predicted with certainty.

19. 19 Experiment Example: Consider the experiment of simultaneously rolling a die and tossing a coin. What are all of the possible outcomes of this experiment? Answer: { 1T, 2T, 3T, 4T, 5T, 6T, 1H, 2H, 3H, 4H, 5H, 6H }

20. 20 Events Definition: A simple event is a single outcome of an experiment. Each of the above is a simple event given our experiment. Definition: The sample space of an experiment is the collection of all of its simple events. Definition: An event is a collection of one or more simple events. Therefore, an event is a subset of the sample space.

21. 21 Events Consider the following events: H: { observe a head }  rule method H: { 1H, 2H, 3H, 4H, 5H, 6H }  roster method E: { observe an even number } E: { 2H, 2T, 4H, 4T, 6H, 6T } R: { observe a three } R: { 3H, 3T } X: { observe a six } X: { 6H, 6T }

22. 22 Venn Diagram 1T1H2H2T 3T3H4H4T 5T5H6H6T R H X E (We will refer to this diagram throughout our discussion.)

23. 23 Probability The probability of an event is a measure of the likelihood of that event. Probability is always between 0 and 1 inclusive; that is, 0  P(x)  1.

24. 24 Probability P(H) = ______ P(E) = ______ P(R) = ______ P(X) = ______

25. 25 Together List the possible gender outcomes for a couple planning to have three children. Assuming that the outcomes listed above are equally likely, find the probability of getting two girls. Find the probability of getting exactly one child of each gender. Repeat as a tree diagram. Construct the complete probability distribution for the random variable X, where X is the number of girls in three children.

26. 26 Odds Don't confuse probability for odds. They are not the same thing! So if P(A) =.4, then odds in favor of A are.4/.6, or 2/3. Makes it sound more likely than it is! Odds are usually expressed in the form a:b, such as 2:3

27. 27 Together We already found that the probability of having two girls in three children is 3/8. What are the odds of having two girls in three children?

28. 28 Union Definition: The union of two events A and B is the event that occurs if A or B or both occur. We denote the union of two events A and B as { A  B } or { A or B } AB

29. 29 Union Consider the following events: If A: { H  E }, then P(A) = ______. If B: { R  X }, then P(B) = ______. If C: { E  X }, then P(C) = ______.

30. 30 Intersection Definition: The intersection of two events A and B is the event that occurs if both A and B occur. We denote the intersection of 2 events A and B as { A  B } or { A and B } AB

31. 31 Intersection Consider the following events: If A: { H  E }, then P(A) = ______. If B: { R  X }, then P(B) = ______. If C: { E  X }, then P(C) = ______. Also known as joint probability.

32. 32 Additive Rule of Probability Note: P(H  E) = P(H) + P(E) - P(H  E) 9/12 = 6/12 + 6/12 - 3/12 Note: P(R  X) = P(R) + P(X) - P(R  X) 4/12 = 2/12 + 2/12 - 0/12 Note: P(E  X) = P(E) + P(X) - P(E  X) 6/12 = 6/12 + 2/12 - 2/12

33. 33 Additive Rule of Probability These examples illustrate the Additive Rule of Probability: P(A  B) = P(A) + P(B) - P(A  B) or P(A or B) = P(A) + P(B) - P(A and B)

34. 34 Complement Definition: The complement of an event A is the event that A does not occur. The complement of A is denoted in several ways: ~A Ac A' A A' is often read as "A prime" or "not A". P(A) + P(A') = 1 therefore P(A') = 1 - P(A)

35. 35 Complement Example: Find the probability of observing a tail. P(tail) = 1 - P(tail') = 1 - P(head) = 1 - 6/12 = 6/12 Example: Let E be defined as the event of observing a 1,2,3,4,or 5. Find the P(E). P(E) = 1 - P(E') = 1 - P(six) = 1 - 2/12 = 10/12

36. 36 Contingency Tables Probabilities are often easier to work with if events and their respective probabilities are represented in tabular form: X X' HH' 1/122/12 6/12 P(H and X') = ____ P(H or X') = ____

37. 37 Together: Hybridization Experiment Hybridization Experiment: Use the figure (next slide) to identify the frequencies in the accompanying table. Green Yellow PurpleWhite Flower Pod

38. 38 Together: Hybridization Experiment GGGGGGGG YYYYYY

39. 39 Together: Hybridization Experiment Assume that one of the peas selected at random. Find the following probabilities: Find P(green pod and purple flower) Find P(green pod or purple flower) Find P(not green pod and not purple flower)

40. 40 Conditional Probability The probability that event A occurs given that event B has occurred (or assuming that B will occur) is referred to as the conditional probability of A given B and is denoted as P(A|B). Example: Find P(H|E). Since E has occurred, that, E, becomes our new sample space: P(H|E) = P(heads | 2H, 2T, 4H, 4T, 6H, 6T) = 3/6

41. 41 Conditional Probability P(R|X) = ______ P(E|X) = ______ P(X|E) = ______

42. 42 Conditional Probability Verify that P(H|E) = 3/6. Verify that P(R|X) = 0. Verify that P(E|X) = 1. Verify that P(X|E) = 2/6.

43. 43 Together Refer again to the hybridization data. If we select one pea at random… What is the probability of getting a purple flower given that the pod was green? What is the probability of getting a green pod given that the flower was purple?

44. 44 Independent Events Definition: Events A and B are independent if the knowledge (or assumption) that B has occurred(or will occur) does not alter the probability that A has occurred (or will occur). Example: P(R) = ______ P(R|H) = ______ Example: P(H) = ______ P(H|R) = ______ In general, if A and B are independent, then P(A|B) = P(A).

