© 2002 prentice-hall, inc.chap 4-1 statistics for managers using microsoft excel (3 rd edition)...

© 2002 Prentice-Hall, Inc. Chap 4-1

Statistics for Managers Using Microsoft Excel

(3rd Edition)

Chapter 4Basic Probability and Discrete Probability

Distributions


Chapter Topics

Basic probability concepts Sample spaces and events, simple

probability, joint probability

Conditional probability Statistical independence, marginal

probability

Bayes’s Theorem


Chapter Topics

The probability of a discrete random variable

Covariance and its applications in finance

Binomial distribution Poisson distribution Hypergeometric distribution

(continued)


Sample Spaces

Collection of all possible outcomes e.g.: All six faces of a die:

e.g.: All 52 cards in a deck:


Events

Simple event Outcome from a sample space with

one characteristic e.g.: A red card from a deck of cards

Joint event Involves two outcomes simultaneously e.g.: An ace that is also red from a

deck of cards


Visualizing Events

Contingency Tables

Tree Diagrams

Red 2 24 26

Black 2 24 26

Total 4 48 52

Ace Not Ace Total

Full Deck of Cards

Red Cards

Black Cards

Not an Ace

Ace

Ace

Not an Ace


Simple Events

The Event of a Triangle

There are 55 triangles in this collection of 18 objects


The event of a triangle AND blue in color

Joint Events

Two triangles that are blue


Special Events

Impossible evente.g.: Club & diamond on one card

draw Complement of event

For event A, all events not in A Denoted as A’ e.g.: A: queen of diamonds

A’: all cards in a deck that are not queen of diamonds

Null Event


Special Events Mutually exclusive events

Two events cannot occur together e.g.: -- A: queen of diamonds; B: queen of clubs

Events A and B are mutually exclusive Collectively exhaustive events

One of the events must occur The set of events covers the whole sample space e.g.: -- A: all the aces; B: all the black cards; C: all the

diamonds; D: all the hearts Events A, B, C and D are collectively exhaustive

Events B, C and D are also collectively exhaustive

(continued)


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


Full Deck of Cards

Tree Diagram

Event Possibilities

Red Cards

Black Cards

Ace

Not an Ace

Ace

Not an Ace


Probability

Probability is the numerical measure of the likelihood that an 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


(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


Computing Joint Probability

The probability of a joint event, A and B:

( and ) = ( )

number of outcomes from both A and B

total number of possible outcomes in sample space

P A B P A B

E.g. (Red Card and Ace)

2 Red Aces 1

52 Total Number of Cards 26

P


P(A1 and B2) P(A1)

TotalEvent

Joint Probability Using Contingency Table

P(A2 and B1)

P(A1 and B1)

Event

Total 1

Joint Probability Marginal (Simple) Probability

A1

A2

B1 B2

P(B1) P(B2)

P(A2 and B2) P(A2)


Computing Compound Probability

Probability of a compound event, A or B:( or ) ( )

number of outcomes from either A or B or both

total number of outcomes in sample space

P A B P A B

E.g. (Red Card or Ace)

4 Aces + 26 Red Cards - 2 Red Aces

52 total number of cards28 7

52 13

P


P(A1)

P(B2)

P(A1 and B1)

Compound Probability (Addition Rule)

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

P(A1 and B2)

TotalEvent

P(A2 and B1)

Event

Total 1

A1

A2

B1 B2

P(B1)

P(A2 and B2) P(A2)

For Mutually Exclusive Events: P(A or B) = P(A) + P(B)


Computing Conditional Probability

The probability of event A given that event B has occurred:

( and )( | )

( )

P A BP A B

P B

E.g.

