topic2 probability

8/3/2019 Topic2 Probability

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PROBABILITY

TOPIC 2


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2.1 Introduction

Random variable

Statistical science deals largely with assessing the likelihood ofoccurrence of uncertain events.

Demands for products is uncertain; times between arrivals to asupermarket are uncertain; stock price returns are uncertain;changes in interest rates are uncertain.


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Introduction

In these examples the uncertain quantitydemand, time between arrivals, stock price

returns, change of interest rate- is a numericalquantity.

Such a numerical quantity is called a randomvariable.

A random variable associates a numerical valuewith each possible random outcome.


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Introduction

A random variable is any quantitative result froma random experiment, that is, an experiment

whose outcomes are uncertain.Probability

Associated with each possible random outcome

is a probability. A probability is a numberbetween 0 and 1 that measures the likelihoodthat some event will occur.


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Introduction

An event with probability 0 cannot occur, whereas anevent with probability 1 is certain to occur.

An event with probability greater than 0 and less thanone is uncertain, but the closer its probability is to 1, themore likely it is to occur.

The probability associated with each value of a randomvariable is found by adding the probabilities for all theoutcomes that are assigned that value.


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Introduction

Probability Distribution

A random variable is determined by specifying

its possible values and the probability associatedwith each value. This specification states theprobability distribution of the random variable.

A probability distribution lists all of the possiblevalues of the random variable and theircorresponding probabilities.


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2.2 Assigning probabilities to Events

Our objective is to determine the probability P(A)that event A will occur.

The case of equally likely events

Use of relative frequencies

The subjective approach


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Given a sample space S={E1,E2,,En}, thefollowing characteristics for the probability P(Ei)

of the individual outcome Eimust hold:

Probability of an event: The probability P(A) ofeventA is the sum of the probabilities assigned

to the individual outcomes contained inA.

n

1i

i

i

1EP.2

ieachfor1EP0.1

Assigning probabilities


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This is a useful device to build a sample space and tocalculate probabilities of simple events and events.

2.3 Probability Trees

Consider a random experiment performed in two stages

with two outcomes S and F at each trial

We assume that P(S)=P(F)=0.5

i.e. the two outcomes are equally likely .


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P(SS)=0.25

P(SF)=0.25

P(FS)=0.25P(FF)=0.25

Origin

Stage 1 Stage 2

First trial

Second trialS

F

SS

FS

FF

SF

S

S

F

FSecond trial

Simple events

Probability Trees - continued


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P(SS)=0.25

P(SF)=0.25

P(FS)=0.25P(FF)=0.25

Example1: Calculate the probability of the eventA thatthe experiment results in at least one outcome ofS

P(A)= P(SS)+P(SF)+ P(FS)=.25+.25+.25=.75

Probability Trees - continued


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2.4 Basic Probability Principles

The Addition Law

It has two forms, depending on whether or not

the events are mutually exclusive. Events aremutually exclusive if they have no outcomes incommon.

If eventsA and B are mutually exclusive, thenP(A or B)=P(A)+ P(B)


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The Addition Law

For any eventsA and B, not necessarily mutuallyexclusive,

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


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Example2Consider the following events involving a companys

annual revenue in 2003.

A: revenue is less than $1 million,

B: revenue is at least $1 million but less than$2million, and

C: revenue is at least $2 million.


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Example2-continued

These events are mutually exhaustive andexclusive and so their probabilities must sum to

one. Suppose these probabilities are P(A)=0.5,P(B)=0.3 and P(C)=0.2

From the addition rule we have

P(revenue is at least $1 million)=P(B)+P(C)=0.5


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The probability of an event when partial knowledgeabout the outcome of an experiment is known, iscalled Conditional probability.

We use the notationP(A|B) = The conditional probability that event A

occurs, given that event B has occurred.

The partial knowledge

is contained in thecondition )B(P

)BandA(P)B|A(P

Conditional Probability


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Two events A and B are said to be independentif P(A|B) = P(A) or P(B|A) = P(B). Otherwise, the

events are dependent.

Note that, if the occurrence of one event does not

change the likelihood of occurrence of the otherevent, the two events are independent

Independent and Dependent Events


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Example3

The personnel department of an insurance companyhas compiled data regarding promotion, classified bygender. Is promotion and gender dependent on one

another?

