2. FORMAL STUDY OF UNCERTAINTY THE ENGINE THAT DRIVES STATISTICS Probability

3. Introduction Nothing in life is certain We gauge the chances of successful outcomes in business, medicine, weather, and other everyday situations such as the lottery (recall the birthday problem)

4. Approaches to Probability Relative frequency event probability = x/n, where x=# of occurrences of event of interest, n=total # of observations Coin, die tossing; nuclear power plants? Limitations repeated observations not practical

5. Approaches to Probability (cont.) Subjective probability individual assigns prob. based on personal experience, anecdotal evidence, etc. Classical approach every possible outcome has equal probability (more later)

6. Basic Definitions Experiment: act or process that leads to a single outcome that cannot be predicted with certainty Examples: 1.Toss a coin 2.Draw 1 card from a standard deck of cards 3.Arrival time of flight from Atlanta to RDU

7. Basic Definitions (cont.) Sample space: all possible outcomes of an experiment. Denoted by S Event: any subset of the sample space S; typically denoted A, B, C, etc. Simple event: event with only 1 outcome Null event: the empty set  Certain event: S

8. Examples 1.Toss a coin once S = {H, T}; A = {H}, B = {T} simple events 2.Toss a die once; count dots on upper face S = {1, 2, 3, 4, 5, 6} A=even # of dots on upper face={2, 4, 6} B=3 or fewer dots on upper face={1, 2, 3}

9. Laws of Probability

10. Laws of Probability (cont.) 3.P(A’ ) = 1 - P(A) For an event A, A’ is the complement of A; A’ is everything in S that is not in A. A A' S

11. Birthday Problem What is the smallest number of people you need in a group so that the probability of 2 or more people having the same birthday is greater than 1/2? Answer: 23 No. of people23304060 Probability.507.706.891.994

12. Example: Birthday Problem A={at least 2 people in the group have a common birthday} A’ = {no one has common birthday}

13. Unions and Intersections S A B A  A 

14. Mutually Exclusive Events Mutually exclusive events-no outcomes from S in common S A B A  = 

15. Laws of Probability (cont.) Addition Rule for Disjoint Events: 4. If A and B are disjoint events, then P(A  B) = P(A) + P(B)

16. 5. For two independent events A and B P(A  B) = P(A) × P(B)

17. Laws of Probability (cont.) General Addition Rule 6. For any two events A and B P(A  B) = P(A) + P(B) – P(A  B)

18. P(A  B)=P(A) + P(B) - P(A  B) S AB A 

19. Example: toss a fair die once S = {1, 2, 3, 4, 5, 6} A = even # appears = {2, 4, 6} B = 3 or fewer = {1, 2, 3} P(A  B) = P(A) + P(B) - P(A  B) =P({2, 4, 6}) + P({1, 2, 3}) - P({2}) = 3/6 + 3/6 - 1/6 = 5/6

20. Laws of Probability: Summary 1. 0  P(A)  1 for any event A 2. P(  ) = 0, P(S) = 1 3. P(A’) = 1 – P(A) 4. If A and B are disjoint events, then P(A  B) = P(A) + P(B) 5. If A and B are independent events, then P(A  B) = P(A) × P(B) 6. For any two events A and B, P(A  B) = P(A) + P(B) – P(A  B)

21. Assigning Probabilities If an experiment has N outcomes, then each outcome has probability 1/N of occurring If an event A 1 has n 1 outcomes, then P(A 1 ) = n 1 /N

22. We Need Efficient Methods for Counting Outcomes

23. Product Rule for Ordered Pairs A student wishes to commute to a junior college for 2 years and then commute to a state college for 2 years. Within commuting distance there are 4 junior colleges and 3 state colleges. How many junior college-state college pairs are available to her?

24. Product Rule for Ordered Pairs junior colleges: 1, 2, 3, 4 state colleges a, b, c possible pairs: (1, a) (1, b) (1, c) (2, a) (2, b) (2, c) (3, a) (3, b) (3, c) (4, a) (4, b) (4, c)

25. THE EQUALLY LIKELY APPROACH (ALSO CALLED THE CLASSICAL APPROACH) Probability Models

26. Product Rule for Ordered Pairs junior colleges: 1, 2, 3, 4 state colleges a, b, c possible pairs: (1, a) (1, b) (1, c) (2, a) (2, b) (2, c) (3, a) (3, b) (3, c) (4, a) (4, b) (4, c) 4 junior colleges 3 state colleges total number of possible pairs = 4 x 3 = 12 4 junior colleges 3 state colleges total number of possible pairs = 4 x 3 = 12

27. Product Rule for Ordered Pairs junior colleges: 1, 2, 3, 4 state colleges a, b, c possible pairs: (1, a) (1, b) (1, c) (2, a) (2, b) (2, c) (3, a) (3, b) (3, c) (4, a) (4, b) (4, c) In general, if there are n 1 ways to choose the first element of the pair, and n 2 ways to choose the second element, then the number of possible pairs is n 1 n 2. Here n 1 = 4, n 2 = 3. In general, if there are n 1 ways to choose the first element of the pair, and n 2 ways to choose the second element, then the number of possible pairs is n 1 n 2. Here n 1 = 4, n 2 = 3.

