Statistical Experiments The set of all possible outcomes of an experiment is the Sample Space, S. Each outcome of the experiment is an element or member or sample point. If the set of outcomes is finite, the outcomes in the sample space can be listed as shown: S = {H, T} S = {1, 2, 3, 4, 5, 6} in general, S = {e1, e2, e3, …, en} where ei = each outcome of interest other examples: gender of SOE students, S = {M, F} specialization of BSE students in MUSE, S = {BME, CPE, ECE, EVE, ISE, MAE} JMB Chapter 2 Lecture 1 v3 EGR 252.003 Fall 2018

Tree Diagram If the set of outcomes is finite sometimes a tree diagram is helpful in determining the elements in the sample space. The tree diagram for students enrolled in the School of Engineering by gender and degree: The sample space: S = {MEGR, MIDM, MTCO, FEGR, FIDM, FTCO} other examples: gender of SOE students, S = {M, F} specialization of BSE students in MUSE, S = {BME, CPE, ECE, EVE, ISE, MAE} JMB Chapter 2 Lecture 1 v3 EGR 252.003 Fall 2018

Your Turn: Sample Space Your turn: The sample space of gender and specialization of all BSE students in the School of Engineering is … or 2 genders, 6 specializations, 12 outcomes in the entire sample space S = {FECE, MECE, FEVE, MEVE, FISE, MISE, FMAE, etc} S = {BMEF, BMEM, CPEF, CPEM, ECEF, ECEM, ISEF, ISEM… } S = {BMEF, BMEM, CPEF, CPEM, ECEF, ECEM, EVEF, EVEM, ISEF, ISEM, MAEF, MAEM}, SEE DIAGRAM … JMB Chapter 2 Lecture 1 v3 EGR 252.003 Fall 2018

Definition of an Event A subset of the sample space reflecting the specific occurrences of interest. Example: In the sample space of gender and specialization of all BSE students in the School of Engineering, the event F could be “the student is female” F = {BMEF, CPEF, ECEF, EVEF, ISEF, MAEF} F = {BMEF, CPEF, ECEF, EVEF, ISEF, MAEF} other examples, all female students: BME = {BMEF, BMEM} all male MAE students, MM = {MAEM} JMB Chapter 2 Lecture 1 v3 EGR 252.003 Fall 2018

Operations on Events Complement of an event, (A’, if A is the event) If event F is students who are female, F’ = {BMEM, EVEM, CPEM, ECEM, ISEM, MAEM} Intersection of two events, (A ∩ B) If E = environmental engineering students and F = female students, (E ∩ F) = {EVEF} Union of two events, (A U B) If E =environmental engineering students and I = industrial engineering students, F’ = {BMEM, EVEM, CPEM, ECEM, ISEM, MAEM} V int. F = {EVEF} mutually exclusive example, students who are EVE and male MAE. Union example, students who are EVE OR students who are male MAE (E U I) = {EVEF, EVEM, ISEF, ISEM} JMB Chapter 2 Lecture 1 v3 EGR 252.003 Fall 2018

Venn Diagrams Mutually exclusive or disjoint events Male Female Intersection of two events Let Event E be EVE students (green circle) Let Event F be female students (red circle) E ∩ F is the overlap – brown area JMB Chapter 2 Lecture 1 v3 EGR 252.003 Fall 2018

Other Venn Diagram Examples Five non-mutually exclusive events Subset – The green circle is a subset of the beige circle mutually exclusive example, EVE and MAE students subset example, female students and female ISE students JMB Chapter 2 Lecture 1 v3 EGR 252.003 Fall 2018

Subset Examples Students who are male Students who are ECE Students who are on the ME track in ECE Female students who are required to take ISE 428 to graduate Female students in this room who are wearing jeans Printers in the engineering building that are available for student use JMB Chapter 2 Lecture 1 v3 EGR 252.003 Fall 2018

Sample Points Multiplication Rule If event A can occur n1 ways and event B can occur n2 ways, then an event C that includes both A and B can occur n1 n2 ways. Example, if there are 6 different female students and 6 different male students in the room, then there are 6 * 6 = 36 ways to choose a team consisting of a female and a male student . JMB Chapter 2 Lecture 1 v3 EGR 252.003 Fall 2018

Permutations Definition: an arrangement of all or part of a set of objects. The total number of permutations of the 6 engineering specializations in MUSE is … 6*5*4*3*2*1 = 720 In general, the number of permutations of n objects is n! First position has 6 options, second has 5, 3rd has 4, etc., so 6*5*4*3*2*1 = 720 by definition, 1! = 1 and 0! = 1. NOTE: 1! = 1 and 0! = 1 JMB Chapter 2 Lecture 1 v3 EGR 252.003 Fall 2018

