basic prob. spring 09

8/6/2019 Basic Prob. Spring 09

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Basic Probability And

Probability Distributions

Business Statistics


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Chapter Topics

Basic Probability Concepts:

Sample Spaces and Events, Simple

Probability, and Joint Probability,

Conditional Probability

Bayes Theorem

The Probability Distribution for a

Discrete Random Variable


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Chapter Topics

Binomial and Poisson Distributions

Covariance and its Applications inFinance

The Normal Distribution

Assessing the Normality Assumption


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Sample Spaces

Collection of all Possible Outcomes

e.g. All 6 faces of a die:

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


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Events

Simple Event: Outcome from a Sample Space

with 1 Characteristic

e.g. ARed CardRed Cardfrom a deck of cards.

Joint Event: Involves 2 OutcomesSimultaneously

e.g. An AceAce which is also aRed CardRed Cardfrom a

deck of cards.


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Visualizing Events

Contingency Tables

Tree Diagrams

Ace Not Ace Total

Red 2 24 26Black 2 24 26

Total 4 48 52


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Simple Events

The Event of a Happy Face

Th

ere are 55h

appy faces in th

is collection of 18 objects


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Joint Events

The Event of a Happy Face ANDAND Light Colored

3 Happy Faces wh

ich

are ligh

t in color


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ii

Special Events

Null event

Club & diamond on1 card draw

Complement of event

For event A,All events not In A:

Null Event

'A


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3 Items: 3 Happy Faces Given they are Light Colored

Dependent or

Independent EventsThe Event of a Happy Face GIVEN it is Light ColoredE = Happy FaceLight Color


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Contingency Table

A Deck of 52 Cards

Ace Not anAce Total

Red

Black

Total

2 24

2 24

26

26

4 48 52

Sample Space

Red Ace


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2500

Contingency Table

2500 Employees of Company ABC

Agree Neutral Opposed | Total

MALE

FEMALE

Total

900 200

300 100

400 | 1500

600 | 1000

1200 300 1000 |

Sample Space


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Tree Diagram

Event Possibilities

Red

Cards

Black

Cards

Ace

Not an Ace

Ace

Not an Ace

FullDeck

ofCards


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Probability

Probability is the numerical

measure of the likelihood

that the event will occur.

Value is between 0 and 1.

Sum of the probabilities of

all mutually exclusive and

collective exhaustive events

is 1.

Certain

Impossible

.5

1

0


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Computing Probability

The Probability of an Event, E:

Each of the Outcome in the Sample Space

equally likely to occur.

e.g. P( ) = 2/36

(There are 2 ways to get one 6 and the other 4)

P(E) =Number of Event Outcomes

Total Number of Possible Outcomes in the Sample Space

=X

T


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Computing

Joint ProbabilityThe Probability of a JointEvent, A andB:

e.g. P(RedC

ard and Ace)

P(A andB)

Number of Event Outcomes from both A and B

Total Number of Possible Outcomes in Sample Space

=

=

2 ed Aces 1

52 Total Number of Cards 26!


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P(A2 and B1)

P(A1 and B1)

Event

Event Total

Total 1

Joint Probability Using

Contingency Table

Joint Probability Marginal (Simple) Probability

P(A1)A1

A2

B1 B2

P(B1) P(B2)

P(A1 and B2)

P(A2 and B2) P(A2)


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Computing

Compound Probability

The Probability of a CompoundEvent, A orB:

e.g.

P(Red Card orAce)

4 Aces + 26 ed Cards 2 ed Aces 28 7

52 Total Number of Cards 52 13

! ! !

Numer of Event Outcomes from EitherA or B

P A or B Total Outcomes in the Sample Space!


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The pervious table refers to 2500 employees of ABCCompany, classified by genderand by opinion on a company proposal to emphasize fringe benefits rather than wage

increases in an impending contract discussion

Calculate the probability that an employee selected (at random) from this group will be:

1. A female opposed to the proposal


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1. A female opposed to the proposal 600/2500 = 0.24