28. Counting in “Either-Or” Situations NCAA Basketball Tournament: how many ways can the “bracket” be filled out? 1. How many games? 2. 2 choices for each game 3. Number of ways to fill out the bracket: 2 63 = 9.2 × 10 18 Earth pop. about 6 billion; everyone fills out 1 million different brackets Chances of getting all games correct is about 1 in 1,000

29. Counting Example Pollsters minimize lead-in effect by rearranging the order of the questions on a survey If Gallup has a 5-question survey, how many different versions of the survey are required if all possible arrangements of the questions are included?

30. Solution There are 5 possible choices for the first question, 4 remaining questions for the second question, 3 choices for the third question, 2 choices for the fourth question, and 1 choice for the fifth question. The number of possible arrangements is therefore 5  4  3  2  1 = 120

31. Efficient Methods for Counting Outcomes Factorial Notation: n!=1  2  …  n Examples 1!=1; 2!=1  2=2; 3!= 1  2  3=6; 4!=24; 5!=120; Special definition: 0!=1

32. Factorials with calculators and Excel Calculator: non-graphing: x ! (second function) graphing: bottom p. 9 T I Calculator Commands (math button) Excel: Paste: math, fact

33. Factorial Examples 20! = 2.43 x 10 18 1,000,000 seconds? About 11.5 days 1,000,000,000 seconds? About 31 years 31 years = 10 9 seconds 10 18 = 10 9 x 10 9 31 x 10 9 years = 10 9 x 10 9 = 10 18 seconds 20! is roughly the age of the universe in seconds

34. Permutations A B C D E How many ways can we choose 2 letters from the above 5, without replacement, when the order in which we choose the letters is important? 5  4 = 20

35. Permutations (cont.)

36. Permutations with calculator and Excel Calculator non-graphing: nPr Graphing p. 9 of T I Calculator Commands (math button) Excel Paste: Statistical, Permut

37. Combinations A B C D E How many ways can we choose 2 letters from the above 5, without replacement, when the order in which we choose the letters is not important? 5  4 = 20 when order important Divide by 2: (5  4)/2 = 10 ways

38. Combinations (cont.)

39. ST 101 Powerball Lottery From the numbers 1 through 20, choose 6 different numbers. Write them on a piece of paper.

40. Chances of Winning?

41. Prior to Jan. 1, 2009 After Jan. 1, 2009 North Carolina Powerball Lottery

42. Visualize Your Lottery Chances How large is 195,249,054? $1 bill and $100 bill both 6” in length 10,560 bills = 1 mile Let’s start with 195,249,053 $1 bills and one $100 bill … … and take a long walk, putting down bills end-to- end as we go

43. Raleigh to Ft. Lauderdale… … still plenty of bills remaining, so continue from …

44. … Ft. Lauderdale to San Diego … still plenty of bills remaining, so continue from…

45. … San Diego to Seattle

46. … still plenty of bills remaining, so continue from … … Seattle to New York

47. … still plenty of bills remaining, so … … New York back to Raleigh

48. Go around again! Lay a second path of bills Still have ~ 5,000 bills left!!

49. Chances of Winning NC Powerball Lottery? Remember: one of the bills you put down is a $100 bill; all others are $1 bills Your chance of winning the lottery is the same as bending over and picking up the $100 bill while walking the route blindfolded.

50. Example: Illinois State Lottery

51. Virginia State Lottery

52. A GRAPHICAL METHOD FOR COMPLICATED PROBABILITY PROBLEMS Probability Trees

53. Example: AIDS Testing V={person has HIV}; CDC: P(V)=.006 +: test outcome is positive (test indicates HIV present) -: test outcome is negative clinical reliabilities for a new HIV test: 1. If a person has the virus, the test result will be positive with probability.999 2. If a person does not have the virus, the test result will be negative with probability.990

54. Question 1 What is the probability that a randomly selected person will test positive?

55. Probability Tree Approach A probability tree is a useful way to visualize this problem and to find the desired probability.

56. Probability Tree clinical reliability

57. Probability Tree Multiply branch probs clinical reliability

58. Question 1 Answer What is the probability that a randomly selected person will test positive? P(+) =.00599 +.00994 =.01593

59. Question 2 If your test comes back positive, what is the probability that you have HIV? (Remember: we know that if a person has the virus, the test result will be positive with probability.999; if a person does not have the virus, the test result will be negative with probability.990). Looks very reliable

60. Question 2 Answer Answer two sequences of branches lead to positive test; only 1 sequence represented people who have HIV. P(person has HIV given that test is positive) =.00599/(.00599+.00994) =.376

61. Summary Question 1: P(+) =.00599 +.00994 =.01593 Question 2: two sequences of branches lead to positive test; only 1 sequence represented people who have HIV. P(person has HIV given that test is positive) =.00599/(.00599+.00994) =.376

62. Recap We have a test with very high clinical reliabilities: 1. If a person has the virus, the test result will be positive with probability.999 2. If a person does not have the virus, the test result will be negative with probability.990 But we have extremely poor performance when the test is positive: P(person has HIV given that test is positive) =.376 In other words, 62.4% of the positives are false positives! Why? When the characteristic the test is looking for is rare, most positives will be false.

63. examples 1. P(A)=.3, P(B)=.4; if A and B are mutually exclusive events, then P(A  B)=? A  B = , P(A  B) = 0 2. 15 entries in pie baking contest at state fair. Judge must determine 1 st, 2 nd, 3 rd place winners. How many ways can judge make the awards? 15 P 3 = 2730