1. 1  Event - any collection of results or outcomes from some procedure  Simple event - any outcome or event that cannot be broken down into simpler components  Compound event – an event made up of two or more other events  Sample space - all possible simple events Definitions

2. 2 Notation P - denotes a probability A, B,... - denote specific events P (E) - denotes the probability of event E occurring

3. 3 Basic Rules for Computing Probability Rule 1: Relative Frequency Approximation Conduct (or observe) an experiment a large number of times, and count the number of times event E actually occurs, then an estimate of P(E) is

4. 4 Basic Rules for Computing Probability Rule 1: Relative Frequency Approximation Conduct (or observe) an experiment a large number of times, and count the number of times event A actually occurs, then an estimate of P(E) is P(E) = Number of times E occurred Total number of possible outcomes

5. 5 Basic Rules for Computing Probability Rule 2: Classical approach (requires equally likely outcomes) If a procedure has n different simple events, each with an equal chance of occurring, and s is the number of ways event E can occur, then

6. 6 Basic Rules for Computing Probability Rule 2: Classical approach (requires equally likely outcomes) If a procedure has n different simple events, each with an equal chance of occurring, and s is the number of ways event E can occur, then P(E) = number of ways E can occur number of different simple events s n =

7. 7 Basic Rules for Computing Probability Rule 3: Subjective Probabilities P(E), the probability of E, is found by simply guessing or estimating its value based on knowledge of the relevant circumstances.

8. 8 Rule 1 The relative frequency approach is an approximation.

9. 9 Rule 2 The classical approach is the actual probability.

10. 10 Law of Large Numbers As a procedure is repeated again and again, the relative frequency probability (from Rule 1) of an event tends to approach the actual probability.

11. 11 Law of Large Numbers Flip a coin 20 times and record the number of heads after each trial. In L1 list the numbers 1-20, in L2 record the number of heads. In L3, divide L2 by L1. Get a scatter plot with L1 and L3. What can you conclude?

12. 12 The sample space consists of two simple events: the person is struck by lightning or is not. Because these simple events are not equally likely, we can use the relative frequency approximation (Rule 1) or subjectively estimate the probability (Rule 3). Using Rule 1, we can research past events to determine that in a recent year 377 people were struck by lightning in the US, which has a population of about 274,037,295. Therefore, P(struck by lightning in a year)  377 / 274,037,295  1 / 727,000 Example: Find the probability that a randomly selected person will be struck by lightning this year.

13. 13 Example: On an ACT or SAT test, a typical multiple-choice question has 5 possible answers. If you make a random guess on one such question, what is the probability that your response is wrong? There are 5 possible outcomes or answers, and there are 4 ways to answer incorrectly. Random guessing implies that the outcomes in the sample space are equally likely, so we apply the classical approach (Rule 2) to get: P(wrong answer) = 4 / 5 = 0.8

14. 14 Probability Limits  The probability of an impossible event is 0.  The probability of an event that is certain to occur is 1.

15. 15 Probability Limits  The probability of an impossible event is 0.  The probability of an event that is certain to occur is 1. 0  P(A)  1

16. 16 Probability Limits  The probability of an impossible event is 0.  The probability of an event that is certain to occur is 1. 0  P(A)  1 Impossible to occur Certain to occur

17. 17 Possible Values for Probabilities Certain Likely 50-50 Chance Unlikely Impossible 1 0.5 0

18. 18 Complementary Events

19. 19 Complementary Events The complement of event E, denoted by E c, consists of all outcomes in which event E does not occur.

20. 20 P(E) Complementary Events The complement of event E, denoted by E c, consists of all outcomes in which event E does not occur. P(E C ) (read “not E”)

21. 21 Example: Testing Corvettes The General Motors Corporation wants to conduct a test of a new model of Corvette. A pool of 50 drivers has been recruited, 20 or whom are men. When the first person is selected from this pool, what is the probability of not getting a male driver?

22. 22 Because 20 of the 50 subjects are men, it follows that 30 of the 50 subjects are women so, P(not selecting a man) = P(man) c = P(woman) = 30 = 0.6 50 Example: Testing Corvettes The General Motors Corporation wants to conduct a test of a new model of Corvette. A pool of 50 drivers has been recruited, 20 or whom are men. When the first person is selected from this pool, what is the probability of not getting a male driver?

23. 23 Using a Tree Diagram Flipping a coin is an experiment and the possible outcomes are heads (H) or tails (T). One way to picture the outcomes of an experiment is to draw a tree diagram. Each outcome is shown on a separate branch. For example, the outcomes of flipping a coin are H T

24. 24 A Tree Diagram for Tossing a Coin Twice There are 4 possible outcomes when tossing a coin twice. H T H T H T First TossSecond TossOutcomes HH HT TH TT

25. 25 Rules of Complementary Events P(A) + P(A) c = 1

26. 26 P(A) c Rules of Complementary Events P(A) + P(A) c = 1 = 1 - P(A)

