basics of probability theory.pptx

5/19/2018 Basics of Probability Theory.pptx

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Basics of Probability Theory

Dr. Gita A. Kumta


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Probability as a Numerical Measure

of the Likelihood of Occurrence

0 1.5

Increasing Likelihood of Occurrence

Probability:

The event

is very

unlikelyto occur.

The occurrence

of the event is

just as likely asit is unlikely.

The event

is almost

certainto occur.


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Probability

Experiment of chance: a phenomena whose outcome isuncertain.

Probabilities Chances

Probability Model

Sample Space

Events

Probability of Events

Sample Space: Set of all possible outcomesEvent: A set of outcomes (a subset of the sample space). An

eventEoccurs if any of its outcomes occurs.

Probability: The likelihood that an event will produce a

certain outcome.


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An Experiment and Its Sample Space

An experimentis any process that generateswell-defined outcomes.

The sample space for an experiment is the

set of all experimental outcomes.

An experimental outcome is also called a

sample point.


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A Counting Rule for

Multiple-Step Experiments

If an experiment consists of a sequence of ksteps

in which there are n1possible results for the first step,

n2possible results for the second step, and so on,

then the total number of experimental outcomes is

given by (n1)(n2) . . . (nk).

A helpful graphical representation of a multiple-step

experiment is a tree diagram.


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Govind Investments can be viewed as a

two-step experiment. It involves two stocks,each with a set of experimental outcomes.

Mukund Oil: n1= 4

Collins Mining: n2= 2

Total Number of

Experimental Outcomes: n1n2= (4)(2) = 8

A Counting Rule for

Multiple-Step Experiments


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A second useful counting rule enables us to count

the number of experimental outcomes when nobjects are to be selected from a set ofNobjects.

Counting Rule for Combinations

CN

n

N

n N nn

N

!

!( )!

Number of Combinations ofNObjects Taken nat a Time

where: N! =N(N- 1)(N- 2) . . . (2)(1)

n! = n(n- 1)(n- 2) . . . (2)(1)

0! = 1


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Number of Permutations ofN

Objects Takenn

at a Time

where: N! =N(N- 1)(N- 2) . . . (2)(1)

n! = n(n- 1)(n- 2) . . . (2)(1)

0! = 1

P nN

n

N

N nn

N

!

!

( )!

Counting Rule for Permutations

A third useful counting rule enables us to count the

number of experimental outcomes when nobjects are tobe selected from a set of Nobjects, where the order of

selection is important.


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Assigning Probabilities

Classical Method

Relative Frequency Method

Subjective Method

Assigning probabilities based on the assumption

of equally likely outcomes

Assigning probabilities based on experimentation

or historical data

Assigning probabilities based on judgment


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Classical Method

If an experiment has npossible outcomes, this method

would assign a probability of 1/nto each outcome.

Experiment: Rolling a dieSample Space: S= {1, 2, 3, 4, 5, 6}

Probabilities: Each sample point has a

1/6 chance of occurring

Example


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Number of

Polishers Rented

Number

of Days

0

12

3

4

4

618

10

2

Lucas Tool Rental would like to assign probabilities to the number

of car polishers it rents each day. Office records show the

following frequencies of daily rentals for the last 40 days.

Example: Lucas Tool Rental


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Each probability assignment is given by dividing the

frequency (number of days) by the total frequency(total number of days).


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Probability

Number of

Polishers Rented

Number

of Days012

34

46

18

10240

.10

.15

.45

.25.051.00


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Subjective Method

When economic conditions and a companys circumstanceschange rapidly it might be inappropriate to assign probabilities

based solely on historical data.

We can use any data available as well as our experience and

intuition, but ultimately a probability value should express our

degree of belief that the experimental outcome will occur.

The best probability estimates often are obtained by combining

the estimates from the classical or relative frequency approachwith the subjective estimate.


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Events and Their Probabilities

The process of making an observation or recording a

measurement under a given set of conditions is a trialor

experiment.

Outcomes of an experiment are called events.An eventis

a collection of sample points.