Manager Promoted NotPromoted Total

Male 46 184 230

Female 8 32 40

total 54 216 270

Events of interest:M: A manager is a male A: A manager is promoted

M: A manager is a female A: A manager is not promoted


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Let us check if P(A|M)=P(A). If this equality holds,there is no difference in probability of promotionbetween a male and a female manager.

Manager Promoted NotPromoted Total

Male 46 184 230

Female 8 32 40

Total 54 216 270

P(A) = Number of promotions / total number of managers

= 54 /270 = .20P(A|M) = Number of promotions | Only male managers are observed

= 46 / 230 = .20.

Conclusion: there is no discriminationin awarding promotions.

46 230

54 270


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Note that independent events andmutually exclusive events are not the same!!

A B

A and B are two mutually exclusive events,and A can take place, that is P(A)>0.

Can A and B be independent?

Lets assume event B has occurred.

B

However, P(A)>0, thus, A and B cannot be independent

Then, the conditional probability that A occursgiven that B has occurred is zero, that is P(A|B) = 0,because P(A and B) = 0.


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2.4 Basic Principles of Probability-continued

Complement rule

Each simple event must belong to either A or .

Since the sum of the probabilities assigned to asimple event is one, we have for any event A

P(A) = 1 - P(A)

A


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Multiplication rule

For any two events A and B

When A and B are independent

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

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


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A stock market analyst feels that the probability that a certain mutual fund will receive

increased contributions from investors is 0.6.

the probability of receiving increased contributions from

investors becomes 0.9 if the stock market goes up. the probability of receiving increased contributions from

investors drops below 0.6 if the stock market drops.

there is a probability of 0.5 that the stock market rises.

The events of interest are:A: The stock market rises;

B: The company receives increased contribution.

Example4


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Calculate the following probabilities

The probability that both A and B will occur isP(A and B). [Sharp increase in earnings].

The probability that either A or B will occur isP(A or B). [At least moderate increase in earning].

Solution

P(A) = 0.5; P(B) = 0.6; P(B|A) = 0.9

P(A and B) = P(A)P(B|A) = (.5)(.9) = 0.45

P(A or B) = P(A) + P(B) - P(A and B) = .5 + .6 - .45 =0.65


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Suppose we are interested in the condition of amachine that produces a particular item.

Information From experience it is known that the machine is in

good conditions 90% of the time.

When in good conditions, the machine produces a

defective item 1% of the time. When in bad conditions, the machine produces a

defective 10% of the time.

An item selected at random from the current

production run was found defective.

With this additional informationwhat is the probability that themachine is in good conditions?

Example5 (Probability trees revisited)


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Solution

Let us define the two events of interest:A: The machine is in good conditionsB: The item is defective

The prior probability that the machine is in goodconditions is P(A) = 0.9.

With the new information, (the selected item isdefective, or, event B has occurred) we can

reevaluate this probability by calculating P(A|B).

)B(P

)BandA(P)B|A(P


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A

A

B

B

B

B

Prior probabilities Conditional probabilitiesSimpleevents

Jointprobabilities

P(A and B) = 0.009

A and B

A and BA: The machine is in good conditionB: Item is defective

P( ) = 0.010A and B

P(B) = 0.019

A and B

A and B


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2.5 Random Variables and

Probability Distributions There are two types of random variables

Discrete random variable

Continuous random variable.


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A random variable is discrete if it can assume only acountable number of values. A random variable iscontinuous if it can assume an uncountable number ofvalues.

0 11/21/41/16

Continuous random variableAfter the first value is definedthe second value, and any valuethereafter are known.

Therefore, the number ofvalues is countable

After the first value is defined,any number can be the next one

Discrete random variable

Therefore, the number ofvalues is uncountable

0 1 2 3 ...

Discrete and Continuous Random Variables


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A table, formula, or graph that lists all possiblevalues a discrete random variable can assume,together with associated probabilities, is called a

discrete probability distribution..

To calculate P(X = x), the probability that the random

variable X assumes the value x, add the probabilitiesof all the simple events for which X is equal to x.

Discrete Probability Distribution


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Example6

Find the probability distribution of the randomvariable describing the number of female children ina randomly selected family with two children.