27. 27 (1,1) (1,2) (1,3) (1,4) (1,5) (1,6) (2,1) (2,2) (2,3) (2,4) (2,5) (2,6) (3,1) (3,2) (3,3) (3,4) (3,5) (3,6) (4,1) (4,2) (4,3) (4,4) (4,5) (4,6) (5,1) (5,2) (5,3) (5,4) (5,5) (5,6) (6,1) (6,2) (6,3) (6,4) (6,5) (6,6) Possible outcomes for two rolls of a die

28. 28 1.Find the probability that the sum is a 2 2.Find the probability that the sum is a 3 3.Find the probability that the sum is a 4 4.Find the probability that the sum is a 5 5.Find the probability that the sum is a 6 6.Find the probability that the sum is a 7 7.Find the probability that the sum is a 8 8.Find the probability that the sum is a 9 9.Find the probability that the sum is a 10 10. Find the probability that the sum is a 11 11.Find the probability that the sum is a 12 1/36 2/36 3/36 4/36 5/36 6/36 5/36 4/26 3/36 2/36 1/36 Find the following probabilities

29. 29

30. 30 Rounding Off Probabilities  give the exact fraction or decimal or

31. 31 Rounding Off Probabilities  give the exact fraction or decimal or  round off the final result to three significant digits

32. 32 How many ways are there to answer a two question test when the first question is a true-false question and the second question is a multiple choice question with five possible answers? Tree Diagram of Test Answers

33. 33 Ta Tb Tc Td Te Fa Fb Fc Fd Fe abcdeabcdeabcdeabcde TFTF Tree Diagram of Test Answers

34. 34 What is the probability that the first question is true and the second question is c? Tree Diagram of Test Answers

35. 35 Ta Tb Tc Td Te Fa Fb Fc Fd Fe abcdeabcdeabcdeabcde TFTF Tree Diagram of Test Answers

36. 36 Ta Tb Tc Td Te Fa Fb Fc Fd Fe abcdeabcdeabcdeabcde TFTF P(T) = FIGURE 3-9 Tree Diagram of Test Answers 1 2

37. 37 Ta Tb Tc Td Te Fa Fb Fc Fd Fe abcdeabcdeabcdeabcde TFTF P(T) = P(c) = Tree Diagram of Test Answers 1 2 1 5

38. 38 Ta Tb Tc Td Te Fa Fb Fc Fd Fe abcdeabcdeabcdeabcde TFTF P(T) = P(c) = P(T and c) = FIGURE 3-9 Tree Diagram of Test Answers 1 2 1 5 1 10

39. 39 P (both correct) = P (T and c)

40. 40 R RPS P RPS S RPS Rock – Paper – Scissors Tree Diagram- 2 Players T B A A T B B A T

41. 41 Rock – Paper – Scissors Tree Diagram- 3 Players A B B B B C B C B B B C B A B C B B B C B C B B B B A

42. 42 1/16 5/8 13/14 1/3 3/4 1/3 Pg 189 #9

43. 43  Compound Event Any event combining 2 or more simple events Definition

44. 44

45. 45  Notation P(A or B) = P (event A occurs or event B occurs or they both occur) Definition

46. 46 General Rule When finding the probability that event A occurs or event B occurs, find the total number of ways A can occur and the number of ways B can occur, but find the total in such a way that no outcome is counted more than once. Compound Event

47. 47 Formal Addition Rule P(A or B) = P(A) + P(B) - P(A and B) where P(A and B) denotes the probability that A and B both occur at the same time. Compound Event

48. 48 Definition Events A and B are mutually exclusive if they cannot occur simultaneously.

49. 49 Venn Diagrams Total Area = 1 P(A) P(B) Non-overlapping Events P(A) P(B) P(A and B) Overlapping Events Total Area = 1

50. 50 Venn Diagrams

51. 51 A  (B  C)A c  B(A  B ) c A  (B  C)(A  B)  C A - B (A  B)  (A  C) U - A c A c  B c  C c

52. 52 A c  (B c  C)A c  (B  C) c (A  B) c (A  B)  (A  C) A - (B  C) A  (B  A c ) (A  B)  (A  C) B - A (A c  B c )  C c

53. 53 A poll was taken of 100 students to find out how they arrived at school. 28 used car pools; 31 used buses; and 42 said they drove to school alone.In addition, 9 used both car pools and buses; 10 used car pools and drove alone; only 6 used buses and their own car and 4 used all three methods. a. Complete the Venn diagram. b. How many used none of the methods? c. How many used only car pools? d. How many used buses exclusively? 4 65 2 30 20 13 20 13 20 P(A U B U C) = 42 + 31 + 28 – 6 – 10 -9 + 4 = 80

54. 54 A survey of 500 television watchers produced the following information: 285 watch football 190 watch hockey 115 watch basketball 45 watch football and basketball 70 watch football and hockey 50 watch hockey and basketball 50 do not watch any sports. a. How many watch all three games? b. How many watch exactly one of the three games? 500 = 285 + 190 + 115 - 45 - 70 - 50 + P(A  B  C) + 50 500 = 475 + P(A  B  C) P(A  B  C) = 25 25 45 20 45 195 95 25 50