(Red Card given that it is an Ace)

2 Red Aces 1

4 Aces 2

P


Conditional Probability Using Contingency Table

BlackColor

Type Red Total

Ace 2 2 4

Non-Ace 24 24 48

Total 26 26 52

Revised Sample Space

(Ace and Red) 2 / 52 2(Ace | Red)

(Red) 26 / 52 26

PP

P


Conditional Probability and Statistical Independence

Conditional probability:

Multiplication rule:

( and )( | )

( )

P A BP A B

P B

( and ) ( | ) ( )

( | ) ( )

P A B P A B P B

P B A P A


Conditional Probability and Statistical Independence

Events A and B are independent if

Events A and B are independent when the probability of one event, A, is not affected by another event, B

(continued)

( | ) ( )

or ( | ) ( )

or ( and ) ( ) ( )

P A B P A

P B A P B

P A B P A P B


Bayes’s Theorem

1 1

||

| |

and

i ii

k k

i

P A B P BP B A

P A B P B P A B P B

P B A

P A

Adding up the parts of A in all the B’s

Same Event


Bayes’s Theorem Using Contingency Table

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?

.50 | .4 | .10P R P C R P C R

| ?P R C


Bayes’s Theorem Using Contingency Table

||

| |

.4 .5 .2 .8

.4 .5 .1 .5 .25

P C R P RP R C

P C R P R P C R P R

(continued)

Repay

Repay

CollegeCollege 1.0.5 .5

.2

.3

.05.45

.25.75

Total

Total


Random Variable

Random Variable Outcomes of an experiment expressed

numerically e.g.: Toss a die twice; count the number of

times the number 4 appears (0, 1 or 2 times)


Discrete Random Variable

Discrete random variable Obtained by counting (1, 2, 3, etc.) Usually a finite number of different values e.g.: Toss a coin five times; count the

number of tails (0, 1, 2, 3, 4, or 5 times)


Probability Distribution Values Probability

0 1/4 = .25

1 2/4 = .50

2 1/4 = .25

Discrete Probability Distribution Example

T

T

T T

Event: Toss two coins

Count the number of tails


Discrete Probability Distribution

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

Xj = value of random variable

P(Xj) = probability associated with value

Mutually exclusive (nothing in common)

Collectively exhaustive (nothing left out)

0 1 1j jP X P X


Summary Measures

Expected value (the mean) Weighted average of the probability

distribution

e.g.: Toss 2 coins, count the number of tails, compute expected value

j jj

E X X P X

0 2.5 1 .5 2 .25 1

j jj

X P X


Summary Measures

Variance Weight average squared deviation about the

mean

e.g. Toss two coins, count number of tails, compute variance

(continued)

222j jE X X P X

22

2 2 2 0 1 .25 1 1 .5 2 1 .25 .5

j jX P X


Covariance and its Application

1

th

th

th

: discrete random variable

: outcome of

: discrete random variable

: outcome of

: probability of occurrence of the

outcome of an

N

XY i i i ii

i

i

i i

X E X Y E Y P X Y

X

X i X

Y

Y i Y

P X Y i

X

thd the outcome of Yi


Computing the Mean for Investment Returns

Return per $1,000 for two types of investments

P(XiYi) Economic condition Dow Jones fund X Growth Stock Y

.2 Recession -$100 -$200

.5 Stable Economy + 100 + 50

.3 Expanding Economy + 250 + 350

Investment

100 .2 100 .5 250 .3 $105XE X

200 .2 50 .5 350 .3 $90YE Y


Computing the Variance for Investment Returns


.2 Recession -$100 -$200



Investment

2 2 22 100 105 .2 100 105 .5 250 105 .3

14,725 121.35X

X

2 2 22 200 90 .2 50 90 .5 350 90 .3

37,900 194.68Y

Y


Computing the Covariance for Investment Returns


.2 Recession -$100 -$200



Investment

The Covariance of 23,000 indicates that the two investments are positively related and will vary together in the same direction.