Solutionx p(x)

0 1/41 1/22 1/4

Simple event x ProbabilityFF 2 1/4FM 1 1/4MF 1 1/4

MM 0 1/4

1/4 if x=0 or 2p(x) =

1/2 if x=1

0 1 2X


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Requirements of discrete probability distribution

If a random variable can take values xi, then thefollowing must be true:

1)x(p.2

xallfor1)p(x0.1

ixall

i

ii

The probability distribution can be used to calculate

probabilities of different events. Example continued:

4

3

4

1

2

1)2X(P)1X(P)2X1(P


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Probabilities as relative frequencies

In practice, often probabilities are estimated fromrelative frequencies

Example7

The number of cars a dealer is selling daily were recorded inthe last 100 days. This data was summarized as follows:

Daily sales Frequency0 5

1 152 353 254 20

100

Estimate the probability

distribution. State the probability of

selling more than 2 cars a day.


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Solution

From the table of frequencies we can calculate therelative frequencies, which becomes our estimatedprobability distribution

Daily sales Relative Frequency0 5/100=.051 15/100=.152 35/100=.353 25/100=.25

4 20/100=.201.00

The probability of sellingmore than 2 a day is

0 1 2 3 4

.05

.15

.35.25

.20

X

P(X>2) = P(X=3) + P(X=4)= .25 + .20 = .45


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2.6 Expected Value and Variance

The expected value

Given a discrete random variable X with values xi,

that occur with probabilities p(xi), the expected valueofX is

i

xall

ii )x(px)X(E

The expected value of a random variable X is theweighted average of the possible values it canassume, where the weights are the corresponding

probabilities of each xi.


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E(c) = c E(cX) = cE(X)

E(X + Y) = E(X) + E(Y)

E(X - Y) = E(X) - E(Y) E(XY) = E(X)E(Y) ifX andY are independent

random variables.

Laws of Expected Value


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Let X be a discrete random variable with possiblevalues xi that occur with probabilities p(xi), and letE(xi) = m. The variance ofX is defined to be

mmixall

i2

i22 )x(p)x()X(E

The variance is the weighted average of the squared

deviations of the values ofX from their mean m,where the weights are the correspondingprobabilities of each xi.

Variance


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Standard deviation

The standard deviation of a random variable X,denoted , is the positive square root of the varianceofX.

Example8 The total number of cars to be sold next week is

described by the following probability distribution

Determine the expected value and standarddeviation ofX, the number of cars sold.

x 0 1 2 3 4p(x) .05 .15 .35 .25 .20


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11.124.1

24.1)20)(.4.24()25)(.4.23(

)35)(.4.22()15)(.4.21()05)(.4.20(

)x(p)4.2x()X(V

40.2

)20.0(4)25.0(3)35.0(2)15.0(1)05.0(0

)x(px)X(E

5

1i

i2

i2

5

1i

ii

m

x 0 1 2 3 4

p(x) .05 .15 .35 .25 .20


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Example9

With the probability distribution of cars sold per week(Example8), assume a salesman earns a fixed weeklywages of $150 plus $200 commission for each car sold.

What is his expected wages and the variance of thewages for the week?

Solution

The weekly wages is Y = 200X + 150

E(Y) = E(200X+150) = 200E(X)+150= 200(2.4)+150=630 $. V(Y) = V(200X+150) = 2002V(X) = 2002(1.24) = 49,600 $2


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To consider the relationship between tworandom variables, the bivariate (or joint)

distribution is needed.

Bivariate probability distribution

The probability that X assumes the value x, and Yassumes the value y is denoted

p(x,y) = P(X=x, Y = y)

2.7 Bivariate Distributions

1y )p(x,2.

1y )p(x,01.

:conditionsfollow ingthesatisfies

functioniesprobabilitjointThe

xa ll yall


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Example10

Xavier and Yvette are two real estate agents. Let Xand Y denote the number of houses that Xavier andYvette will sell next week, respectively.

The bivariate (joint) probability distribution

XY 0 1 2 p(y)0 .12 .42 .06 .60

1 .21 .06 .03 .302 .07 .02 .01 .10p(x) .40 .50 .10 1.00

p(0,0)

p(0,1)p(0,2)

P(X=0)

The marginal probability

P(Y=1), the marginalprobability.


