lecture prob

8/8/2019 Lecture Prob

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A Very Brief Introduction to ProbabilityA Very Brief Introduction to Probability

s Experiments and Assigning ProbabilitiesExperiments and Assigning Probabilities

s Events and Their ProbabilityEvents and Their Probability


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Probability of an eventProbability of an event

s ProbabilityProbability

is a numerical measure of theis a numerical measure of the

likelihood that an event will occur.likelihood that an event will occur.


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Probability of an eventProbability of an event

Getting Heads in a coin tossing, Getting even number inGetting Heads in a coin tossing, Getting even number in

a die throwing, Getting two defectives in a sample ofa die throwing, Getting two defectives in a sample of

10 items from a large lot containing 5% defectives,10 items from a large lot containing 5% defectives,

Delay of a flight, Opening a bank account within twoDelay of a flight, Opening a bank account within two

days, Rain in Ahmedabad on a specific day in Junedays, Rain in Ahmedabad on a specific day in June


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The First Law of ProbabilityThe First Law of Probability

s Probability values are always assigned on aProbability values are always assigned on a

scale from 0 to 1.scale from 0 to 1.

s A probability near 0 indicates an event isA probability near 0 indicates an event is

virtually certain not to occur.virtually certain not to occur.

s A probability near 1 indicates an event isA probability near 1 indicates an event is

almost certain to occur.almost certain to occur.

s A probability of 0.5 indicates the occurrence ofA probability of 0.5 indicates the occurrence of

the event is just as likely as it is unlikely.the event is just as likely as it is unlikely.


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Probability as a Numerical MeasureProbability as a Numerical Measure

of the Likelihood of Occurrenceof the Likelihood of Occurrence

00 11..

55

Increasing Likelihood of OccurrenceIncreasing Likelihood of Occurrence

ProbabilitProbability:y:

The occurrence of theThe occurrence of theevent isevent is

just as likely as it isjust as likely as it isunlikely.unlikely.


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Assigning ProbabilitiesAssigning Probabilities

s Classical MethodClassical Method

Assigning probabilities based on theAssigning probabilities based on the

assumption ofassumption ofequally likely outcomeseq

ually likely outcomes..

s Relative Frequency MethodRelative Frequency Method

Assigning probabilities based onAssigning probabilities based on

experimentation or historical dataexp

erimentation or historical data..

s Subjective MethodSubjective Method

Assigning probabilities based on theAssigning probabilities based on the

assignors judgmentassig

nors judgment..


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Classical MethodClassical Method

If an experiment hasIf an experiment has nn possible outcomes, thispossible outcomes, this

methodmethod

would assign a probability of 1/would assign a probability of 1/nn to eachto each

outcome.outcome.

s ExampleExample

Experiment: Rolling a dieExperiment: Rolling a die

Sample Space:Sample Space: SS = {1, 2, 3, 4, 5, 6}= {1, 2, 3, 4, 5, 6}

Probabilities: Each sample point has a 1/6Probabilities: Each sample point has a 1/6

chancechanceof occurring.of occurring.


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Example: Delay in FlightsExample: Delay in Flights

s Relative Frequency MethodRelative Frequency Method

Jet Airways have 10 flights originating fromJet Airways have 10 flights originating from

Ahmedabad daily. The following is theAhmedabad daily. The following is the

frequency distribution of number of delayedfrequency distribution of number of delayed

flights observed over last 60 daysflights observed over last 60 days

Number ofNumber of NumberNumberDelayed FlightsDelayed Flights of Daysof Days

44 44

55 66

66 181877 1010

88 99

9 79 7

10 610 6


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Subjective MethodSubjective Method

s Assigning probability to future events : it might beAssigning probability to future events : it might be

inappropriate to assign probabilities based solely oninappropriate to assign probabilities based solely onhistorical data. We need to use our experience andhistorical data. We need to use our experience and

intuition as well.intuition as well.

s Ultimately a probability value should express ourUltimately a probability value should express our degreedegree

of beliefof beliefthat the experimental outcome will occur.that the experimental outcome will occur.

s The best probability estimates often are obtained byThe best probability estimates often are obtained bycombining the estimates from the classical or relativecombining the estimates from the classical or relative

frequency approach with the subjective estimates.frequency approach with the subjective estimates.