Permutation Subsets In general, where n = the total number of distinct items and r = the number of items in the subset Given that there are 6 specializations, if we take the number of specializations 3 at a time (n = 6, r = 3), the number of permutations is JMB Chapter 2 Lecture 1 v3 EGR 252.003 Fall 2018

Permutation Example A new group, the MUSE Ambassadors, is being formed and will consist of two students (1 male and 1 female) from each of the BSE specializations. If a prospective student comes to campus, he or she will be assigned one Ambassador at random as a guide. If three prospective students are coming to campus on one day, how many possible selections of Ambassador are there? If the outcome is defined as ‘ambassador assigned to student 1, ambassador assigned to student 2, ambassador assigned to student 3’ Outcomes are : A1,A2,A3 or A2,A4,A12 or A2, A1,A3 etc Total number of outcomes is 12P3 = 12!/(12-3)! = 1320 12P3 = 12!/(12-3)! = 479,001,600/362,880 = 1320 or, 12*11*10 = 1320 JMB Chapter 2 Lecture 1 v3 EGR 252.003 Fall 2018

Combinations Selections of subsets without regard to order. Example: How many ways can we select 3 guides from the 12 Ambassadors? Outcomes are : A1,A2,A3 or A2,A4,A12 or A12, A1,A3 but not A2,A1,A3 Total number of outcomes is 12C3 = 12! / [3!(12-3)!] = 220 Note: difference between permutations and combinations Perm. = “how many ways can I get an Ambassador given I have 12 choices to start, then 11, then 10?” Comb. = “If I’m going to divide the Ambassadors into groups of 4, how many different groups can I have?” (12 choose 3) = 12!/3!(12-3)! = 479,001,600/(6*362,880) = 220 JMB Chapter 2 Lecture 1 v3 EGR 252.003 Fall 2018

Introduction to Probability The probability of an event, A is the likelihood of that event given the entire sample space of possible events. P(A) = target outcome / all possible outcomes 0 ≤ P(A) ≤ 1 P(ø) = 0 P(S) = 1 For mutually exclusive events, P(A1 U A2 U … U Ak) = P(A1) + P(A2) + … P(Ak) JMB Chapter 2 Lecture 1 v3 EGR 252.003 Fall 2018

Calculating Probabilities Examples: There are 26 students enrolled in a section of EGR 252, 3 of whom are BME students. The probability of selecting a BME student at random off of the class roll is: P(BME) = 3/26 = 0.1154 2. The probability of drawing 1 heart from a standard 52-card deck is: P(heart) = 13/52 = 1/4 P = 3/26 = 0.1154 JMB Chapter 2 Lecture 1 v3 EGR 252.003 Fall 2018

Additive Rules Experiment: Draw one card at random from a standard 52 card deck. What is the probability that the card is a heart or a diamond? Note that hearts and diamonds are mutually exclusive. Your turn: What is the probability that the card drawn at random is a heart or a face card (J,Q,K)? These are mutually exclusive. P(A U B) = P(A) + P(B) = 0.25 + .25 = 0.5 P(H U F) = P(H) + P(F) – P(H∩F)= 13/52 + 12/52 – 3/52= 22/52 JMB Chapter 2 Lecture 1 v3 EGR 252.003 Fall 2018

Your Turn: Solution P(H U F) = P(H) + P(F) – P(H∩F) Experiment: Draw one card at random from a standard 52 card deck. What is the probability that the card drawn at random is a heart or a face card (J,Q,K)? Note that hearts and face cards are not mutually exclusive. P(H U F) = P(H) + P(F) – P(H∩F) = 13/52 + 12/52 – 3/52 = 22/52 These are mutually exclusive. P(A U B) = P(A) + P(B) = 0.25 + .25 = 0.5 P(H U F) = P(H) + P(F) – P(H∩F)= 13/52 + 12/52 – 3/52= 22/52 JMB Chapter 2 Lecture 1 v3 EGR 252.003 Fall 2018

Card-Playing Probability Example P(A) = target outcome / all possible outcomes Suppose the experiment is being dealt 5 cards from a 52 card deck Suppose Event A is 3 kings and 2 jacks K J K J K K K K J J (combination or perm.?) P(A) = = 9.23E-06 combinations(3 kings) = combinations(2 jacks) = number of ways of being dealt 2 jacks from 4 is: (4 choose 2) = 4!/2!2! = 6 number of ways of dealing 3 kings from 4 is: 4!/3!1! = 4 number of ways of getting 2 jacks and 3 kings is: 6*4 = 24 total number of 5 card poker hands is: N = (52 choose 5) = 52!/5!47! = 2,598,960 Therefore, the probability P(2J & 3K) = 24/2,598,960 = 9.23*10-6 JMB Chapter 2 Lecture 1 v3 EGR 252.003 Fall 2018