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X

Y

X=0 X=2X=1

y=1

y=2

y=0

0.42

0.12

0.21

0.07

0.06

0.02

0.06

0.03

0.01

p(x,y) x p(x) y p(y)0 .4 0 .61 .5 1 .32 .1 2 .1

E(X) = .7 E(Y) = .5V(X) = .41 V(Y) = .45

X


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)yY(P

)yYandxX(P)yY|xX(P

1.30.

03.

)1Y(P

)1Yand2X(P)1Y|2X(P

2.30.

06.

)1Y(P

)1Yand1X(P)1Y|1X(P

7.30.

21.

)1Y(P

)1Yand0X(P)1Y|0X(P

Example 10 - continued

The sum isequal to 1.0

Y 0 1 2 p(y)0 .12 .42 .06 .601 .21 .06 .03 .302 .07 .02 .01 .10

p(x) .40 .50 .10 1.00

Calculating ConditionalProbability


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Two random variables are said to be independentwhen

This leads to the following relationship forindependent variables


Since P(X=0|Y=1)=.7 but P(X=0)=.4, The variables X andY are not independent.

P(X=x|Y=y)=P(X=x) or P(Y=y|X=x)=P(Y=y).

P(X=x and Y=y) = P(X=x)P(Y=y)

Conditions for independence


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Additional example 11

The table below represent the joint probabilitydistribution of the variable X and Y. Are thevariables X and Y independent

XY 1 21 .28 .42

2 .12 .18

Find the marginal probabilitiesof X and Y.Then apply the multiplication rule.

p(y).7.3

P(y) .40 .60

P(X=1)P(Y=1) = .40(.70) = .28

P(X=1 and Y=1) = .28

P(X=1)P(Y=2) = .40(.30) = .12

P(X=1 and Y=2) = .12

Compare the other two pairs.Yes, the two variables are

independent


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The sum of two variables

To calculate the probability distribution for a sumof two variables X and Y observe the examplebelow.


Find the probability distribution of the total number ofhouses sold per week by Xavier and Yvette.

Solution

X+Y is the total number of houses sold.X+Y can have thevalues 0, 1, 2, 3, 4.

We find the distribution of X+Y as demonstrated next.


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..

..X

Y 0 1 2 p(y)

0 .12 .42 .06 .601 .21 .06 .03 .302 .07 .02 .01 .10p(x) .40 .50 .10 1.00

P(X+Y=0) = P(X=0 and Y=0) = .12

P(X+Y=1) = P(X=0 and Y=1)+ P(X=1 and Y=0) =.21 + .42 = .63

P(X+Y=2) = P(X=0 and Y=2)+ P(X=1 and Y=1)+ P(X=2 and Y=0)= .07 + .06 + .06 = .19

x +y 0 1 2 3 4p(x+y) .12 .63 .19 .05 .01

The probabilities P(X+Y)=3 andP(X+Y) =4 are calculated thesame way. The distribution follows


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Expected value and variance of X+Y

When the distribution of X+Y is known (see the previousexample) we can calculate E(X+Y) and V(X+Y) directlyusing their definitions.

An alternative is to use the relationships

E(aX+bY) =aE(X) + bE(Y);

V(aX+bY) = a2V(X) + b2V(Y) if X and Y are independent.

When X and Y are not independent, (see the previous example)

we need to incorporate the covariance in the calculations of the

variance V(aX+bY).


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The covariance is a measure of the degree to whichtwo random variables tend to move together.

COV(X,Y) = S(X-mx)(y-my)p(x,y)

=E[(X - mx)(Y - my)]=E(XY) - mxmy

Over all x,y

yx

)Y,X(COVncorrelatiooftcoefficienThe

Covariance


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Solution

Calculation of the expected values:mx = Sxip(xi) = 0(.4)+1(.5)+2(.1)=.7my = Syip(yi) = 0(.6)+1(.3)+2(.1)=.5

There is a negative relationship between the two variables


Find the covariance of the sales variables X and Y, thencalculate the coefficient of correlation.

