1. CHAPTER 1 Probability Chapter 1 1

2. Introduction o Definition ~ the measure of any possibilities occurrence of a result in an event. o Question : With reference to the sample space describe the event B that the total number of points rolled with the pair of dice is 7. o Answer : B = (1,6),(2,5),(3,4),(4,3),(5,2),(6,1) Chapter 1 2

3. Exercise A box contains 4 red balls, 6 green balls and 5 yellow balls. Two balls are drawn at random. Determine the sample space for this experiment. Given the following events: A : all balls are different colors B : at least one yellow ball is chosen C : no green balls are chosen. Chapter 1 3

4. Sample space The set of all possible outcomes of a random experiment. The sample space is denoted as e.g. Sample without replacement using example if the batch consists of three items and you have to select only two items : Chapter 1 4

5. Sample space Sample with replacement Can be described graphically with tree diagrams Chapter 1 5

6. Sample space : Tree Diagrams Chapter 1 6

7. 7 Sample Space : Experiment Toss a coin Experiment Head Tail or Experiment outcomes Sample space

8. Chapter 1 8 Sample space : Example The die toss: The die toss: Simple events: Sample space: 1 1 2 2 3 3 4 4 5 5 6 6 E1E2E3E4E5E6E1E2E3E4E5E6 S ={E 1, E 2, E 3, E 4, E 5, E 6 }S E 1 E 6 E 2 E 3 E 4 E 5

9. Events Subset of the sample space of a random experiment. Union ~ consists of all outcomes that are contained in either events. Intersection ~ consists of all outcomes that contained in all events. Complement ~ set of outcomes in the sample space that are not in the event. Chapter 1 9

10. 10 Events : Union union The union of two events, A and B, is the event that consists of all outcomes that are contained in either of the two events. We write A  B S AB

11. Chapter 1 11 Events : Intersection The intersection of two events, A and B, is the event that both A and B occur when the experiment is performed. We write S AB If two events A and B are mutually exclusive, then P(A B) = 0.

12. Chapter 1 12 Events : Complement The complement of an event A consists of all outcomes of the experiment that do not result in event A. We write A C. S A ACAC

13. Chapter 1 13 Events : Mutually Exclusive Events that cannot occur together

14. Exercise S = 1,2,3,4,5,6 A = 1,2,3 B = 2,4,5,6 Find the union of AB, intersection of AB and the complement of A Chapter 1 14

15. Events : Counting techniques Counts of the numbers of outcomes in the sample space and various events are used to analyze the random experiments. Multiplication rule Chapter 1 15 If an operation can be described as a sequence of k steps, therefore the total number of ways of completing the operation is :

16. Example In the design of a casing for a gear housing, we can use four different types of fasteners, three different bolt lengths, and three different bolt locations. From the multiplication rule ; ~ 36 different designs are possible Chapter 1 16

17. Events : Permutations Variation of the order or choice from a set of things Example : Thus, the permutations of are ~ Chapter 1 17 The number of permutations of n different elements is n ! where

18. Events : Permutations Chapter 1 18 The number of permutations of subsets of r elements selected from a set of n different elements is Sometimes we are interested in the number of arrangements of only some of the elements of a set.

19. Example A printed circuit board has eight different locations in which a component can be placed. If four different components are to be placed on the board, how many designs are possible? Each design consists of selecting a location from the 8th locations for the first component, a location from the remaining seven for the second component, a location from the remaining six for the third component, and a location from the remaining five for the fourth component. Therefore, Chapter 1 19 *1680 different designs are possible

20. Events : Permutations Chapter 1 20 The number of permutations of n = n 1 + n 2 +…+ n r objects of which n 1 are of one type, n 2 are of second type,…and n r are of an r th type is Sometimes we are interested in counting the number of ordered sequences for objects that are not all different. The following result is a useful, general calculation.

21. Example (Bar Codes) A part is labeled by printing with four thick lines, three medium lines, and two thin lines. If each ordering of the nine lines represents a different label, how many different labels can be generated by using the scheme? Chapter 1 21 *1260 different part labels

22. Events : Permutations Another counting problem of interest is the number of subsets of r elements that can be selected from a set of n elements. In this case, order is not important. These are called combinations. Chapter 1 22 The number of combinations, subsets of size r that can be selected from a set of n elements, is denoted as or and

23. Example A printed circuit board has eight different locations in which a component can be placed. If five identical components are to be placed on the board, how many different designs are possible? Each design is a subset of the eight locations that are to contain the components. From the equation, the number of possible designs is Chapter 1 23

24. Chapter 1 24 What is Probability? Probability the study of chance associated with the occurrence of events Used to represent risk or uncertainty in engineering applications Can be interpreted as our degree of belief

25. Chapter 1 25 The Probability of an Event The probability of an event A measures “Chance” we think A will occur. P(A). Denoted by P(A).