The probability of any event is equal to the sum of the

probabilities of the sample points in the event.

We denote events by capital letters A, B, C,

The probability of an event A, denoted by P(A), in general,

is the chance A will happen.


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Probability

Consider a deck of playing cards

Sample Space: Set of 52 cards

Event: R: The card is red. F:The card is a face card

.

A:The card is a heart. B:The card is a 3.

Probability: P(R) = 26/52 P(F) = 12/52

P(A) = 13/52 P(B) = 3/52


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Events and variables

Can be described as random or deterministic:

The outcome of a random event cannot be predicted:

The sum of two numbers on two rolled dice.

The time of emission of the i

th

particle fromradioactive material.

The measured length of a table to the nearest cm. Motion of macroscopic objects (projectiles, planets, space

craft) as predicted by classical mechanics.

The outcome of a deterministic event can be predicted:


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Random variables

Can be described as discrete or continuous:

A discrete variable has a countable number of values.Number of customers who enter a store before one purchases

a product.

The values of a continuous variable can not be listed:

Distance between two oxygen molecules in a room.

Random Variable Possible Values

Gender Male, Female

Class Fresh, Soph, Jr, Sr

Height (inches) # in interval {30,90}

College Arts, Education, Engineering, etc.

Shoe Size 3, 3.5 18

Consider data collected for undergraduate students:

Is the height a discrete or continuous variable?


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Tree Diagram

A tree diagram is a way of describing all the

possible outcomes from a series of events

A tree diagram is a way of calculating theprobability of all the possible outcomes from a

series of events


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Examplea fair coin is flipped twice

H

H

H

T

T

T

HH

HT

TH

TT

2nd1st

Possible

Outcomes


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Outcome Table

If you flip a coin twice, you can model also model

the results with an outcome table

Flip 1 Flip 2 SimpleEvent

H H HH

H T HTT H TH

T T TT


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Tree DiagramsFor flipping a coin

Probability of two or more events

1st Throw 2ndThrow

THHHHH TTTT 1/21/21/21/21/21/21/2

OUTCOMES

H,H

H,T

T,H

T,T

P(Outcome)

P(H,H) =1/4=1/2x1/2

P(H,T) =1/4=1/2x1/2

P(T,H) =1/4=1/2x1/2

P(T,T) =1/4=1/2x1/2

Total P(all outcomes) = 1

Total=4 (2x2)


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Rules for calculating probability

The addition rule

for pairwise mutually exclusive events

P(A1+ A2+ ...+An)= P(A1)+P(A2)+ ...+P(An)

for two non-mutually exclusive events A and B

P(A+B) = P(A) + P(B)P(AB).

Multiplicative rule

P(AB) = P(A) P(B|A) = P(B) P(A|B).

Formula of total probabilityP(B)= P(A1)P(B|A1)+P(A2)P(B|A2)+ ...+P(An)P(B|An).


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Some Basic Relationships of Probability

There are some basic probability relationships that can be used to

compute the probability of an event without knowledge of all thesample point probabilities.

Complement of an Event

Intersection of Two Events

Mutually Exclusive Events

Union of Two Events


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The complement ofAis denoted byAc.

The complement of eventA is defined to be the eventconsisting of all sample points that are not inA.

Complement of an Event

EventA AcSampleSpace S

VennDiagram


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The union of eventsAand Bis denoted byA B

The union of eventsAand Bis the event containingall sample points that are inA orB or both.

Union of Two Events

SampleSpace SEventA Event B


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The intersection of eventsAand Bis denoted byA

The intersection of eventsAand Bis the set of allsample points that are in bothA and B.

SampleSpace SEventA Event B

Intersection of Two Events

Intersection ofAandB


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The addition law provides a way to compute theprobability of eventA,or B,or bothAand B occurring.

Addition Law

The law is written as:

P(A B) = P(A) + P(B) P(AB


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Two events are said to be mutually exclusive if theevents have no sample points in common.