100 105 200 90 .2 100 105 50 90 .5

250 105 350 90 .3 23,300

XY


Important Discrete Probability Distributions

Discrete Probability Distributions

Binomial Hypergeometric Poisson


Binomial Probability Distribution

‘n’ identical trials e.g.: 15 tosses of a coin; ten light bulbs

taken from a warehouse Two mutually exclusive outcomes on

each trials e.g.: Head or tail in each toss of a coin;

defective or not defective light bulb Trials are independent

The outcome of one trial does not affect the outcome of the other


Binomial Probability Distribution

Constant probability for each trial e.g.: Probability of getting a tail is the same

each time we toss the coin Two sampling methods

Infinite population without replacement Finite population with replacement

(continued)


Binomial Probability Distribution Function

Tails in 2 Tosses of Coin

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

1 2/4 = .50

2 1/4 = .25

!1

! !

: probability of successes given and

: number of "successes" in sample 0,1, ,

: the probability of each "success"

: sample size

n XXnP X p p

X n X

P X X n p

X X n

p

n


Binomial Distribution Characteristics

Mean E.g.

Variance and Standard Deviation

E.g.

E X np 5 .1 .5np

n = 5 p = 0.1

0.2.4.6

0 1 2 3 4 5

X

P(X)

1 5 .1 1 .1 .6708np p

2 1

1

np p

np p


Binomial Distribution in PHStat

PHStat | probability & prob. Distributions | binomial

Example in excel spreadsheet

Microsoft Excel Worksheet


Poisson Distribution

Poisson Process: Discrete events in an “interval”

The probability of One Successin an interval is stable

The probability of More thanOne Success in this interval is 0

The probability of success isindependent from interval to interval

e.g.: number of customers arriving in 15 minutes

e.g.: number of defects per case of light bulbs

P X x

x

x

( |

!

e-


Poisson Probability Distribution Function

e.g.: Find the probability of 4 customers arriving in 3 minutes when the mean is 3.6.

3.6 43.6

.19124!

eP X

!

: probability of "successes" given

: number of "successes" per unit

: expected (average) number of "successes"

: 2.71828 (base of natural logs)

XeP X

XP X X

X

e


Poisson Distribution in PHStat

PHStat | probability & prob. Distributions | Poisson




Poisson Distribution Characteristics

Mean

Standard Deviationand Variance

1

N

i ii

E X

X P X

= 0.5

= 6

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)

2


Hypergeometric Distribution

“n” trials in a sample taken from a finite population of size N

Sample taken without replacement Trials are dependent Concerned with finding the probability of

“X” successes in the sample where there are “A” successes in the population


Hypergeometric Distribution Function

E.g. 3 Light bulbs were selected from 10. Of the 10 there were 4 defective. What is the probability that 2 of the 3 selected are defective?

4 6

2 12 .30

10

3

P

: probability that successes given , , and

: sample size

: population size

: number of "successes" in population

: number of "successes" in sample

0,1,2,

A N A

X n XP X

N

n

P X X n N A

n

N

A

X

X

,n


Hypergeometric Distribution Characteristics

Mean

Variance and Standard Deviation

AE X n

N

22

2

1

1

nA N A N n

N N

nA N A N n

N N

FinitePopulationCorrectionFactor


Hypergeometric Distribution in PHStat

PHStat | probability & prob. Distributions | Hypergeometric …




Chapter Summary

Discussed basic probability concepts Sample spaces and events, simple

probability, and joint probability

Defined conditional probability Statistical independence, marginal

probability

Discussed Bayes’s theorem


Chapter Summary

Addressed the probability of a discrete random variable

Defined covariance and discussed its application in finance

Discussed binomial distribution Addressed Poisson distribution Discussed hypergeometric distribution

(continued)

© 2002 prentice-hall, inc.chap 4-1 statistics for managers using microsoft excel (3 rd edition)...

Documents

event b

compound event

queen of diamonds b

event e

exhaustiveevents b

aces b

ace simple eventsthe

objectsjoint eventsthe