26. Chapter 1 26 The Probability of an Event P(A) must be between 0 and 1. If event A can never occur, P(A) = 0. If event A always occurs when the experiment is performed, P(A) =1. The sum of the probabilities for all simple events in S equals 1.

27. Chapter 1 27 Probability : Example Experiment: Toss a coin once S = {Head, Tail} Let A be the event that a head is obtained

28. Chapter 1 28 Example Experiment: Roll a die once S = {1, 2, 3, 4, 5, 6} Let M be the event that an even number is obtained. Probability : Example

29. Chapter 1 29 Probability : Example Toss a fair coin twice. What is the probability of observing at least one head? H H 1st Coin 2nd Coin E i P(E i ) H H T T T T H H T T HH HT TH TT 1/4 P(at least 1 head) = P(E 1 ) + P(E 2 ) + P(E 3 ) = 1/4 + 1/4 + 1/4 = 3/4 P(at least 1 head) = P(E 1 ) + P(E 2 ) + P(E 3 ) = 1/4 + 1/4 + 1/4 = 3/4

30. Chapter 1 30 Probability : Example A bowl contains three M&Ms ®, one red, one blue and one green. A child selects two M&Ms at random. What is the probability that at least one is red? 1st M&M 2nd M&M E i P(E i ) RB RG BR BG 1/6 P(at least 1 red) = P(RB) + P(BR)+ P(RG) + P(GR) = 4/6 = 2/3 P(at least 1 red) = P(RB) + P(BR)+ P(RG) + P(GR) = 4/6 = 2/3 m m m m m m m m m GB GR

31. Chapter 1 31 Probability : Union For any two events, A and B, the probability of their union, P(A  B), is A B

32. Chapter 1 32 Example : Union Suppose that there were 120 students in the classroom, and that they could be classified as follows: BrownNot Brown Male2040 Female30 A: brown hair P(A) = 50/120 B: female P(B) = 60/120 P(A  B) = P(A) + P(B) – P(A  B) = 50/120 + 60/120 - 30/120 = 80/120 = 2/3 P(A  B) = P(A) + P(B) – P(A  B) = 50/120 + 60/120 - 30/120 = 80/120 = 2/3 Check: P(A  B) = (20 + 30 + 30)/120 Check: P(A  B) = (20 + 30 + 30)/120

33. Chapter 1 33 Example : Special Case mutually exclusive, When two events A and B are mutually exclusive, P(A  B) = 0 and P(A  B) = P(A) + P(B). BrownNot Brown Male2040 Female30 A : male with brown hair P(A) = 20/120 B: female with brown hair P(B) = 30/120 P(A  B) = P(A) + P(B) = 20/120 + 30/120 = 50/120 P(A  B) = P(A) + P(B) = 20/120 + 30/120 = 50/120 A and B are mutually exclusive, so that

34. Chapter 1 34 Probability : Complement A: We know that for any event A: P(A ∩ A C ) = 0 Since either A or A C must occur, P(A  A C ) =1 so thatP(A A C ) = P(A)+ P(A C ) = 1 P(A C ) = 1 – P(A) A ACAC

35. Chapter 1 35 Example : Complement BrownNot Brown Male2040 Female30 A: male P(A) = 60/120 B: female P(B) = 1- P(A) = 1- 60/120 = 60/120 = 1/2 P(B) = 1- P(A) = 1- 60/120 = 60/120 = 1/2 A and B are complementary, so that Select a student at random from the classroom. Define:

36. Conditional probability The probability of an event B under the knowledge that the outcome will be in event A is denoted as The conditional probability of an event B given an event A, denoted as for Chapter 1 36

37. Chapter 1 37 Conditional Pr : Example A jar contains black and white marbles. Two marbles are chosen without replacement. The probability of selecting a black marble and a white marble is 0.34, and the probability of selecting a black marble on the first draw is 0.47. What is the probability of selecting a white marble on the second draw, given that the first marble drawn was black? Solution: P(White|Black) = P(Black and White) P(Black) = = 0.72 = 72%

