Statistics for Managers Using Microsoft Excel

(3rd Edition)

Chapter Basic 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

Properties of Probability

If A is an event and A’ is its complement then P(A) = 1-P(A’)

For any two events A and B P(AUB) = P(A) + P(B)-P(AB)

A B

A=(AB)U(AB’)P(A)=P(AB)+P(AB’)P(AUB)=P(B)+P(AB’) = P(B)+P(A)-P(AB)

Properties of Probability

If A subset of B then P(A)≤P(B)

AA

B

)()(

)'()()(

'

APBP

BAPAPBP

BAAB

Properties of Probability

)())()((

)()()()(

))()((

)()()()(

))(()()(

))(()(

CBAPCBPCAP

CPCAPBPAP

CBCAP

CPCAPBPAP

CBAPCPBAP

CBAPCBAP

Properties of Probability

)(1

....)(

)()()(

1

1

1

n

i in

kjikji

jiji

ii

n

i i

AP

AAAP

AAPAPAP

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

Joint Probability Using Contingency Table

Joint Probability

Marginal (Simple) Probability

n(A1 and B2) n(A1)

TotalEvent

n(A2 and B1)

n(A1 and B1)

Event

Total N(S)

A1

A2

B1 B2

n(B1) n(B2)

n(A2 and B2) n(A2)

Joint Probability Using Contingency Table

Joint Probability

Marginal (Simple) Probability

P(A1 and B2) P(A1)

TotalEvent

P(A2 and B1)

P(A1 and B1)

Event

Total 1

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

Compound Probability (Addition Rule)

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

P(A1)

P(B2)

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

P(A)

P(B’)

P(A and B)

Conditional Probability

P(A1 and B’)

TotalEvent

P(A’ and B)

Event

Total 1

A

A’

B B’

P(B)

P(A’ and B’) P(A)

( and )( | )

( )

P A BP A B

P B

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

Example

A family has two children. What is the conditional probability that both are boys given that at least one of them is a boy ? Assume that the sample space S is given by S={(b,b),(b,g),(g,b),(g,g)}, and all outcomes are equally likely. [(b,g) means for instance that the older child is boy and the younger child is a girl.]

Solution

Letting E denote the event that both children are boys, and F the event that at least one of them is a boy, then the desired probability is given by

3/1

)|(

4/34/1

)}),(),,(),,({()}),({(

)()(

bggbbbPbbP

FPEFPFEP

Example

Bety can either take a course in mathematics or in statistics. If She takes the statistic course, then she will receive an A grade with probability ½ , while if she takes the math course then she will receive an A grade with prob. 1/3 . Bety decides to base her decision on the flip of fair coin. What is the prob that Bety will get an A in math ?

Solution

If we let F be the event that Bety takes math and E denote the event that she receives an A in whatever course she takes, then the prob is P(EF) = P(E|F)P(F) = 1/3.1/2 = 1/6. P(F) =1/2 , because Bety decides to base

her decision on the flip of fair coin.

Example

Suppose that each of three men at the party throws his hat into the center of the room. The hats are first mixed up and then each man randomly selects a hat. What is the probability that none of the three men selects his own hat ?

Solution

Let us denote by Ei ,i=1,2,3, the event that the ith man selects his own hat. The probability that none selects his own hat is

Now we compute

)(1 321 EEEP

)( 321 EEEP

32

61

61

313

1

61

61

31

21

31

33)(

.1

)()|()(

)()|()(

3,2,1,)(

i i

jijikkji

jjiji

i

EP

EEPEEEPEEEP

EPEEPEEP

iEP

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

Example A series system of two components, C1 and C2.

The probability C1 fail is 0.1 and C2 fail is 0.2 and both of them are independent.

The probability that the system fails is P(C1 fail U C2 fail) =P(C1 fail) + P(C2 fail) -

P(C1,C2 fail) = P(C1 fail) + P(C2 fail) - P(C1 fail)xP(C2 fail)

C1 C2

Example A paralel system of two components, C1 and C2.

The probability C1 fail is 0.1 and C2 fail is 0.2 and both of them are independent.

The probability that the system fails is P(C1 fail and C2 fail) =P(C1 fail).P(C2 fail)

=0.1x0.2 = 0.02

C1

C2

Total Probability

Let E and F be events. We may express E as E = EF U EF’ , since both of them are abviously mutually exclusive, we have that P(E) = P(E|F)P(F) + P(E|F’)P(F’)

If F can be separated by F1 , F2 , …, F k and each of them mutually exclusive then P(E) = P(E|F1)P(F1) + …+ P(E|Fk)P(Fk

)

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?