Calculation of the covariance:

COV(X,Y) = S(x - mx)(y - my)p(x,,y)= (0-.7)(0-.5)(.12)+(0-.7)(1-.5)(.21)+ (0-.7)(2-5)(.07)++(2-..7)(2-.5)(.01)= -.15


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Calculation of the standard deviations of X and YV(X) = S(xi-mx)

2p(xi) = (0-.7)2(.4)+(1-.7)2(.5)+(2-.7)2(.1)=.41

x = [V(X)]1/2 = .64

In a similar manner we have V(Y) = .45y = [.45]

1/2=.67

Calculation of r

35.)67)(.64(.

15.)Y,X(COV

yx

There is a relatively weaknegative relationship betweenX and Y .

To find how strong the relationship between X and Y is

we need to calculate the coefficient of correlation.


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The variance of the sum of two variables X and

Y can now be calculated using

V(aX + bY) = a2V(X) + b2V(Y) + 2abCOV(X,Y)

= a2

V(X) + b2

V(Y) + 2abyx


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Example12

Investment portfolio diversification An investor has decided to invest equal amounts of

money in two investments.

Find the expected return on the portfolio

If = 1, .5, 0 find the standard deviation of theportfolio.

Mean return tandard devInvestment 1 15% 25%

Investment 2 27% 40%


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Solution

The return on the portfolio can be represented byRp = w1R1 + w2R2 = .5R1 + .5R2

The relative weights are proportional to the amounts invested.

Thus, E(Rp) = w1E(R1) + w2E(R2)

=.5(.15) + .5(.27) = .21

The variance of the portfolio return isV(Rp) = w1

2V(R1) + w22V(R1) +2w1w212


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Substituting the required coefficient of correlation

we have: For = 1 : V(Rp) = .1056 = .3250

For = .5: V(Rp) = .0806 = .2839

For = 0: V(Rp) = .0556 = .2358

Larger diversification is expressed bysmaller correlation.

As the correlation coefficient decreases,the standard deviation decreases too.

p

p

p


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2.8 The Binomial Distribution

The binomial experiment can result in only oneout of two outcomes.

Typical cases where the binomial experimentapplies:

A coin flipped results in heads or tails

An election candidate wins or loses An employee is male or female

A car uses 87octane gasoline, or another gasoline.


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There are n trials (n is finite and fixed). Each trial can result in a success or a failure.

The probability p of success is the same for all the

trials. All the trials of the experiment are independent.

Binomial Random Variable

The binomial random variable counts the number ofsuccesses in n trials of the binomial experiment.

By definition, this is a discrete random variable.

Binomial experiment


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Developing the Binomial probability distribution

S1

S2 S3

S3

S2S3

S3

F2

F3

F3

F1

F2

F3

F3

P(SSS)=p3

P(SSF)=p2(1-p)

P(SFS)=p(1-p)p

P(SFF)=p(1-p)2

P(FSS)=(1-p)p2

P(FSF)=(1-p)P(1-p)

P(FFS)=(1-p)2p

P(FFF)=(1-p)3

Since the outcome of each trial isindependent of the previous outcomes,we can replace the conditional probabilitieswith the marginal probabilities.

P(S2|S1


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P(SSS)=p3

P(SSF)=p2(1-p)

P(SFS)=p(1-p)p

P(SFF)=p(1-p)2

P(FSS)=(1-p)p2

P(FSF)=(1-p)P(1-p)

P(FFS)=(1-p)2p

P(FFF)=(1-p)3

Let X be the number of successesin three trials. Then,

X = 3

X =2

X = 1

X = 0

P(X = 3) = p3

P(X = 2) = 3p2(1-p)

P(X = 1) = 3p(1-p)2

P(X = 0) = (1- p)3

This multiplier is calculated in the following formula

SSS

SS

S S

SS


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xnxnx )p1(pC)x(p)xX(P

In general, The binomial probability is calculated by:

)!xn(!x!nCwhere nx

3)21)(1(

321

)!13(!1

!3C:1x

1)321)(1(

321

)!03(!0

!3C:0x

31

30

Example: For n = 3

Calculating the Binomial Probability


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Example 13

5% of a catalytic converter production run isdefective.

A sample of 3 converter s is drawn. Find the

probability distribution of the number of defectives. Solution

A converter can be either defective or good.

There is a fixed finite number of trials (n=3)

We assume the converter state is independent on oneanother.

The probability of a converter being defective does notchange from converter to converter (p=.05).