A movie will be a hit, The new product on introductionA movie will be a hit, The new product on introduction

will capture a good marketwill capture a good market


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Brief Introduction to Discrete ProbabilityBrief Introduction to Discrete Probability

DistributionsDistributions

s Random VariablesRandom Variables

s Discrete Probability DistributionsDiscrete Probability Distributions

s Mean Value and VarianceMean Value and Variance

s Binomial Probability DistributionBinomial Probability Distribution

.10.10

.20.20

.30.30

.40.40

0 1 2 3 40 1 2 3 4


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Random VariablesRandom Variables

s AA random variablerandom variable is a numerical description ofis a numerical description of

the outcome of an experiment.the outcome of an experiment.


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Example: JSL AppliancesExample: JSL Appliances

s Random variable with a finite number of valuesRandom variable with a finite number of values

LetLetxx= number of TV sets sold at the store in one day= number of TV sets sold at the store in one daywherewherexxcan take at most 5 values (0, 1, 2, 3, 4)can take at most 5 values (0, 1, 2, 3, 4)

s Random variable with no fixed upper limit on the numberRandom variable with no fixed upper limit on the number

of valuesof values

LetLetxx= number of customers arriving at a mall on a given= number of customers arriving at a mall on a givendayday

wherewherexxcan take on the values 0, 1, 2, . . .can take on the values 0, 1, 2, . . .

We can count the customers arriving, but there is no fixedWe can count the customers arriving, but there is no fixed

upper limit on the number that might arrive.upper limit on the number that might arrive.


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s Using past data on TV sales (below left), aUsing past data on TV sales (below left), a

tabular representation of the probabilitytabular representation of the probabilitydistribution for TV sales (below right) wasdistribution for TV sales (below right) was

developed.developed.

NumberNumberUnits SoldUnits Sold of Daysof Days xx P(x)P(x)

00 8080 00 .40.40

11 5050 11 .25.25

22 4040 22 .20.2033 1010 33 .05.05

44 2020 44 .10.10

TotalTotal 200200 1.001.00



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s Graphical Representation of the ProbabilityGraphical Representation of the Probability

DistributionDistribution

.10.10

.20.20

.30.30

.40.40

.50.50

0 1 2 3 40 1 2 3 4

Values of Random VariableValues of Random Variablexx(TV sales)(TV sales)Values of Random VariableValues of Random Variablexx(TV sales)(TV sales)

Pro

ba

bilit

y

Pro

ba

bilit

y

Pro

ba

bilit

y

Pro

ba

bilit

y


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s Expected Value or Mean of a Discrete RandomExpected Value or Mean of a Discrete Random

VariableVariable

xx PP((xx)) xPxP((xx))

00 .40.40 .00.00

11 .25.25 .25.25

22 .20.20 .40.4033 .05.05 .15.15

44 .10.10 .40.40

Total =Total = MeanMean((xx) = 1.20) = 1.20

The expected number of TV sets sold in a day is 1.2The expected number of TV sets sold in a day is 1.2


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s Variance and Standard DeviationVariance and Standard Deviation

of a Discrete Random Variableof a Discrete Random Variable

xx x -x - ((x -x - ))22 PP((xx)) ((xx-- ))22PP((xx))

00 -1.2-1.2 1.441.44 .40.40 .576.576

11 -0.2-0.2 0.040.04 .25.25 .010.010

22 0.80.8 0.640.64 .20.20 .128.12833 1.81.8 3.243.24 .05.05 .162.162

44 2.82.8 7.847.84 .10.10 .784.784

1.660 =1.660 = 22

The variance of daily sales is 1.66 TV setsThe variance of daily sales is 1.66 TV sets squaredsquared..The standard deviation of sales is 1.2884 TV sets.The standard deviation of sales is 1.2884 TV sets.



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Example: Attrition ProblemExample: Attrition Problem


Navin is concerned about low retention rate ofNavin is concerned about low retention rate ofmgt trainees in his organization. On the basis of pastmgt trainees in his organization. On the basis of past

experience, he has seen a turnover of 10% mgt traineesexperience, he has seen a turnover of 10% mgt trainees

annually. Thus, for any trainee chosen at random, heannually. Thus, for any trainee chosen at random, he

estimates a probability of 0.1 that the person will not beestimates a probability of 0.1 that the person will not be

with the company next year.with the company next year.

