Probability

Learning ObjectivesConcepts: Events, Sample Space,

Intersection & Union

Tools: Venn Diagram, Tree Diagram, Contingency Table

Probability, Conditional Probability, Addition Rule & Multiplication Rule

Use Bayes’ Theorem

Summarize the Rules of Permutation and Combination

Basic Concepts

Random Experiment is a doing something that render to at least two possible outcomes with uncertainty as to which will occur. A coin is thrown A consumer is asked which of two

products he or she prefers The daily change in an index of stock

market prices is observed

Sample Spaces

Collection of all possible outcomes

e.g.: All six faces of a die:

e.g.: All 52 cardsa deck of bridge cards

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

Simple Event

A: Male

B: Over age 20

C: Has 3 credit cards

D: Red card from a deck of cards

E: Ace card from a deck of cards

Joint Event

D and E, (DE) :

Red, ace card from a deck

A and B, (AB):

Male, over age 20

among a group of

survey respondents

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 or null set.

Compound Event

Ace or Red card from deck of cards

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

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

Special Events

Null EventClub & Diamond on

1 Card Draw

Complement of EventFor Event A, All

Events Not In A:

A' or A

Null Event

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

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 BID employee survey if satisfied with his or her job.

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.

(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

Presenting Probability & Sample Space

1. Listing

S = {Head, Tail}

2. Venn Diagram

3. Tree Diagram

4. Contingency

Table

S

ĀA

Venn Diagram

Example:Survey on Job Satisfation

Event: A = Satisfied, Ā = Dissatisfied

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

Employee

Tree Diagram

Satisfied

Not Satisfied

Advanced(.545)Not Advanced (.455)Advanced(.318) Not Advanced(.682)

P(A)=.89

P(Ā)=.11

.485

.405

.035

.075

Example:Employee Survey JointProbability

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

Joint Probability Using Contingency Table

Employee Survey

Satisfied Not Satisfied

Total

AdvancedNot Advanced

Total

.485 .035

.405 .075

.52

.48

.89 .11 1.00

Joint Probability

Simple Probability

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

That is; sum of the joint probability is always marginal

probability

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)

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 (AorB) = 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

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)

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.

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?

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

Statistical Independence

1. Event Occurrence Does Not Affect Probability of Another Event

e.g. Toss 1 Coin Twice, Throw 3 Dice

2. Tests For Independence

P(A | B) = P(A), or P(B | A) = P(B),

or P(A and B) = P(A)P(B)

Statistical Independence

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)

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) For dep.event

3. For Independent Events:P(A and B) = P(AB) = P(A)P(B)

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

P(B | A) = P(A | B P(B )

P(A | B P(B ) + P(A | B P(B )

P(B A)

P(A)

ii i

1 k k

i

1

)) )

Bayes’ Theorem Formula

Same Event

Bayes’s Theorem Example Fifty percent of borrowers repaid their loans. Out of those who repaid, 40% had a college degree. Ten percent of those who defaulted had a college degree. What is the probability that a randomly selected borrower who has a college degree will repay the loan?B1= repay, B2= default, A=college degree

P(B1) = .5, P(A|B1) = .4, P(A|B2) = .1, P(B1|A) =?

)()|()()|(

)()|()|(

2211

111 BPBAPBPBAP

BPBAPABP

8.25.

2.

)5)(.1(.)5)(.4(.

)5)(.4(.

Event PriorProb

Cond.Prob

JointProb

Post.Prob

Bi P(Bi) P(A|Bi) P(Bi A) P(Bi |A)

B1 .5 .4 .20 .20/.25 = .8

B2 .5 .1 .05 .05/.25 = .2

1.0 P(A) = 0.25 1.0

Bayes’ Theorem Example Table Solution

DefaultRepay P(College)

X =