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2. Neutral


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2. Neutral 300/2500 = 0.12


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2. Neutral 300/2500 = 0.12

3. Opposed to th

e proposal, GIVEN th

atthe employee selected is a female


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2. Neutral 300/2500 = 0.12

3. Opposed to th


atthe employee selected is a female 600/1000 = 0.60


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2. Neutral 300/2500 = 0.12

3. Opposed to th


atthe employee selected is a female 600/1000 = 0.60

4. Either a female or opposed to the

proposal


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2. Neutral 300/2500 = 0.12

3. Opposed to the proposal, GIVEN that

the employee selected is a female 600/1000 = 0.60


proposal .. 1000/2500 + 1000/2500 - 600/2500 =

1400/2500 = 0.56


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2. Neutral 300/2500 = 0.12




proposal .. 1000/2500 + 1000/2500 - 600/2500 =

1400/2500 = 0.56

5. Are Gender and Opinion (statistically) independent?


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2. Neutral 300/2500 = 0.12




proposal .. 1000/2500 + 1000/2500 - 600/2500 =

1400/2500 = 0.56

5. Are Gender and Opinion (statistically) independent?

For Opinion and Gender to be independent, the joint probability of each pair ofA events (GENDER) and B events (OPINION) should equal the product of the

respective unconditional probabilities.clearly this does not hold..check, e.g.,

the prob. Of MALE and IN FAVOR against the prob. of MALE times the prob.

of IN FAVOR they are not equal.900/2500 does not equal 1500/2500 *

1200/2500


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P(A1 and B1)

P(B2)P(B1)

P(A2 and B2)P(A2 and B1)

Event

Event Total

Total 1

Compound Probability

Addition Rule

P(A1 and B2) P(A1)A1

A2

B1 B2

P(A2)

P(A1 or B1 ) = P(A1) +P(B1) - P(A1 and B1)

ForMutually Exclusive Events: P(A or B) = P(A) + P(B)


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Computing


The Probability ofEvent A given that Event B has

occurred:

P(A B) =

e.g.

P(Red Card giventhatitis an Ace) =

P A B

P B

and

2 Red Aces 1

4 Aces 2!


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BlackColorType Red Total

Ace 2 2 4

Non-Ace 24 24 48

Total 26 26 52


UsingC

ontingency TableConditional Event: Draw 1 Card. Note Kind & Color

26

2

5226

522!!

/

/

P(R

RDP(=R|P(

RevisedSampleSpace


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Conditional Probability and

Statistical Independence

)B(P

)BandA(P

Conditional Probability:

P(AB) =

P(A and B) = P(A B) P(B)Multiplication Rule:

= P(B %) P(A)


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Conditional Probability and

Statistical Independence(continued)

Events are Independent:

P(A

B) = P(A)

Or, P(A and B) = P(A) P(B)

Events A and B are Independentwhen the

probability of one event, A is notaffectedby

another event, B.

Or, P(B A) = P(B)


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Bayes Theorem

)B(P)BA(P)B(P)BA(P

)B(P)BA(P

kk

ii

yyyyy

y

11

)A(

)AandB( i!

P(BiA) =

Adding up

the parts

of A in allthe Bs

SameEvent


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A manu acturer o V Rs purchases a particular microchip, called the LS-24, rom three suppliers: Hall

Electronics, Spec Sales, and rown omponents. Thirty percent o the LS-24 chips are purchased rom

Hall, 20% rom Spec, and the remaining 50% rom rown. The manu acturer has extensive past records

or the three suppliers and knows that there is a 3% chance that the chips rom Hall are de ective, a 5%

chance that chips rom Spec are de ective and a 4% chance that chips rom rown are de ective. WhenLS-24 chips arrive at the manu acturer, they are placed directly into a bin and not inspected or otherwise

identi ied as to supplier. A worker selects a chip at random.

What is the probability that the chip is de ective?

A worker selects a chip at random or installation into a V R and inds it is de ective. What is the

probability that the chip was supplied by Spec Sales?


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What are the chances of repaying a loan,given a college education?

Bayes Theorem: Contingency Table

Loan Status

Education Repay Default Prob.

College .2 .05 .25

No College

Prob. 1

P(Repay College) =

? ? ?

??

P(College and Repay)

P(College and Repay) + P(College and Default)

= .80=.20.25


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Random Variable: outcomes of an

experiment expressed numerically

e.g.

Throw a die twice: Count the number of times 4

comes up (0, 1, or 2 times)


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Discrete Random Variable:

Obtained by Counting (0, 1, 2, 3, etc.)

Usually finite by number of differentvalues

e.g.

Toss a coin 5 times. Count the number oftails. (0, 1, 2, 3, 4, or 5 times)


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Discrete Probability

Distribution Example

Probability distributionValues probability

0 1/4 = .25

1 2/4 = .502 1/4 = .25

Event: Toss 2 Coins. Count # Tails.