38. Chapter 1 38 Multiplication and Total Probability Rules Multiplication Rule Total Probability Rule Assume E 1,E 2,…,E k are k mutually exclusive and exhaustive sets. Then,

39. Chapter 1 39 Example : Multiplication Rule The probability that an automobile battery subject to high engine compartment temperature suffers low charging current is 0.7. The probability that a battery is subject to high engine compartment temperature is 0.05. Let C denote the event that a battery suffers low charging current, and let T denote the event that a battery is subject to high engine compartment temperature. The probability that a battery is subject to low charging current and high engine compartment temperature is

40. Chapter 1 40 Example : Total Probability Rule Semiconductor contamination Let F denote the event that the product fails, and let H denote the event that the chip is exposed to high levels of contamination. The requested probability is P(F), and the information provided can be represented as Probability of FailureLevel of ContaminationProbability of Level 0.1High0.2 0.005Not High0.8 P(F) = 0.10(0.20) + 0.005(0.80) = 0.024 Which can be interpreted as just the weighted average of the two probabilities of failure.

41. Chapter 1 41 Independence Event Definition: Two events are independent if any one of the following equivalent statements is true: (1)P(A ∩ B)=P(A)P(B) (2)P(A │ B)=P(A) (3)P(B │ A)=P(B)

42. Example : Independence Event The following circuit operates only if there is a path of functional devices from left to right. The probability that each device functions is shown on the graph. L and R are independent. What is the probability that the circuit operates? Chapter 1 42 0.80.9

43. Solution Let L and R denote the event that the left and right devices operate, respectively. There is only a path if both operate. The probability the circuit operates is Chapter 1 43

44. Bayes’ Theorem If E 1,E 2,…,E k are k mutually exclusive and exhaustive events and B is any event, Chapter 1 44

45. Chapter 1 45 WHEN TO APPLY? Part of the challenge in applying Bayes‘ theorem involves recognizing the types of problems that warrant its use. You should consider Bayes' theorem when the following conditions exist. The sample space is partitioned into a set of mutually exclusive eventssample spaceset mutually exclusive { A1, A2,..., An }. Bayes’ Theorem

46. Chapter 1 46 Bayes’ Theorem Prior Probabilities New information Application of Bayes’ Theorem Posterior Probabilities

47. Chapter 1 47 Example : Rain Marie is getting married tomorrow, at an outdoor ceremony in the desert. In recent years, it has rained only 5 days each year. Unfortunately, the weatherman has predicted rain for tomorrow. When it actually rains, the weatherman correctly forecasts rain 90% of the time. When it doesn't rain, he incorrectly forecasts rain 10% of the time. What is the probability that it will rain on the day of Marie's wedding?

48. Chapter 1 48 The sample space is defined by two mutually- exclusive events - it rains or it does not rain. Additionally, a third event occurs when the weatherman predicts rain. Notation for these events appears below. Event A1. It rains on Marie's wedding. Event A2. It does not rain on Marie's wedding Event B. The weatherman predicts rain. Solution : Rain

49. Chapter 1 49 In terms of probabilities, we know the following: P( A1 ) = 5/365 =0.0136985 [It rains 5 days out of the year.] P( A2 ) = 360/365 = 0.9863014 [It does not rain 360 days out of the year.] P( B | A1 ) = 0.9 [When it rains, the weatherman predicts rain 90% of the time.] P( B | A2 ) = 0.1 [When it does not rain, the weatherman predicts rain 10% of the time.] Solution : Rain

50. Chapter 1 50 We want to know P( A1 | B ), the probability it will rain on the day of Marie's wedding, given a forecast for rain by the weatherman. The answer can be determined from Bayes' theorem, as shown below. Solution : Rain

51. Chapter 1 51 Note the somewhat unintuitive result. When the weatherman predicts rain, it actually rains only about 11% of the time. Despite the weatherman's gloomy prediction, there is a good chance that Marie will not get rained at her wedding. Conclusion : Rain

52. Chapter 1 52 A proposed shopping center will provide strong competition for downtown businesses like L. S. Clothiers. If the shopping center is built, the owner of L. S. Clothiers will need to take few actions so that his business can keep going. The shopping center cannot be built unless a zoning change is approved by the town council. The planning board must first make a recommendation, for or against the zoning change, to the council. Example : L.S. Clothiers

53. Chapter 1 53 (1)Prior Probabilities Let: A 1 = town council approves the zoning change A 2 = town council disapproves the change Using subjective judgment: P(A 1 ) = 0.7, P(A 2 ) = 0.3 Solution : L.S. Clothiers

54. Chapter 1 54 Solution : L.S. Clothiers (2)New Information The planning board has recommended against the zoning change. Let B denote the event of a negative recommendation by the planning board. Given that B has occurred, should L.S. Clothiers revise the probabilities that the town council will approve or disapprove the zoning change?