R = Repaid ; C = College

.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

Example

In answering a question on a multiple choice test, a student either knowns the answer of he guesses . Let p be the prob that she knows the answer. There are m multiple-choice alternatives. What is the conditional that a student knew the answer to a question given that she answered it correctly ?

Solution

Let C and K denote respectively the event that the student answers the question correctly and the event that she actually knows the answer. Now

pmpp

KPKCPKPKCPKPKCPCKP

1/11

)'()'|()()|()()|()|(

Example

A laboratory blood test is 95 percent effective in detecting a certain disease when it is, in fact present. However, the test also yields a “ false positive” result for 1 percent of the healthy persons tested. If 0.5 percent of the population actually has the disease, what is the prob a person has the disease given that his test result is positive ?

Solution

Let D be the event that the tested person has the disease, and E the event that his test result is positive.

323.0

)|(

995.01.0005.095.0005.095.0

)'()'|()()|()()|(

DPDEPDPDEPDPDEPEDP

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

Example

Suppose we toss a coin having a prob p of coming up heads, until the first head appears. Letting N denote the number of flips required, then assuming that the outcome of successive flips are independent, N is a random variable taking on one of the values 1,2,3,…with respective probabilities

Solution

P(N=1) = P(H) = p; P(N=2) = P({T,H}) = (1-p)p ; : P(N=n) = P({T,…,T,H})= (1-p)n-1 p, n>=1

As a check, note that

1

)1(

}{}{

1

1

11

n

n

nn

pp

nNPnNP

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

Correlation

The correlation coefficient of X and Y is

YX

XY

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

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

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

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

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

Correlation

The correlation coefficient of X and Y is

If the value of X increase, then the value of Y increase too.

986.068.19435.121300,23

Cumulative Distribution Function

The cumulative distribution function of a random variable X is defined for any real x by

xx

i

dttf

xfxXPxF i

)(

)()()(

Example

Consider the distribution of lifetimes , X (in months), of a particular type of component. We will assume that the CDF has the form

The median lifetime is

0;1)(2

3 )( xexFx

monthsm

e

mF

m

m

m

498.2

)5.0ln(

)5.0ln(

5.01

5.0)(

2/13

2

3

23

It is desired to find the time t such that 10% of the component fail before t. This is the 10th percentile :

Thus if the components are guaranteed for one month, slightly more than 10% will need to be replaced

monthsx

x

e

xF

x

x

974.0

)]5.0ln([3

)9.0ln(

1.01

1.0)(

2/1

2

3

23

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

Proof of the Probability

Note that, by the binomial theorem, the probabilities sum to one, that is

n

xnxn

x

pp

ppx

nxp

))1((

)1()(0

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

Expectation

np

ppnp

ppnp

pp

ppx

nxXE

n

k

knkkkn

n

n

x

xnxxxn

n

n

x

xnxxxn

n

n

x

xnx

1

0

1)!()!1(

)!1(

1

1)!1()!(

)!1(

1)!1()!(

!

0

)1(

)1(

)1(

)1()(

Variance

2

2

0

2)!()!2(

)!2(2

2

2)!2()!(

)!2(2

2)!2()!(

!

0

)1(

)1()1(

)1()1(

)1(

)1()1())1((

pnn

pppnn

pppnn

pp

ppx

nxxXXE

n

k

knkkkn

n

n

x

xnxxxn

n

n

x

xnxxxn

n

n

x

xnx

Binomial Distribution in PHStat

PHStat | probability & prob. Distributions | binomial

Example in excel spreadsheet

Microsoft Excel Worksheet

Example

S uppose that an airplane engine will fall, when in flight, with prob 1-p independently from engine to engine; suppose that the airplane will make a succesful flight if at least 50 percent of its engines remain operative. For what values of p is a four-engine plane preferable to a two-engine plane ?

Solution

The probaility that a four-engine plane makes a successful flight is

Whereas the corresponding probability for a two-engine plane is

4

2

43224 )1(4)1(614

x

xx pppppppx

2

1

22 )1(212

x

xx pppppx

Solution

Hence the four-engine is safer if

Hence, the four-engine plane is safer when the engine success probability is at least as large as 2/3 , whereas the two-engine plane is safer when this probability falls below 2/3

32

24322

023

)1(2)1(4)1(6

p

p

pppppppp

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

Example in excel spreadsheet

Microsoft Excel Worksheet

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

Approximate a binomial to poisson

An important property of the poisson random variable is that it may be used to approximate a binomial random variabel when the binomial parameter n is large and p is small. To see this, suppose that X is a binomial r.v. with parameters (n,p), and let µ = np. Then

Proof

1;1)/1(;/1

1)(

)1)...(1(

!