T

T

T T


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Discrete

Probability Distribution

List of all possible [ xi,p(xi) ] pairs

Xi= value of random variable

P(xi) = probability associated with value

Mutually exclusive (nothing in common)

Collectively exhaustive (nothing left out)

0 ep(xi) e 1

7 P(xi) = 1


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Summary Measures

Expected value (The mean)Weighted average of the probability distribution

Q = E(X) = xip(xi)

E.G. Toss 2 coins, count tails, compute expected value:

Q= 0 v .25 + 1 v.50 + 2 v .25 = 1

Number of Tails


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Summary Measures

Variance

Weighted average squared deviation about mean

W = E[ (xi- Q )2]=7 (xi- Q )2p(xi)

E.G. Toss 2 coins, count tails, compute variance:W = (0 - 1)2(.25) + (1 - 1)2(.50) + (2 - 1)2(.25)

= .50


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Important Discrete Probability

Distribution Models

Discrete Probability

Distributions

Binomial Poisson


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Binomial Distributions

N identical trials

E.G. 15 tosses of a coin, 10 light bulbs

taken from a warehouse

2 mutually exclusive outcomes on each

trial

E.G. Heads or tails in each toss of a coin,

defective or not defective light bulbs


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Binomial Distributions

Constant Probability for each Trial e.g. Probability of getting a tail is the same

each time we toss the coin and each light bulb

has the same probability of being defective

2 Sampling Methods:

Infinite Population Without Replacement

Finite Population With Replacement

Trials are Independent:

The Outcome of One Trial Does Not Affect the

Outcome of Anoth

er


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Binomial Probability

Distribution Function

P(X) = probability that X successes given a knowledge of nand p

X = number ofsuccesses insample, (X = 0, 1, 2, ..., n)

p = probability of eachsuccess

n = sample size

P(X)n

X! n X

p pX n X!

( )!

( )!

1

Tails in 2 Tosses of Coin

X P(X)0 1/4 = .25

1 2/4 = .50

2 1/4 = .25


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Binomial Distribution

Characteristics

n = 5 p = 0.1

n = 5 p = 0.5

Mean

Standard Deviation

Q

W

EXnp

np p

! !

!

( )

( )

0.2

.4

.6

0 1 2 3 4 5

X

P(X)

.2

.4

.6

0 1 2 3 4 5

X

P(X)

e.g. Q = 5(.1) = .5

e.g. W = 5(.5)(1 - .5)

= 1.118

0


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Poisson Distribution

Poisson process:

Discrete events in an interval The probability ofone success in

an interval is stable

The probability ofmore than onesuccess in this interval is 0

Probability of success is

Independent from interval toInterval

E.G. # Customers arriving in 15 min

# Defects per case of light

bulbs

P X x

x

x

( |

!

! P

P

P

e

-


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Function

P(X) = probability ofXsuccesses given PP = expected (mean) number of successes

e = 2.71828 (base of natural logs)

X = number of successes per unit

PXX

X

( )!

! PP

e

e.g. Findthe probability of4

customers arrivingin 3

minutes whenthe meanis 3.6.P(X) e

-3.6

3.64!

4

= .1912


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Characteristics

P= 0.5

P= 6

Mean

Standard Deviation

Q P

W P

ii

N

i

EX

X P X

! !

!

!

!

( )

( )1

0.2

.4

.6

0 1 2 3 4 5

X

P(X)

0

.2

.4

.6

0 2 4 6 8 10

X

P(X)


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Covariance

X=discreteran

domvariableX

Xi = valueoftheith outcomeofX

P(xiyi) =probabilityofoccurrenceofthe

ithoutc

omeofXan

dith

outc

omeofY

Y=discreterandom variableY

Yi =valueoftheith outcomeofY

I=1, 2, , N

? A ? A )YX(P)Y(EY)X(EX iiiN

iiXY yy !

!1

W


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C i h V i f


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Computing the Variance forInvestment Returns

P(XiYi) Economic condition Dow Jones fund X GrowthStock Y

.2 Recession -$100 -$200

.5 Stable Economy + 100 + 50

.3 Expanding Economy + 250 + 350

Investment

Var(X) = = (.2)(-100 -105)2 + (.5)(100 - 105)2 + (.3)(250 - 105)2

= 14,725, WX = 121.35

Var(Y) = = (.2)(-200 - 90)2 + (.5)(50 - 90)2 + (.3)(350 - 90)2

= 37,900, WY = 194.68

2

XW

2

YW

C ti th C i f


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Computing the Covariance forInvestment Returns

P(XiYi) Economic condition Dow Jones fund X GrowthStock Y

.2 Recession -$100 -$200

.5 Stable Economy + 100 + 50

.3 Expanding Economy + 250 + 350

Investment

WXY= (.2)(-100 - 105)(-200 - 90) + (.5)(100 - 105)(50 - 90)

+ (.3)(250 -105)(350 - 90) = 23,300

The Covariance of 23,000 indicates that the two investments are

positively related and will vary together in the same direction.


