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1
Basic Probability
2
Sample Spaces
Collection of all possible outcomes
e.g.: All six faces of a die:
e.g.: All 52 cardsa deck of bridge cards
3
Events and Sample Spaces
An event is a set of basic outcomes from the sample space, and it is said to occur if the random experiment gives rise to one of its constituent basic outcomes.
Simple Event Outcome With 1 Characteristic
Joint Event 2 Events Occurring Simultaneously
Compound Event One or Another Event Occurring
4
Simple Event
A: Male
B: Over age 20
C: Has 3 credit cards
D: Red card from a deck of bridge cards
E: Ace card from a deck of bridge cards
5
Joint Event
D and E, (DE):
Red, ace card from a bridge deck
A and B, (AB):
Male, over age 20
among a group of
survey respondents
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Intersection• Let A and B be two events in the sample
space S. Their intersection, denoted AB, is the set of all basic outcomes in S that belong to both A and B.
• Hence, the intersection AB occurs if and only if both A and B occur.
• If the events A and B have no common basic outcomes, their intersection AB is said to be the empty set.
7
Compound Event
D or E, (DE): Ace or Red card from bridge deck
8
Union
• Let A and B be two events in the sample space S. Their union, denoted AB, is the set of all basic outcomes in S that belong to at least one of these two events.
• Hence, the union AB occurs if and only if either A or B or both occurs
9
Event Properties
Mutually Exclusive Two outcomes that cannot occur
at the same time
E.g. flip a coin, resulting in head and tail
Collectively ExhaustiveOne outcome in sample space
must occur
E.g. Male or Female
10
Special Events
Null EventClub & Diamond on
1 Card Draw
Complement of EventFor Event A, All
Events Not In A:
A' or A
Null Event
11
What is Probability?
1.Numerical measure of likelihood that the event will occur
Simple EventJoint EventCompound
2.Lies between 0 & 1
3.Sum of events is 1
1
.5 0
Certain
Impossible
12
Concept of ProbabilityA Priori classical probability, the probability of
success is based on prior knowledge of the process involved.
i.e. the chance of picking a black card from a deck of bridge cards
Empirical classical probability, the outcomes are based on observed data, not on prior knowledge of a process.
i.e. the chance that individual selected at random from the Kardan employee survey if satisfied with his or her job. (.89)
13
Concept of Probability
Subjective probability, the chance of occurrence assigned to an event by a particular individual, based on his/her experience, personal opinion and analysis of a particular situation.
i.e. The chance of a newly designed style of mobile phone will be successful in market.
14
(There are 2 ways to get one 6 and the other 4)
e.g. P( ) = 2/36
Computing Probabilities
• The probability of an event E:
• Each of the outcomes in the sample space is equally likely to occur
number of event outcomes( )
total number of possible outcomes in the sample space
P E
X
T
15
Concept of Probability
Experi-Number Number Relativementer of Trials of heads Frequency
AITBAR 2048 1061 0.5181
ABBASI 4040 2048 0.5069
WAHIDI 12000 6019 0.5016
TARIQ 24000 12012 0.5005
What conclusion can be drawn from the observations?
16
Presenting Probability & Sample Space
1. Listing
S = {Head, Tail}
2. Venn Diagram
3. Tree Diagram
4. Contingency
Table
17
S
ĀA
Venn Diagram
Example: Kardan Employee Survey
Event: A = Satisfied, Ā = Dissatisfied
P(A) = 356/400 = .89, P(Ā) = 44/400 = .11
18
Kardan Employee
Tree Diagram
Satisfied
Not Satisfied
Advanced
Not Advanced
Advanced
Not Advanced
P(A)=.89
P(Ā)=.11
.485
.405
.035
.075
Example: Kardan Employee Survey JointProbability
19
EventEvent B1 B2 Total
A1 P(A1 B1) P(A1 B2) P(A1)
A2 P(A2 B1) P(A2 B2) P(A2)
Total P(B1) P(B2) 1
Joint Probability Using Contingency Table
Joint Probability Marginal (Simple) Probability
20
Joint Probability Using Contingency Table
Kardan Employee Survey
Satisfied Not Satisfied
Total
AdvancedNot Advanced
Total
.485 .035
.405 .075
.52
.48
.89 .11 1.00
Joint Probability
Simple Probability
21
Use of Venn Diagram
Fig. 3.1: AB, Intersection of events A & B, mutually exclusive
Fig. 3.2: AB, Union of events A & B
Fig. 3.3: Ā, Complement of event A
Fig. 3.4 and 3.5:
The events AB and ĀB are mutually exclusive, and their union is B.
(A B) (Ā B) = B
22
Use of Venn Diagram
Let E1, E2,…, Ek be K mutually exclusive and
collective exhaustive events, and let A be
some other event. Then the K events E1 A,
E2 A, …, Ek A are mutually exclusive,
and their union is A.
(E1 A) (E2 A) … (Ek A) = A
(See supplement Fig. 3.7)
23
Compound ProbabilityAddition Rule
1. Used to Get Compound Probabilities for Union of Events
2. P(A or B) = P(A B) = P(A) + P(B) P(A B)
3. For Mutually Exclusive Events: P(A or B) = P(A B) = P(A) + P(B)
4. Probability of Complement
P(A) + P(Ā) = 1. So, P(Ā) = 1 P(A)
24
Addition Rule: ExampleA hamburger chain found that 75% of all customers
use mustard, 80% use ketchup, 65% use both. What is the probability that a particular customer will use at least one of these?A = Customers use mustardB = Customers use ketchupAB = a particular customer will use at least one of these
Given P(A) = .75, P(B) = .80, and P(AB) = .65, P(AB) = P(A) + P(B) P(AB)
= .75 + .80 .65= .90
25
Conditional Probability
1. Event Probability Given that Another Event Occurred
2. Revise Original Sample Space to Account for New Information
Eliminates Certain Outcomes
3. P(A | B) = P(A and B) , P(B)>0 P(B)
26
ExampleRecall the previous hamburger chain example, what is the probability that a ketchup user uses mustard?
