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

6

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

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Special Events

Null EventClub & Diamond on

1 Card Draw

Complement of EventFor Event A, All

Events Not In A:

A' or A

Null Event

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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

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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.

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(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

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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?

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Presenting Probability & Sample Space

1. Listing

S = {Head, Tail}

2. Venn Diagram

3. Tree Diagram

4. Contingency

Table

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S

ĀA

Venn Diagram

Example: Kardan Employee Survey

Event: A = Satisfied, Ā = Dissatisfied

P(A) = 356/400 = .89, P(Ā) = 44/400 = .11

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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

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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

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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

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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

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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)

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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)

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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

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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)

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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.

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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?

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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

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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?

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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)

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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)

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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.

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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

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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

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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