© 1999 Prentice-Hall, Inc. Chap. 4 - 1 Statistics for Managers Using Microsoft Excel/SPSS Chapter 4 Basic Probability And Discrete Probability Distributions
© 1999 Prentice-Hall, Inc. Chap. 4 - 1
Statistics for Managers
Using Microsoft Excel/SPSS
Chapter 4
Basic Probability And Discrete Probability Distributions
© 1999 Prentice-Hall, Inc. Chap. 4 - 2
Chapter Topics
• Basic Probability Concepts:
Sample Spaces and Events, Simple
Probability, and Joint Probability,
• Conditional Probability
• Bayes’ Theorem
• The Probability of a Discrete Random Variable
• Binomial, Poisson, and Hypergeometric
Distributions
• Covariance and its Applications in Finance
© 1999 Prentice-Hall, Inc. Chap. 4 - 3
Sample Spaces
Collection of all Possible Outcomes
e.g. All 6 faces of a die:
e.g. All 52 cards of a bridge deck:
© 1999 Prentice-Hall, Inc. Chap. 4 - 4
Events
• Simple Event: Outcome from a Sample Space
with 1 Characteristic
e.g. A Red Card from a deck of cards.
• Joint or Compound Event: Involves 2 Outcomes
Simultaneously
e.g. An Ace which is also a Red Card from a
deck of cards.
An Ace given that it is a Red Card.
© 1999 Prentice-Hall, Inc. Chap. 4 - 5
Visualizing Events
•Contingency Tables
•Tree Diagrams
Ace Not Ace Total
Red 2 24 26
Black 2 24 26
Total 4 48 52
© 1999 Prentice-Hall, Inc. Chap. 4 - 6
Simple Events
The Event of a Happy Face
There are 5 happy faces in this collection of 18 objects
© 1999 Prentice-Hall, Inc. Chap. 4 - 7
Joint Events
The Event of a Happy Face AND Light Colored
3 Happy Faces which are light in color
© 1999 Prentice-Hall, Inc. Chap. 4 - 8
12 Items, 5 happy faces and 7 other light objects
Compound Events
The Event of Happy Face OR Light Colored
© 1999 Prentice-Hall, Inc. Chap. 4 - 9
Special Events
Null Event
Club & Diamond on
1 Card Draw
Complement of Event
For Event A,
All Events Not In A: A’
Null Event
© 1999 Prentice-Hall, Inc. Chap. 4 - 10
3 Items, 3 Happy Faces Given they are Light Colored
Dependent or
Independent Events The Event of a Happy Face GIVEN it is Light Colored
E = Happy Face Light Color
© 1999 Prentice-Hall, Inc. Chap. 4 - 11
Contingency Table
A Deck of 52 Cards
Ace Not an
Ace Total
Red
Black
Total
2 24
2 24
26
26
4 48 52
Sample Space
Red Ace
© 1999 Prentice-Hall, Inc. Chap. 4 - 12
Tree Diagram
Event Possibilities
Red
Cards
Black
Cards
Ace
Not an Ace
Ace
Not an Ace
Full Deck
of Cards
© 1999 Prentice-Hall, Inc. Chap. 4 - 13
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
events is 1.
Certain
Impossible
.5
1
0
© 1999 Prentice-Hall, Inc. Chap. 4 - 14
Computing Probability
•The Probability of an Event, E:
•Each of the Outcome in the Sample Space
equally likely to occur.
SpaceSampleinOutcomesTotal
OutcomesEventofNumber)E(P
T
X
e.g. P( ) = 2/36
(There are 2 ways to get one 6 and the other 4)
© 1999 Prentice-Hall, Inc. Chap. 4 - 15
Computing
Joint Probability
The Probability of a Joint Event, A and B:
e.g. P(Red Card and Ace) =
CardsofNumberTotal
AcesdRe
52
2
26
1
P(A and B) =
Number of Event Outcomes from both A and B
Total Outcomes in Sample Space
© 1999 Prentice-Hall, Inc. Chap. 4 - 16
P(A2 and B2)
P(A1 and B2)
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(A2)
P(B1) P(B2)
© 1999 Prentice-Hall, Inc. Chap. 4 - 17
Computing
Compound Probability
The Probability of a Compound Event, A or B:
SpaceSampleinOutcomesTotal
BorAeitherfromOutcomesEventofNumber)BorA(P
e.g.