Two events are mutually exclusive if, when one eventoccurs, the other cannot occur.

SampleSpace S

EventA

EventB


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If eventsAand Bare mutually exclusive, P(AB= 0.

The addition law for mutually exclusive events is:

P(A

B

) =P

(A

) +P

(B

)

theres no need toinclude P(AB


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The probability of an event given that another eventhas occurred is called a conditional probability.

A conditional probability is computed as follows :

The conditional probability ofAgiven Bis denotedby P(A|B).

Conditional Probability

( )

( | ) ( )

P A BP A B

P B


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Multiplication Law

The multiplication law provides a way to compute theprobability of the intersection of two events.


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


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Independent Events

If the probability of eventA

is not changed by theexistence of event B, we would say that eventsAand Bare independent.

Two eventsAand Bare independent if:

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

Multiplication Law


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The multiplication law also can be used as a test to seeif two events are independent.


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

Multiplication Law

for Independent Events


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Bayes Theorem

NewInformation

Applicationof BayesTheorem

PosteriorProbabilities

PriorProbabilities

Often we begin probability analysis with initial or

prior probabilities.

Then, from a sample, special report, or a producttest we obtain some additional information.

Given this information, we calculate revised orposterior probabilities.

Bayes theoremprovides the means for revising theprior probabilities.


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A proposed shopping center will provide strong

competition for suburban businesses like

L. S. Clothiers. If the shopping center is built, the owner of L.S. Clothiers feels it would be best to relocate to the center.

Bayes Theorem

Example: L. S. Clothiers

The shopping center cannot be built unless a zoning changeis approved by the town council. The planning board mustfirst make a recommendation, for or against the zoning

change, to the council.


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Prior ProbabilitiesLet:

Bayes Theorem

A1= town council approves the zoning change

A2= town council disapproves the change

P(A1) = .7, P(A2) = .3

Using subjective judgment:


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New Information

The planning board has recommended

against the zoning change. LetBdenote the

event of a negative recommendation by the

planning board.

Given thatBhas occurred, should L. S.

Clothiers revise the probabilities that the

town council will approve or disapprove the

zoning change?

Bayes Theorem

B Th


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Conditional ProbabilitiesPast history with the planning board and the towncouncil indicates the following:

Bayes Theorem

P(B|A1) = .2 P(B|A2) = .9

P(BC|A1) = .8 P(BC|A2) = .1

Hence:

B Th


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P(Bc|A1) = .8

P(A1) = .7

P(A2) = .3

P(B|A2) = .9

P(Bc|A2) = .1

P(B|A1) = .2 P(A1B) = .14

P(A2B) = .27

P(A2Bc) = .03

P(A1Bc) = .56

Bayes Theorem

Tree Diagram

Town Council Planning Board ExperimentalOutcomes

B Th


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Bayes Theorem

1 1 2 2

( ) ( | )( | )

( ) ( | ) ( ) ( | ) ... ( ) ( | )

i i

i

n n

P A P B AP A B

P A P B A P A P B A P A P B A

To find the posterior probability that eventA

i will occur giventhat eventB has occurred, we apply Bayes theorem.

Bayes theorem is applicable when the events for

which we want to compute posterior probabilities

are mutually exclusive and their union is the entire

sample space.

B Th


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Posterior Probabilities

Given the planning boards recommendation not to

approve the zoning change, we revise the prior

probabilities as follows:

1 11

1 1 2 2

( ) ( | )( | )

( ) ( | ) ( ) ( | )

P A P B AP A B

P A P B A P A P B A

(. )(. )

(. )(. ) (. )(. )

7 2

7 2 3 9

Bayes Theorem

= .34

B Th


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Conclusion

The planning boards recommendation is good newsfor L. S. Clothiers. The posterior probability of the

town council approving the zoning change is .34compared to a prior probability of .70.

Bayes Theorem

basics of probability theory.pptx

Documents

basics of probability

set of outcomes

set of experimental

npossible outcomes

useful counting rule

generateswelldefined

atwostep experiment

set ofnobjects