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

JMB Chapter 2 Lecture 1 v3 EGR 252 Spring 2014 Slide 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


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 F

EGR IDM TCO EGR IDM TCO


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… }


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}


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}


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


Other Venn Diagram Examples Five non-mutually exclusive events

Subset – The green circle is a subset of the beige circle


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

Let’s Try It

MDH Chapter 2 Lecture 1 v1 EGR 252 Fall 2015 Slide 9

AB

C

7

2

14

3

6

5

A U C = ?B’∩ A = ?A ∩ B ∩ C = ?(A U B) ∩ C’ = ?


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 .


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


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

!!

rn

nPrn

120!36

!636

P


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 12P3 = 12!/(12-3)! = 1320

Order matters


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

12C3 = 12! / [3!(12-3)!] = 220

!!

!

rnr

n

r

n !!

!

rnr

nCrn

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

n1 * n2 = 18 ways

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 matters7 faculty members selected 4 at a time:

840!47

!747

P


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)


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


Additive RulesExperiment: 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)?

5.052

26

52

13

52

13)()()( 2121 APAPAAP


Your Turn: SolutionExperiment: 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


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) N

n

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

2

4combinations(2 aces) = →

combinations(3 jacks) = →

3

4

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.

6!24!2

!4

2

4

4!14!3

!4

3

4

Card Playing Probability Example Con’t


Likely Outcomes (N) = →

5

52

The total number of 5-card poker hands are equally likely therefore N =

960,598,2!552!5

!52

5

52

Per rule 2.3: P(A) = 5109.0960,598,2

24 XN

n

The probability of getting 2 aces and 3 jacks in a 5-card poker hand is 0.9 X 10-5

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.


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


Homework Reading Read section 2.6 and Chapter 3 of your textbook


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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