He has chosen 5 trainees at random for a very specialHe has chosen 5 trainees at random for a very specialtraining of whom 3 would be required next year. If ittraining of whom 3 would be required next year. If it

becomes less 3 he will be in a problem. what is thebecomes less 3 he will be in a problem. what is the

probability that he wont have a problem next year?probability that he wont have a problem next year?


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Binomial Probability DistributionBinomial Probability Distribution

s Binomial Probability FunctionBinomial Probability Function

where:where:

PP((xx) = the probability of) = the probability ofxxsuccesses insuccesses in nntrialstrials

nn = the number of trials= the number of trials

pp = the probability of success on any one= the probability of success on any one

trialtrial

)()1()()( xnx ppxchoosenxP =


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s Using the Binomial Probability FunctionUsing the Binomial Probability Function

= (1)(1)(0.59)= (1)(1)(0.59)

= 0.59= 0.59

)()1(x)choose()( xnx ppnxf =

50 )9.0()1.0(0)choose5()0( =f


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)9.0()1.0(1)choose5()1( =f32 )9.0()1.0(2)choose5()2( =f

=0.32805=0.32805

=0.0729=0.0729


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Binomial Probability DistributionBinomial Probability Distribution

s Expected ValueExpected Value

MeanMean((xx) =) = == npnp

s VarianceVariance

Var(Var(xx) =) = 22

== npnp(1 -(1 -pp))s Standard DeviationStandard Deviation

SD( ) ( ) x np p= = 1


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Expected ValueExpected ValueMeanMean((xx) =) = = 5(.1) = .5 employees out= 5(.1) = .5 employees out

of 5of 5

VarianceVarianceVar(x) =Var(x) = 22 = 5(.1)(.9) = .45= 5(.1)(.9) = .45 Standard DeviationStandard Deviation

employees67.)9)(.1(.5)(SD ===x


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Poisson DistributionPoisson Distribution

Number of events occurring over a given time intervalNumber of events occurring over a given time interval

or a given space follows Poisson distribution.or a given space follows Poisson distribution.

Ex: Number of e-mails received on a dayEx: Number of e-mails received on a day

Number of telephone calls received on a dayNumber of telephone calls received on a day

Number of bacteria in 100 cc of waterNumber of bacteria in 100 cc of water

Number of customers arriving at a railway bookingNumber of customers arriving at a railway bookingcounter in an hourcounter in an hour

Number of cars entering a parking lot in an hourNumber of cars entering a parking lot in an hour

Number of cars passing through a crossing betweenNumber of cars passing through a crossing between

10 pm to 11 pm on a week day10 pm to 11 pm on a week day


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== ,...,1,0,!/)exp()( xxxPx

: Mean of the Poisson distribution > 0: Mean of the Poisson distribution > 0

Variance of the Poisson distribution =Variance of the Poisson distribution =

0! = 1, 2! = 1 X 2 = 2, 3! = 1 X 2 X 3 = 6,0! = 1, 2! = 1 X 2 = 2, 3! = 1 X 2 X 3 = 6,

4! = ? 5! = ?, 6! = ?4! = ? 5! = ?, 6! = ?

exp(-1) = (2.7182)exp(-1) = (2.7182)-1-1 = 0.3679= 0.3679

exp (2) = (2.7182)exp (2) = (2.7182)22 = 7.3890= 7.3890


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As an approximation to binomial distribution:As an approximation to binomial distribution:

If n is large and p is small thenIf n is large and p is small then

Suppose from a large lot a random sample of 20 itemsSuppose from a large lot a random sample of 20 items

are chosen for inspection. It is known that the lotare chosen for inspection. It is known that the lot

contains 1% defective items. What is the probability thatcontains 1% defective items. What is the probability that

there will be more than 2 defectives in the sample?there will be more than 2 defectives in the sample?

!/))(exp()1(x)choose()( )( xnpnpppnxf xxnx =

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