P(Red Card or Ace)
CardsNumberofTotal
AcesdReCardsdReAces
52
2264
13
7
52
28
© 1999 Prentice-Hall, Inc. Chap. 4 - 18
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)
For Mutually Exclusive Events: P(A or B) = P(A) + P(B)
© 1999 Prentice-Hall, Inc. Chap. 4 - 19
Computing
Conditional Probability
The Probability of the Event:
Event A given that Event B has occurred
P(A B) =
e.g.
P(Red Card given that it is an Ace) =
)B(P
)BandA(P
2
1
4
2
Aces
AcesdRe
© 1999 Prentice-Hall, Inc. Chap. 4 - 20
Black Color
Type Red Total
Ace 2 2 4
Non-Ace 24 24 48
Total 26 26 52
Conditional Probability
Using Contingency Table
Conditional Event: Draw 1 Card. Note Kind & Color
26
2
5226
522
/
/
P(Red)
Red)AND P(Ace = Red) |P(Ace
Revised
Sample
Space
© 1999 Prentice-Hall, Inc. Chap. 4 - 21
Conditional Probability and
Statistical Independence
)B(P
)BandA(PConditional Probability: P(AB) =
P(A and B) = P(A B) • P(B)
Events are
Independent:
P(A B) = P(A)
Or, P(A and B) = P(A) • P(B)
Events A and B are Independent when the probability
of one event, A is not affected by another event, B.
Multiplication Rule:
© 1999 Prentice-Hall, Inc. Chap. 4 - 22
Bayes’ Theorem
)B(P)BA(P)B(P)BA(P
)B(P)BA(P
kk
ii
11
)A(P
)AandB(P i
P(Bi A) =
Adding up
the parts
of A in all
the B’s Same
Event
© 1999 Prentice-Hall, Inc. Chap. 4 - 23
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) = 08.
)DefaultandCollege(P)payReandCollege(P
)payReandCollege(P
? ? ?
? ?
© 1999 Prentice-Hall, Inc. Chap. 4 - 24
Discrete Random Variable
• Random Variable: represents outcomes of an
experiment.
e.g. Throw a die twice:
Count the number of times 4 comes up (0, 1, or 2 times)
• Discrete Random Variable:
• Obtained by Counting (0, 1, 2, 3, etc.)
• Usually finite by number of values
e.g. Toss a coin 5 times. Count the number of tails.
(0, 1, 2, 3, 4, or 5 times)
© 1999 Prentice-Hall, Inc. Chap. 4 - 25
Discrete Probability
Distribution Example
Probability Distribution
Values Probability
0 1/4 = .25
1 2/4 = .50
2 1/4 = .25
Event: Toss 2 Coins. Count # Tails.
T
T
T T
© 1999 Prentice-Hall, Inc. Chap. 4 - 26
Discrete
Probability Distribution
• List of All Possible [ Xi, P(Xi) ] Pairs
Xi = Value of Random Variable (Outcome)
P(Xi) = Probability Associated with Value
• Mutually Exclusive (No Overlap)
• Collectively Exhaustive (Nothing Left
Out) 0 P(Xi) 1
S P(Xi) = 1
© 1999 Prentice-Hall, Inc. Chap. 4 - 27
Discrete Random Variable
Summary Measures
Expected Value
The Mean of the Probability Distribution Weighted Average
m = E(X) = Xi P(Xi)
e.g. Toss 2 coins, Count tails, Compute Expected Value:
m = 0 .25 + 1 .50 + 2 .25 = 1
Variance
Weighted Average Squared Deviation about Mean
s2 = E [ (Xi - m )2]=S (Xi - m )2P(Xi)
e.g. Toss 2 coins, Count tails, Compute Variance:
s2 = (0 - 1)2(.25) + (1 - 1)2(.50) + (2 - 1)2(.25) = .50
Number of Tails
© 1999 Prentice-Hall, Inc. Chap. 4 - 28
Important Discrete Probability
Distribution Models
Discrete Probability
Distributions
Binomial Hypergeometric Poisson
© 1999 Prentice-Hall, Inc. Chap. 4 - 29
Binomial Probability Distributions
• ‘n’ Identical Trials, e.g. 15 tosses of a coin,
10 light bulbs taken from a warehouse
• 2 Mutually Exclusive Outcomes,
e.g. heads or tails in each toss of a coin,
defective or not defective light bulbs
• 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
© 1999 Prentice-Hall, Inc. Chap. 4 - 30
Binomial Probability
Distributions
• 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 Another
© 1999 Prentice-Hall, Inc. Chap. 4 - 31
Binomial Probability
Distribution Function
P(X) = probability that X successes given a knowledge of n
and p
X = number of ‘successes’ in
sample, (X = 0, 1, 2, ..., n)
p = probability of ‘success’
n = sample size
P(X) n
X ! n X p p
X n X !
