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 = {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
EGR Spring 2008

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}
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
EGR Spring 2008

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… }
S = {BMEF, BMEM, CPEF, CPEM, ECEF, ECEM, EVEF, EVEM, ISEF, ISEM, MAEF, MAEM}, SEE DIAGRAM …
JMB Chapter 2 Lecture 1
EGR Spring 2008

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}
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
EGR Spring 2008

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,
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
EGR Spring 2008

6. Venn Diagrams Mutually exclusive or disjoint events Male Female
Intersection of two events
Events E (EVE students) and F (female students) E ∩ F
JMB Chapter 2 Lecture 1
EGR Spring 2008

7. Other Venn Diagram Examples
Five non-mutually exclusive events
Subsets
mutually exclusive example, EVE and MAE students
subset example, female students and female ISE students
JMB Chapter 2 Lecture 1
EGR Spring 2008

8. 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
EGR Spring 2008

9. 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
EGR Spring 2008

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!
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 and 0! = 1
JMB Chapter 2 Lecture 1
EGR Spring 2008

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
EGR Spring 2008

12. 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
EGR Spring 2008

13. 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
EGR Spring 2008

14. 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) ≤ P(ø) = 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
EGR Spring 2008

15. 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 =
2. The probability of drawing 1 heart from a standard 52-card deck is:
P(heart) = 13/52 = 1/4
P = 3/26 =
number of ways of being dealt 2 aces from 4 is: (4 choose 2) = 4!/2!2! = 6
number of ways of dealing 3 jacks from 4 is: 4!/3!1! = 4
number of ways of getting 2 aces and 3 jacks is: n = 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(2A & 3J) = 24/2,598,960 = 9.23*10-6
JMB Chapter 2 Lecture 1
EGR Spring 2008

16. 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 face card (J,Q,K)?
These are mutually exclusive.
P(A U B) = P(A) + P(B) = = 0.5
P(H U F) = P(H) + P(F) – P(H∩F)= 13/ /52 – 3/52= 22/52
JMB Chapter 2 Lecture 1
EGR Spring 2008

17. Your Turn: Additive Rules
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 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 – 3/52 = 22/52
These are mutually exclusive.
P(A U B) = P(A) + P(B) = = 0.5
P(H U F) = P(H) + P(F) – P(H∩F)= 13/ /52 – 3/52= 22/52
JMB Chapter 2 Lecture 1
EGR Spring 2008

18. 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) =
6*4 /
JMB Chapter 2 Lecture 1
EGR Spring 2008