45. 45 Independent Events But we know that …so if A and B are independent, then...therefore

46. 46 Together Earlier we constructed a probability distribution for the random variable X, where X is the number of girls in three children. Repeat as a tree diagram. B (.5) G (.5) B (.5) G (.5) BBB (.5*.5*.5) BBG (.5*.5*.5) BGB (.5*.5*.5) BGG (.5*.5*.5) GBB (.5*.5*.5) GBG (.5*.5*.5) GGB (.5*.5*.5) GGG (.5*.5*.5)

47. 47 Mutually Exclusive Events Definition: Two events, A and B, are mutually exclusive if the two events cannot occur at the same time. If two events are mutually exclusive, then they have no intersection; P(A  B) = 0. Also known as disjoint sets. Example: Events R (three) and X (six) are mutually exclusive. Events H (heads) and E (even) are not mutually exclusive.

48. 48 Mutually Exclusive Events If A and B are complements, then they are mutually exclusive. However, if A and B are mutually exclusive, they are not necessarily complements!

49. 49 Components in Series vs. Parallel Two components are known to fail 10% of the time, each independent of the other. Find the probability that the system will not fail in each of the following configurations: AB Series A B Parallel The system fails if either component fails. The system fails if both components fail.

50. 50 Application (1 of 3) 100 shoppers at a large suburban mall were asked two questions -- (1) Did you see a television ad for the sale at department store X during the past two weeks, and (2) Did you shop at department store X during the past two weeks. The results of this survey are shown on the following slide…

51. 51 Application (2 of 3) Did not Shopped shop Totals Saw ad 42 28 70 Not see ad 18 12 30 Totals 60 40 100

52. 52 Application (3 of 3) Let A: {Saw ad} and S: {Shopped at X}. Answer… 1.Find P(A) 2.Find P(S) 3.Find P(A  S) 4.Find P(A  S) 5.Find P(S | A) 6.Find P(S | A’) 7.Are A and S independent events? 8.Comment on the effectiveness of the advertising.

53. 53 How to calculate

54. 54 An Interesting Example (Time Permitting) A test for a particular type of cancer returns a correct positive result in 98% of the cases in which the cancer is actually present, and a correct negative result in 97% of the cases in which the cancer is not present. Only 0.8% of the entire population have this cancer. A patient takes a lab test and the result comes back positive. What is the probability that the patient has this cancer?

55. 55 Counting Factorials, Permutations and Combinations

56. 56 Introduction Consider an athletic club consisting of four people -- Alice, Betty, Carol, and Donna -- which we will conveniently abbreviate as A, B, C and D.

57. 57 Introduction Question: In how many ways can the four finish in a race? Answer: 24 ABCD BACD CABD DABC ABDC BADC CADB DACB ACBD BCAD CBAD DBAC ACDB BCDA CBDA DBCA ADBC BDAC CDAB DCAB ADCB BDCA CDBA DCBA

58. 58 Introduction Question: In how many ways can a president and vice president be chosen? Answer: 12 AB BA CA DA AC BC CB DB AD BD CD DC

59. 59 Introduction Question: In how many ways can a committee of two be chosen? Answer: 6 AB (same as BA) AC (same as CA) AD (same as DA) BC (same as CB) BD (same as DB) CD (same as DC)

60. 60 Factorial The factorial symbol ! denotes the product of decreasing positive whole numbers. For example 4! = 4 * 3 * 2 * 1. By special definition, 0! = 1.

61. 61 Factorial Rule A collection of n different items can be arranged in order n ! different ways. This factorial rule reflects the fact that the first item may be selected n different ways, the second item may be selected n - 1 ways, and so on.

62. 62 Permutation Rule (When all items are different) There are n different items available. We select r of the n items (without replacement). We consider rearrangement of the same items to be different sequences. (The permutation of ABC is different from CBA and is counted separately.)

63. 63 Permutation Rule (When all items are different) If the preceding requirements are satisfied, the number of permutations (or sequences) of r items selected from n different items (without replacement) is

64. 64 Permutation Rule (When all items are different) Find 4 P 2 without the calculator: Find 4 P 2 with the calculator: 4 [MATH] [PRB] [nPr] 2 [ENTER]

65. 65 Permutation Rule (When some items are identical) If the preceding requirements are satisfied, and if there are n 1 alike, n 2 alike,..., n k alike, the number of permutations (or sequences) of r items selected from n different items (without replacement) is (Shown here only for completeness, we will not use this.)

66. 66 Combinations Rule There are n different items available. We select r of the n items (without replacement). We consider rearrangement of the same items to be the same. (The combination of ABC is the same as CBA.)

67. 67 Combinations Rule If the preceding requirements are satisfied, the number of combinations of r items selected from n different items is

68. 68 Combinations Rule Find 4 C 2 without the calculator: Find 4 C 2 with the calculator: 4 [MATH] [PRB] [nCr] 2 [ENTER]

69. 69 When to use which... A very simple mnemonic: Finish (all)  Factorial President  Permutation Committee  Combination

70. 70 Together A company with five female employees and ten male employees forms a committee of four employees consisting entirely of males. Assuming the employees were chosen at random, what is the probability of this happening? Is there reason to suspect discrimination? If the employees were chosen at random, what is the probability that the committee would consist of three men and one woman? Two men and two women?

71. 71 Excel Solution =HYPGEOM.DIST(A5,$A$9,$B$2,$B$1+$B$2,FALSE)