P(A|B) = P(AB)/P(B)
= .65/.80 = .8125
Please pay attention to the difference from the joint event in wording of the question.
27
S
Black
Ace
Conditional Probability
Black ‘Happens’: Eliminates All Other Outcomes and Thus Increase the Conditional Probability
Event (Ace and Black)
(S)Black
Draw a card, what is the probability of black ace? What is the probability of black ace when black happens?
28
ColorType Red Black Total
Ace 2 2 4
Non-Ace 24 24 48
Total 26 26 52
Conditional Probability Using Contingency Table
Conditional Event: Draw 1 Card. Note Black Ace
Revised Sample Space
P(Ace|Black) = P(Ace and Black)
P(Black)= = 2/26 2/52
26/52
29
Game Show• Imagine you have been selected for a
game show that offers the chance to win an expensive new car. The car sits behind one of three doors, while monkeys reside behind the other two.
• You choose a door, and the host opens one of the remaining two, revealing a monkey.
• The host then offers you a choice: stick with your initial choice or switch to the other, still-unopened door.
• Do you think that switch to the other door would increase your chance of winning the car?
30
Statistical Independence
1. Event Occurrence Does Not Affect Probability of Another Event
e.g. Toss 1 Coin Twice, Throw 3 Dice
2. Causality Not Implied
3. Tests For Independence
P(A | B) = P(A), or P(B | A) = P(B),
or P(A and B) = P(A)P(B)
31
Statistical Independence
Kardan Employee Survey
Satisfied Not Satisfied
Total
AdvancedNot Advanced
Total
.485 .035
.405 .075
.52
.48
.89 .11 1.00
P(A1)
Note: (.52)(.89) = .4628 .485
P(B1
)P(A1and B1)
32
Multiplication Rule
1. Used to Get Joint Probabilities for Intersection of Events (Joint Events)
2. P(A and B) = P(A B) P(A B) = P(A)P(B|A)
= P(B)P(A|B)
3. For Independent Events:P(A and B) = P(AB) = P(A)P(B)
33
Randomized Response
An approach for solicitation of honest answers to sensitive questions in surveys.
• A survey was carried out in spring 1997 to about 150 undergraduate students taking Statistics for Business and Economics at Peking University.
• Each student was faced with two questions.• Students were asked first to flip a coin and
then to answer question a) if the result was the national emblem and b) otherwise.
34
Practice of Randomized Response
Questiona) Is the second last digit of your office phone
number odd?b) Recall your undergraduate course work,
have you ever cheated in midterm or final exams?
Respondents, please do as follows:1. Flip a coin.2. If the result is the head, answer question
a); otherwise answer question b). Please circle “Yes” or “No” below as your answer.
Yes No
35
Bayes’ Theorem
1. Permits Revising Old Probabilities Based on New Information
2. Application of Conditional Probability
3. Mutually Exclusive Events
NewInformation
RevisedProbability
Apply Bayes'Theorem
PriorProbability
36
Permutation and Combination
Counting Rule 1
Example : In TV Series: Kangxi Empire, Shilang tossed 50 coins, the number of outcomes is 2·2 ·… ·2 = 250. What is the probability of all coins with heads up?
If any one of n different mutually exclusive and collectively exhaustive events can occur on each of r trials, the number of possible outcomes is equal to : n·n ·… ·n = nr
37
Permutation and Combination
Application: Lottery Post Card (Post of China)
Six digits in each group (2000 version)
Winning the first prize: No.035718
Winning the 5th prize: ending with number 3
If there are k1 events on the first trial, k2 events
on the second trial, and kr events on the rth
trial, then the number of possible outcomes is:
k1· k2 ·… · kr
38
Permutation and Combination
Applicatin: If a license plate consists of 3 letters followed by 3 digits, the total number of outcomes would be? (most states in the US)
Application: China License PlatesHow many licenses can be issued?Style 1992: one letter or digit plus 4 digits.Style 2002: 1) three letters + three digits
2) three digits + three letters 3) three digits + three digits
39
Permutation and Combination
Counting Rule 2
Example: The number of ways that 5 books could
be arranged on a shelf is: (5)(4)(3)(2)(1) = 120
The number of ways that all n objects can be arranged in order is :
= n(n -1)(n -2)(2)(1) = n!
Where n! is called factorial and 0! is defined as 1.
nnP
40
Permutation and Combination
Counting Rule 3: Permutation
Example : What is the number of ways of arranging 3 books selected from 5 books in order? (5)(4)(3) = 60
The number of ways of arranging r objects selected from n objects in order is :
)!(
!
rn
nPn
r
41
Permutation and CombinationCounting Rule 4: Combination
Example : The number of combinations of 3 books selected from 5 books is
(5)(4)(3)/[(3)(2)(1)] = 10
Note: 3! possible arrangements in order are irrelevant
The number of ways that arranging r objects selected from n objects, irrespective of the order, is equal to
)!(!
!
rnr
n
r
nC nr