)/1(

/1!

)1)...(1(

!)!(!

i

i

i

ni

i

n

innnin

i

n

nin

innn

in

n

i

niinn

nen

e

iXP

Expectation

ee

e

XE

xx

xx

e

xx

xe

x

xxx

1)!1(

1)!1(

0!

1

1

)(

Variance

1

1

1

1))1((

2)!2(

2)!2(

0!

)1(

2

2

ee

e

XXE

xx

xx

e

xxexx

x

xxx

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 …

Example in excel spreadsheet

Microsoft Excel Worksheet

Expectation

0

)(

)(

22

22

22

2/)(

21

2/)(

21

2/)(

21

dxe

xdex

dxxeXE

x

x

x

Variance

2

)(

2

2/2

2

2/)(2

212

22

2

dyey

deXE

y

xxx

Jointly Distributed Random Variables

The joint probability mass function of X and Y is p(x,y)=P(X=x,Y=y)

The probability mass function of X

The probability mass function of Y

dyyxf

yxpxp y

,

),()(

dxyxf

yxpyp x

,

),()(

Expectation

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

)()(

)()(

),(),(

),()(

11 nn XEXEXXE

YEXE

dyyyfdxxxf

dydxyxfydxdyyxfx

dxdyyxfyxYXE

Example As another example of the usefulness of

equation above, let us use it to obtain the expectation of a binomial r.v.

pqXVpXEfailed

succesX iii

)(;)(;,0

,1

npqXVXVXV

npXEXEXE

XXX

n

n

n

)(...)()(

)(...)()(

...

1

1

1

Example

At a party N men throw their hats into the center of a room. The hats are mixed up and each man randomly selects one. Find the expected number of men that select their own hats

Solution

Letting X denote the number of men that select their own hats, we can best compute E(X) by noting that X = X1+…+XN ; where Xi is indicator function if the ith man select his own hat. So P(Xi = 1) = 1/N. And so E(Xi) = 1/N. Hence We obtain that E(X) = 1. So, no matter how many people are at the party, on the average exactly one of the men will select his own hat.

Independent R.V

X and Y are independent if

)()(),(

)()(),(

yfxfyxf

ypxpyxp

)()(

)()(

)()(

),()(

YEXE

dyyyfdxxxf

dxdyyfxxyf

dxdyyxxyfXYE

Covariance and Variance of Sums of Random Variables

The covariance of any two random variables X and Y, denoted by Cov(X,Y), is defined by

Cov(X,Y) = E[(X-E[X])(Y-E[Y])]

= E(XY)-E(X)E(Y) If X and Y are independent Cov(X,Y) = 0

Properties of Covariance

Cov(x,X) = Var(X) Cov(cX,Y) = c Cov(X,Y) Cov(X,Y+Z)= E[X(Y+Z)]-E[X]E[Y+Z]

= E[XY]-E[X]E[Y] + E[XZ]-E[X]E[Z]

= (Cov(X,Y) + Cov(X,Z)The last property easily generalizes to

give ),(, jiji YXCovYXCov

Variance of Sum Variabel

n

i

n

ijji

n

ii

n

i

n

ijji

n

iii

n

i

n

jji

n

ji

n

ii

n

ii

XXCovXV

XXCovXXCov

XXCov

XXCovXVar

11

11

1 1

111

),(2)(

),(),(

),(

,

Proposition

Suppose that X1,…,Xn are independent and identically distributed with expected value µ and variance σ2. Then

nn

ijjiniin

jij

iin

ii

XVarXXCovXXCov

XVarXXXCov

XXCovXXCovXXXCov

22

0

)()(),(

)(),(

),(),(),(

11

1

Example

Sums of independent Poisson Random Variables : Let A and Y be independent Poisson random variables wirh respective means λ1 and λ2 . Calculate the distribution of X + Y.

Solution : Since the event {X+Y = n} may be written as the union of the disjoint events {X=k,Y=n-k}, 0≤k≤n, we have

nne

n

k

knkknk

nn

e

n

kknk

n

k

n

k

knk

ee

knYPkXP

knYkXPnYXP

21!

021)!(!

!!

0)!(!

0

0

21

21

2211

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