Statistical Experiments The set of all possible outcomes
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 = {e 1, e 2, e 3, …, en} § where ei = each outcome of interest JMB Chapter 2 Lecture 1 v 3 EGR 252 Spring 2014 Slide 1
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: S M EGR IDM F TCO EGR IDM TCO § The sample space: S = {MEGR, MIDM, MTCO, FEGR, FIDM, FTCO} JMB Chapter 2 Lecture 1 v 3 EGR 252 Spring 2014 Slide 2
Your Turn: Sample Space § Your turn: The sample space of gender and specialization of all BSE students in the School of Engineering is … S = {FECE, MECE, FEVE, MEVE, FISE, MISE, FMAE, etc} or S = {BMEF, BMEM, CPEF, CPEM, ECEF, ECEM, ISEF, ISEM… } § 2 genders, 6 specializations, § 12 outcomes in the entire sample space JMB Chapter 2 Lecture 1 v 3 EGR 252 Spring 2014 Slide 3
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} JMB Chapter 2 Lecture 1 v 3 EGR 252 Spring 2014 Slide 4
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, (E U I) = {EVEF, EVEM, ISEF, ISEM} JMB Chapter 2 Lecture 1 v 3 EGR 252 Spring 2014 Slide 5
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 v 3 EGR 252 Spring 2014 Slide 6
Other Venn Diagram Examples § Five non-mutually exclusive events § Subset – The green circle is a subset of the beige circle JMB Chapter 2 Lecture 1 v 3 EGR 252 Spring 2014 Slide 7
Subset Examples § Students who are male § 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 v 3 EGR 252 Spring 2014 Slide 8
Let’s Try It AUC =? B’∩ A = ? A∩B∩C=? (A U B) ∩ C’ = ? B A 2 6 7 4 1 3 5 C MDH Chapter 2 Lecture 1 v 1 EGR 252 Fall 2015 Slide 9
Sample Points § Multiplication Rule § If event A can occur n 1 ways and event B can occur n 2 ways, then an event C that includes both A and B can occur n 1 * n 2 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 v 3 EGR 252 Spring 2014 Slide 10
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! NOTE: 1! = 1 and 0! = 1 JMB Chapter 2 Lecture 1 v 3 EGR 252 Spring 2014 Slide 11
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 v 3 EGR 252 Spring 2014 Slide 12
Permutation Example § Mercer is introducing a new scholarship competition program for computer engineers interested in Big Data analysis. First, second, and third place winners will receive a specified scholarship amount. If 12 students applied for the scholarship, how many ways can the winners be selected? § If the outcome is defined as ‘first place student, second place student, and third place student § Total number of outcomes is 12 P 3 = 12!/(12 -3)! = 1320 § Order matters JMB Chapter 2 Lecture 1 v 3 EGR 252 Spring 2014 Slide 13
Combinations § Selections of subsets without regard to order. § Example: How many ways can we select 3 winners (w/out regard to placing) from the 12 students? § Total number of outcomes is 12 C 3 = 12! / [3!(12 -3)!] = 220 JMB Chapter 2 Lecture 1 v 3 EGR 252 Spring 2014 Slide 14
Let’s Try It § Registrants at a large convention are offered 6 sightseeing tours on each of 3 days. In how many ways can a person arrange to go on a sightseeing tour planned by this convention? Multiplication Rule: On each of 3 days, you have a choice of 6 tours. Event A: The particular day, can occur 3 ways Event B: The specific tour, can occur 6 ways n 1 * n 2 = 18 ways MDH Chapter 2 Lecture 1 v 1 EGR 252 Fall 2015 Slide 15
Let’s Try It § Find the number of ways that 7 faculty members can be assigned to 4 sections of EGR 252 if no faculty member is assigned to more than one section. Permutation: Order matters 7 faculty members selected 4 at a time: MDH Chapter 2 Lecture 1 v 1 EGR 252 Fall 2015 Slide 16
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(A 1 U A 2 U … U Ak) = P(A 1) + P(A 2) + … P(Ak) JMB Chapter 2 Lecture 1 v 3 EGR 252 Spring 2014 Slide 17
Calculating Probabilities § Examples: 1. 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 JMB Chapter 2 Lecture 1 v 3 EGR 252 Spring 2014 Slide 18
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)? JMB Chapter 2 Lecture 1 v 3 EGR 252 Spring 2014 Slide 19
Your Turn: Solution 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 JMB Chapter 2 Lecture 1 v 3 EGR 252 Spring 2014 Slide 20
Card-Playing Probability Example § P(A) = target outcome / all possible outcomes § If an experiment can result in any of N different equally likely outcomes, and if exactly n of those outcomes correspond to event A, then the probability Event A is P(A) JMB Chapter 2 Lecture 1 v 3 EGR 252 Spring 2014 Slide 21
Card Playing Probability Example § In a poker hand consisting of 5 cards, find the probability of holding 2 aces and 3 jacks. Combination…order does not matter. The number of ways of being dealt 2 aces from 4 cards is combinations(2 aces) = → The number of ways of being dealt 3 jacks from 4 cards is combinations(3 jacks) = → Per the multiplication rule, there are n = 6*4 = 24 possible hands with 2 aces and 3 jacks given the number of aces and jacks available in a 52 card deck. MDH Chapter 2 Lecture 1 v 1 EGR 252 Fall 2015 Slide 22
Card Playing Probability Example Con’t The total number of 5 -card poker hands are equally likely therefore N = Likely Outcomes (N) = → Per rule 2. 3: P(A) = The probability of getting 2 aces and 3 jacks in a 5 -card poker hand is 0. 9 X 10 -5 MDH Chapter 2 Lecture 1 v 1 EGR 252 Fall 2015 Slide 23
Your Turn § A box contains 500 envelopes, of which 75 contain $100 in cash, 150 contain $25, and 275 contain $10. An envelop may be purchased for $25. § (a) What is the sample space for the different amounts of money? § (b) Assign probabilities to the sample points § (c) Find the probability that the first envelop purchased will contain less than $100. MDH Chapter 2 Lecture 1 v 1 EGR 252 Fall 2015 Slide 24
Your Turn: Solution § (a) S = {$10, $25, $100} § (b) P($10) = 0. 55, P($25) 0. 3, P($100) = 0. 15 § (c) P($10) + P($25) = 0. 55 + 0. 3 = 0. 85 or, 1 – P($100) = 1 – 0. 15 = 0. 85 MDH Chapter 2 Lecture 1 v 1 EGR 252 Fall 2015 Slide 25
Homework Reading § Read section 2. 6 and Chapter 3 of your textbook MDH Chapter 2 Lecture 1 v 1 EGR 252 Fall 2015 Slide 26
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