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The Normal Distribution

Bell Shaped

Symmetrical

Mean, Median and

Mode are Equal

Middle Spread

Equals 1.33 W

Random Variable has

Infinite Range

MeanMedianMode

X

f(X)

Q


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The Mathematical Model

f(X) = frequency of random variable X

T = 3.14159; e = 2.71828

W = population standard deviation

X = value of random variable (-g


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Varying the Parameters W and Q, we obtainDifferent Normal Distributions.

There are

an Infinite

Number

Many Normal Distributions

N l Di t ib ti


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Normal Distribution:

Finding Probabilities

Probability is thearea under thecurve!

c dX

f(X)

P c X d( ) ?e e !


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Infinitely Many Normal Distributions Means

Infinitely Many Tables to Look Up!

Each distribution

has its own table?

Which Table?


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Z Z

Z = 0.12

Z

.00 .010.0 .0000 .0040 .0080

.0398 .0438

0.2 .0793 .0832 .0871

0.3 .0179 .0217 .0255

Solution (I): The Standardized

Normal Distribution

.0478

.02

0.1 .0478

Standardized Normal Distribution

Table (Portion) Q = 0 and W = 1

Probabilities

Shaded AreaExaggerated

Only One Table is Needed

S l ti (II) Th C l ti


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Z = 0.12

Z

.00 .010.0 .5000 .5040 .5080

.5398 .5438

0.2 .5793 .5832 .5871

0.3 .5179 .5217 .5255

Solution (II): The Cumulative

Standardized Normal Distribution

.5478

.02

0.1 .5478

Cumulative Standardized Normal

Distribution Table (Portion)

Probabilities


Only One Table is Needed

0 and 1Q W> >! !


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ZQ = 0

WZ

= 1

.12

Standardizing Example

NormalDistribution

StandardizedNormal Distribution

XQ = 5

W = 10

6.2

12010

526.

.XZ !

!

!

W

Q

Shaded Area Exaggerated

E l


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0

W = 1

-.21 Z.21

Example:

P(2.9


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Fi di Z V l


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Z .00 0.2

0.0 .0000 .0040 .0080

0.1 .0398 .0438 .0478

0.2 .0793 .0832 .0871

.1179 .1255

ZQ = 0

W = 1

.31

Finding ZValues

for Known Probabilities

.1217 .01

0.3

Standardized NormalProbability Table (Portion)

What Is ZGivenProbability = 0.1217?


.1217

Recovering X Values


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ZQ = 0

W = 1

.31XQ = 5

W = 10

?

Recovering X Values

for Known Probabilities

Normal Distribution Standardized Normal Distribution

.1217 .1217

Shaded Area Exaggerated

X 8.1!Q ZW= 5 + (0.31)(10) =


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Assessing Normality

Compare Data Characteristics to

Properties of Normal Distribution

Put Data into Ordered Array

Find Corresponding Standard Normal

Quantile Values Plot Pairs of Points

Assess by Line Shape


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Assessing Normality

Normal Probability Plot forNormal Distribution

Look forStraigh

t Line!

30

60

90

-2 -1 0 1 2

Z

X


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Normal Probability Plots

Left-Skewed Right-Skewed

Rectangular U-Shaped

30

60

90

-2 -1 0 1 2

Z

X

30

60

90

-2 -1 0 1 2

Z

X

30

60

90

-2 -1 0 1 2

Z

X

30

60

90

-2 -1 0 1 2

Z

X


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Chapter Summary

Discussed Basic Probability Concepts:

Sample Spaces and Events, Simple

Probability, and Joint Probability

Defined Conditional Probability

Discussed Bayes Theorem

Addressed the Probability of a Discrete

Random Variable


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Summary

Discussed Binomial and PoissonDistributions

Addressed Covariance and itsApplications in Finance

Covered Normal Distribution Discussed Assessing the NormalityAssumption

basic prob. spring 09

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