( ) ! ( )
1
Tails in 2 Toss of Coin
X P(X)
0 1/4 = .25
1 2/4 = .50
2 1/4 = .25
© 1999 Prentice-Hall, Inc. Chap. 4 - 32
Binomial Distribution
Characteristics
n = 5 p = 0.1
n = 5 p = 0.5
Mean
Standard Deviation
m
s
E X np
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. m = 5 (.1) = .5
e.g. s = 5(.5)(1 - .5)
= 1.118
0
© 1999 Prentice-Hall, Inc. Chap. 4 - 33
Poisson Distribution
Poisson Process:
• Discrete Events in an ‘Interval’
The Probability of One Success in
Interval is Stable
The Probability of More than One
Success in this Interval is 0
• Probability of Success is
Independent from Interval to
Interval
e.g. # Customers Arriving in 15 min.
# Defects Per Case of Light
Bulbs.
P X x
x
x
( |
!
l
l l e
-
© 1999 Prentice-Hall, Inc. Chap. 4 - 34
Poisson Probability
Distribution Function
P(X ) = probability of X successes given l
l = expected (mean) number of ‘successes’
e = 2.71828 (base of natural logs)
X = number of ‘successes’ per unit
P X X
X
( ) !
l l
e
e.g. Find the probability of 4
customers arriving in 3 minutes
when the mean is 3.6.
P(X) = e -3.6
3.6
4!
4
= .1912
© 1999 Prentice-Hall, Inc. Chap. 4 - 35
Poisson Distribution
Characteristics
l = 0.5
l = 6
Mean
Standard Deviation
m l
s l
i i
N
i
E X
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)
© 1999 Prentice-Hall, Inc. Chap. 4 - 36
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
© 1999 Prentice-Hall, Inc. Chap. 4 - 37
Hypergeometric Distribution
P(X) = probability that X successes given n, N, and A
n = sample size
N = population size
A = number of “successes”
in population
X = number of “successes”
in sample (X = 0, 1, 2, ..., n)
P X ) ( )( ) A
X (
( )
N - A
n - X
N
n
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?
P(2) = ( )( ) 4
2
6
1 10
3 ( )
= .30
(
© 1999 Prentice-Hall, Inc. Chap. 4 - 38
Hypergeometric Characteristics
Mean
Standard Deviation
m
s
E X n
nA N A
( )
N n
N 1
Finite
Population
Correction
A
N
N 2
)(
© 1999 Prentice-Hall, Inc. Chap. 4 - 39
Covariance
X = discrete random variable X
Xi = ith outcome of X
P(XiYi) = probability of occurrence of the
ith outcome of Y
Y = discrete random variable Y
Yi = ith outcome of Y
i = 1, 2, …, N
)YX(P)Y(EY)X(EX iii
N
iiXY
1
s
© 1999 Prentice-Hall, Inc. Chap. 4 - 40
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
E(X) = mX = (-100)(.2) + (100)(.5) + (250)(.3) = $105
E(Y) = mY = (-200)(.2) + (50)(.5) + (350)(.3) = $90
© 1999 Prentice-Hall, Inc. Chap. 4 - 41
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
Var(X) = = (.2)(-100 -105)2 + (.5)(100 - 105)2 + (.3)(250 - 105)2
= 14,725, sX = 121.35
Var(Y) = = (.2)(-200 - 90)2 + (.5)(50 - 90)2 + (.3)(350 - 90)2
= 37,900, sY = 194.68
2Xs
2Ys
© 1999 Prentice-Hall, Inc. Chap. 4 - 42
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
sXY = (.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
strongly related and will vary together in the same direction.
© 1999 Prentice-Hall, Inc. Chap. 4 - 43
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
•Discussed Binomial, Poisson, and Hypergeometric Distributions
• Addressed Covariance and its Applications in Finance