55. 55 Applying the Addition Rule P(A or B) Addition Rule Are A and B mutually exclusive ? P(A or B) = P(A)+ P(B) - P(A and B) P(A or B) = P(A) + P(B) Yes No

56. 56 Find the probability of randomly selecting a man or a boy. Men Women Boys Girls Totals Survived 332 31829 27 706 Died 1360 10435 18 1517 Total 1692 422 64 45 2223 Contingency Table

57. 57 Find the probability of randomly selecting a man or a boy. Men Women Boys Girls Totals Survived 332 31829 27 706 Died 1360 10435 18 1517 Total 1692 422 64 45 2223 Contingency Table

58. 58 Find the probability of randomly selecting a man or a boy. P(man or boy) = Men Women Boys Girls Totals Survived 332 31829 27 706 Died 1360 10435 18 1517 Total 1692 422 64 45 2223 Contingency Table * Mutually Exclusive *

59. 59 Find the probability of randomly selecting a man or someone who survived. Men Women Boys Girls Totals Survived 332 31829 27 706 Died 1360 10435 18 1517 Total 1692 422 64 45 2223 Contingency Table

60. 60 Find the probability of randomly selecting a man or someone who survived. Men Women Boys Girls Totals Survived 332 31829 27 706 Died 1360 10435 18 1517 Total 1692 422 64 45 2223 Contingency Table

61. 61 Find the probability of randomly selecting a man or someone who survived. P(man or survivor) = Men Women Boys Girls Totals Survived 332 31829 27 706 Died 1360 10435 18 1517 Total 1692 422 64 45 2223 Contingency Table * NOT Mutually Exclusive * = 0.929

62. 62 Complementary Events P(A) and P(A) c are mutually exclusive All simple events are either in A or A c

63. 63 Venn Diagram for the Complement of Event A Total Area = 1 P (A) P (A) c = 1 - P (A)

64. 64 Finding the Probability of Two or More Selections  Multiple selections  Multiplication Rule

65. 65 Notation P(A and B) = P(event A occurs in a first trial and event B occurs in a second trial)

66. 66 Conditional Probability  Definition  The conditional probability of event B occurring, given that A has already occurred, can be found by dividing the probability of events A and B both occurring by the probability of event A.

67. 67 Conditional Probability P(A and B) = P(A) P(B | A) The conditional probability of B given A can be found by assuming the event A has occurred and, operating under that assumption, calculating the probability that event B will occur. P(A and B) P(A) P(B | A) =

68. 68  P(B|A) represents the probability of event B occurring after it is assumed that event A has already occurred (read B|A as “B given A”). Notation for Conditional Probability

69. 69 Definitions  Independent Events Two events A and B are independent if the occurrence of one does not affect the probability of the occurrence of the other.  Dependent Events If A and B are not independent, they are said to be dependent.

70. 70 Formal Multiplication Rule  P(A and B) = P(A) P(B|A)  If A and B are independent events, P(B|A) is really the same as P(B)

71. 71 Applying the Multiplication Rule P(A or B) Multiplication Rule Are A and B independent ? P(A and B) = P(A) P(B|A) P(A and B) = P(A) P(B) Yes No

72. 72 Probability of ‘At Least One’  ‘At least one’ is equivalent to ‘one or more’.  The complement of getting at least one item of a particular type is that you get no items of that type. IfP(A) = P(getting at least one), then P(A) = 1 - P(A) c where P(A) c is P(getting none)

73. 73 Probability of ‘At Least One’  Find the probability of a couple have at least 1 girl among 3 children. If P(A) = P(getting at least 1 girl), then P(A) = 1 - P(A) c where P(A) c is P(getting no girls) P(A) c = (0.5)(0.5)(0.5) = 0.125 P(A) = 1 - 0.125 = 0.875

74. 74 If P(B | A) = P(B) then the occurrence of A has no effect on the probability of event B; that is, A and B are independent events. Testing for Independence

75. 75 If P(B | A) = P(B) then the occurrence of A has no effect on the probability of event B; that is, A and B are independent events. or If P(A and B) = P(A) P(B) then A and B are independent events. Testing for Independence

76. 76 Find the probability of randomly selecting a man if you know the person is a survivor Men Women Boys Girls Totals Survived 332 31829 27 706 Died 1360 10435 18 1517 Total 1692 422 64 45 2223 Contingency Table Find the probability of selecting a survivor if you know the person is not a boy

77. 77 Calculate the following probabilities: a.P(A 1 ) e. P(A 2 U B 3 ) b. P(B 3 ) f. P(B 1 U B 4 ) c. P (A 1  B 4 ) g. P( B 2  B 4 ) d. P(B 1 |A 3 ) h. P(A 2 |B 4 ) 754050 95 70 50 215 95/215= 19/43 50/215=10/43 25/215 = 5/43 25/50=1/2 105/215=21/43 125/215=25/43 none 15/50 = 3/10

78. 78 P(A) = 1/3 P(B) = 1/4 P(A U B) = 1/2 1/12 3/122/12 6/12 A B a.1/12 b.1/3 c.1/4 d.1/4 e.1/2 f.2/3 g.2/3 h.11/12