The conditions required for the binomial experiment are met


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Let X be the binomial random variable indicating the

number of defectives. Define a success as a converter is found to be

defective.

0001.)95(.)05(.)!33(!3

!3)3(p)3x(P

0071.)95(.)05(.)!23(!2

!3

)2(p)2x(P

1354.)95(.)05(.)!13(!1

!3)1(p)1x(P

8574.)95(.)05(.)!03(!0

!3

)0(p)0X(P

333

232

131

030

X P(X)0 .85741 .13542 .0071

3 ..0001


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Mean and variance of binomial random

variableE(X) = m = np

V(X) = 2 = np(1-p)

Example 6.10 Records show that 30% of the customers in a shoe

store make their payments using a credit card.

This morning 20 customers purchased shoes.

Answer some questions stated in the next slide.


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pk .01.. 30

0..

11

What is the probability that at least 12 customersused a credit card?

This is a binomial experiment with n=20 and p=.30.

.995

P(At least 12 used credit card)= P(X>=12)=1-P(X


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pk .01.. 30

What is the probability that at least 3 but not morethan 6 customers used a credit card?

.608

P(3


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What is the expected number of customers whoused a credit card?

E(X) = np = 20(.30) = 6

Find the probability that exactly 14 customers did notuse a credit card.

Let Y be the number of customers who did not use a creditcard.P(Y=14) = P(X=6) = P(X


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2.9 Poisson Distribution

The Poisson experiment typically fits cases ofrare events that occur over a fixed amount of

time or within a specified region Typical cases

The number of errors a typist makes per page

The number of customers entering a service stationper hour

The number of telephone calls received by aswitchboard per hour.


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Properties of the Poisson experiment The number of successes (events) that occur in a certain

time interval is independent of the number of successesthat occur in another time interval.

The probability of a success in a certain time interval is the same for all time intervals of the same size,

proportional to the length of the interval.

The probability that two or more successes will occur in

an interval approaches zero as the interval becomessmaller.

Poisson Experiment


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The Poisson Random Variable

The Poisson variable indicates the number ofsuccesses that occur during a given time intervalor in a specific region in a Poisson experiment

Probability Distribution of the PoissonRandom Variable.

m

m

m

)X(V)X(E

...2,1,0x

!x

e)x(p)xX(P

x


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0

0.1

0.2

0.3

0.4

1 2 3 4 5 6 7 8 9 10 11

Poisson probability distribution with m = 1

3678.e!0

1e)0(p)0X(P 1

01

3678.e!1

1e)1(p)1X(P 1

11

1839.2

e

!2

1e)2(p)2X(P

121

0613.6

e

!3

1e)3(p)3X(P

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0 1 2 3 4 5The X axis in ExcelStarts with x=1!!


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0

0.05

0.1

0.15

0.2

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16

0

0.05

0.1

0.15

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0

0.05

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Poisson probabilitydistribution with m =2

Poisson probability

distribution with m =5

Poisson probabilitydistribution with m =7

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0 1 2 3 4 5 6 7 8 9 10

0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15


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Example 13

Cars arrive at a tollbooth at a rate of 360 cars perhour.

What is the probability that only two cars will arriveduring a specified one-minute period? (Use the

formula) The probability distribution of arriving cars for any one-

minute period is Poisson with m = 360/60 = 6 cars perminute. Let X denote the number of arrivals during a one-

minute period.

0446.!2

6e)2X(P

26


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What is the probability that only two cars will arrive

during a specified one-minute period? P(X = 2) = P(X


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What is the probability that at least four cars will

arrive during a one-minute period? P(X>=4) = 1 - P(X


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When n is very large, binomial probability table maynot be available.

If p is very small (p< .05), we can approximate thebinomial probabilities using the Poisson distribution.

Use m = np and make the following approximation:

)xX(P)xX(P PoissonBinomial

With parameters n and p With m = np

Poisson Approximation of the Binomial


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Example 14

A warehouse has a policy of examining 50sunglasses from each incoming lot, and acceptingthe lot only if there are no more than two defectivepairs.

What is the probability of a lot being accepted if, infact, 2% of the sunglasses are defective?

Solution

This is a binomial experiment with n = 50, p = .02. Tables for n = 50 are not available; p

topic2 probability

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