55. Chapter 1 55 Solution : L.S. Clothiers (3)Conditional Probabilities Past history with the planning board and the town council indicates the following: Hence:

56. Chapter 1 56 Solution : L.S. Clothiers

57. Chapter 1 57 Solution : L.S. Clothiers To find the posterior probability that event A i will occur given that event B has occurred, we apply Bayes’ theorem. Bayes’ theorem is applicable when the events for which we want to compute posterior probabilities are mutually exclusive and their union is the entire sample space.

58. Chapter 1 58 Solution : L.S. Clothiers (4)Posterior Probabilities Given the planning board’s recommendation not to approve the zoning change, we revise the prior probabilities as follows:

59. Chapter 1 59 Conclusion : L.S. Clothiers The planning board’s recommendation is good news for L.S. Clothiers. The posterior probability of the town council approving the zoning change is 0.34 compared to a prior probability of 0.7

60. Random Variables A function that assigns a real number to each outcome in the sample space of a random experiment. Chapter 1 60 Discrete ~ rv with a finite (or countably infinite) range. Continuous ~ rv with an interval (either finite or infinite) of real numbers for its range.

61. Chapter 1 61 When the value of a variable is the outcome of a statistical experiment, that variable is a random variable.statistical experiment A random variable Denoted by an uppercase such as X After an experiment is conducted, the measured value of random variable is denoted by a lowercase x. Random Variables

62. Example A voice communication system for a business contains 48 external lines. At a particular time, the system is observed, and some of the lines are being used. Let the random variable X denote the number of lines in use. Then X can assume any of the integer values 0 through 48. When the system is observed, if 10 lines are in use, x = 10. Chapter 1 62

63. Chapter 1 63 A random variable can be classified as being either discrete or continuous depending on the numerical values it assumes. A discrete random variable may assume either a finite number of values or an infinite sequence of values. A continuous random variable may assume any numerical value in an interval or collection of intervals. Random Variables

64. Chapter 1 64 Discrete Random Variable A discrete random variable is one which may take on only a countable number of distinct values such as 0, 1, 2, 3,.. Number of scratches on the surface, proportion of defective parts among 1000 tested. Continuous Random Variable A continuous random variable is one which takes an infinite number of possible values. Continuous random variables are usually measurements. Length, pressure, temperature and time Random Variables

65. Random variables  Outcomes of an experiment expressed numerically  Example : Toss a die twice; Count the number of times the number 4 appears (0, 1 or 2 times) 65 Chapter 1

66. The probability distribution for a random variable describes how probabilities are distributed over the values of the random variable. The probability distribution is defined by a probability function, denoted by P(x), which provides the probability for each value of the random variable. The required conditions for a discrete probability function are: P(x) > 0  P(x) = 1 We can describe a discrete probability distribution with a table, graph, or equation. Discrete Probability Distribution 66 Chapter 1

67. Discrete Probability Distribution Suppose you flip a coin two times. This simple statistical experiment can have four possible outcomes: HH, HT, TH, and TT. Let the variable X represent the number of Heads that result from this experiment. The variable X can take on the values 0, 1, or 2. In this example, X is a random variable; because its value is determined by the outcome of a statistical experiment. 67 Chapter 1

68. Probability Distribution Values Probability 01/4 =.25 12/4 =.50 21/4 =.25 Event: Toss 2 Coins. Count # Tails. T T TT Discrete Probability Distribution 68 Chapter 1

69. Mean and Variance Mean The mean of a discrete probability distribution is computed by the formula: Variance The variance of a discrete probability distribution 69 Chapter 1

70. A random variable X has the probability distribution P(X=x) x p Find (a) the value of p (b) the mean of X. Example

71. 71 Chapter 1

72. References Montgomery, D.C. and Runger G.C. (2007). Applied Statistics and Probability for Engineers, 4 th Ed. John Wiley & Sons, Inc. Miller, I and Miller, M (2004). John E. Freund’s Mathematical Statistics with Applications, 7 th Ed. Pearson Prentice Hall. Chapter 1 72