1. JMB Chapter 2 Lecture 1 v3EGR 252 Spring 2014Slide 1 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, …, e n }  where e i = each outcome of interest

2. JMB Chapter 2 Lecture 1 v3EGR 252 Spring 2014Slide 2 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} S M EGRIDMTCO F EGRIDMTCO

3. JMB Chapter 2 Lecture 1 v3EGR 252 Spring 2014Slide 3 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… }

4. JMB Chapter 2 Lecture 1 v3EGR 252 Spring 2014Slide 4 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}

5. JMB Chapter 2 Lecture 1 v3EGR 252 Spring 2014Slide 5 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}

6. JMB Chapter 2 Lecture 1 v3EGR 252 Spring 2014Slide 6 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

7. JMB Chapter 2 Lecture 1 v3EGR 252 Spring 2014Slide 7 Other Venn Diagram Examples  Five non-mutually exclusive events  Subset – The green circle is a subset of the beige circle

8. JMB Chapter 2 Lecture 1 v3EGR 252 Spring 2014Slide 8 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

9. Let’s Try It MDH Chapter 2 Lecture 1 v1EGR 252 Fall 2015Slide 9 A B C 7 2 1 4 3 6 5 A U C = ? B’∩ A = ? A ∩ B ∩ C = ? (A U B) ∩ C’ = ?

10. JMB Chapter 2 Lecture 1 v3EGR 252 Spring 2014Slide 10 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.

11. JMB Chapter 2 Lecture 1 v3EGR 252 Spring 2014Slide 11 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

12. JMB Chapter 2 Lecture 1 v3EGR 252 Spring 2014Slide 12 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

13. JMB Chapter 2 Lecture 1 v3EGR 252 Spring 2014Slide 13 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

14. JMB Chapter 2 Lecture 1 v3EGR 252 Spring 2014Slide 14 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

15. 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? MDH Chapter 2 Lecture 1 v1EGR 252 Fall 2015Slide 15 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

16. 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. MDH Chapter 2 Lecture 1 v1EGR 252 Fall 2015Slide 16 Permutation: Order matters 7 faculty members selected 4 at a time:

17. JMB Chapter 2 Lecture 1 v3EGR 252 Spring 2014Slide 17 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 A k ) = P(A 1 ) + P(A 2 ) + … P(A k )

18. JMB Chapter 2 Lecture 1 v3EGR 252 Spring 2014Slide 18 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

19. JMB Chapter 2 Lecture 1 v3EGR 252 Spring 2014Slide 19 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)?

20. JMB Chapter 2 Lecture 1 v3EGR 252 Spring 2014Slide 20 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

21. JMB Chapter 2 Lecture 1 v3EGR 252 Spring 2014Slide 21 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)

22. 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. MDH Chapter 2 Lecture 1 v1EGR 252 Fall 2015Slide 22 The number of ways of being dealt 2 aces from 4 cards is combinations(2 aces) = → combinations(3 jacks) = → The number of ways of being dealt 3 jacks from 4 cards is 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.

23. Card Playing Probability Example Con’t MDH Chapter 2 Lecture 1 v1EGR 252 Fall 2015Slide 23 Likely Outcomes (N) = → The total number of 5-card poker hands are equally likely therefore 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

24. 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 v1EGR 252 Fall 2015Slide 24

25. 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 v1EGR 252 Fall 2015Slide 25

26. Homework Reading  Read section 2.6 and Chapter 3 of your textbook MDH Chapter 2 Lecture 1 v1EGR 252 Fall 2015Slide 26