Title: Basic Probability And Probability Distributions
1Business Statistics
Basic Probability And Probability Distributions
2Chapter 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
3Chapter Topics
 Binomial and Poisson Distributions
 Covariance and its Applications in Finance
 The Normal Distribution
 Assessing the Normality Assumption
4Sample Spaces
Collection of all Possible Outcomes
e.g. All 6 faces of a die e.g. All 52 cards
of a bridge deck
5Events
 Simple Event Outcome from a Sample Space
 with 1 Characteristic
 e.g. A Red Card from a deck of cards.
 Joint Event Involves 2 Outcomes
Simultaneously  e.g. An Ace which is also a Red Card from a
 deck of cards.

6Visualizing Events
 Contingency Tables
 Tree Diagrams
Ace Not Ace Total
Black 2 24 26
Red 2 24 26
Total 4 48 52
7Simple Events
The Event of a Happy Face
There are 5 happy faces in this collection of 18
objects
8Joint Events
The Event of a Happy Face AND Light Colored
3 Happy Faces which are light in color
9Special Events
Null Event
 Null event
 Club diamond on 1 card draw
 Complement of event
 For event A,
 All events not In A
10Dependent or Independent Events
The Event of a Happy Face GIVEN it is Light
Colored
E Happy Face???Light Color
3 Items 3 Happy Faces Given they are Light
Colored
11Contingency Table
A Deck of 52 Cards
Red Ace
Not an Ace
Total
Ace
Red
2
24
26
Black
2
24
26
Total
4
48
52
Sample Space
12Contingency Table
2500 Employees of Company ABC
Agree
Neutral
Opposed Total
MALE
200
400 1500
900
300
600 1000
FEMALE
100
2500
1000
Total
1200
300
Sample Space
13Tree Diagram
Event Possibilities
Ace
Red Cards
Not an Ace
Full Deck of Cards
Ace
Black Cards
Not an Ace
14Probability
 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
1
.5
Impossible
0
15Computing Probability
 The Probability of an Event, E
 Each of the Outcome in the Sample Space
 equally likely to occur.
Number of Event Outcomes
P(E)
Total Number of Possible Outcomes in the Sample
Space
X
e.g. P( ) 2/36
T
(There are 2 ways to get one 6 and the other 4)
16 Computing Joint Probability
The Probability of a Joint Event, A and B
P(A and B)
Number of Event Outcomes from both A and B
Total Number of Possible Outcomes in Sample Space
e.g. P(Red Card and Ace)
17Joint Probability Using Contingency Table
Event
Total
B1
B2
Event
P(A1 and B1)
P(A1)
A1
P(A1 and B2)
P(A2 and B1)
A2
P(A2 and B2)
P(A2)
Total
1
P(B1)
P(B2)
Marginal (Simple) Probability
Joint Probability
18 Computing Compound Probability
The Probability of a Compound Event, A or B
e.g. P(Red Card or Ace)
19Contingency Table
2500 Employees of Company ABC
Agree
Neutral
Opposed Total
MALE
200
400 1500
900
300
600 1000
FEMALE
100
2500
1000
Total
1200
300
Sample Space
20The pervious table refers to 2500 employees of
ABC Company, classified by gender and 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
21The pervious table refers to 2500 employees of
ABC Company, classified by gender and 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
600/2500 0.24 
22The pervious table refers to 2500 employees of
ABC Company, classified by gender and 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
600/2500 0.24  2. Neutral

23The pervious table refers to 2500 employees of
ABC Company, classified by gender and 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
600/2500 0.24  2. Neutral
300/2500 0.12 

24The pervious table refers to 2500 employees of
ABC Company, classified by gender and 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
600/2500 0.24  2. Neutral
300/2500 0.12  3. Opposed to the proposal, GIVEN that
 the employee selected is a female

25The pervious table refers to 2500 employees of
ABC Company, classified by gender and 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
600/2500 0.24  2. Neutral
300/2500 0.12  3. Opposed to the proposal, GIVEN that
 the employee selected is a female
600/1000 0.60 
26The pervious table refers to 2500 employees of
ABC Company, classified by gender and 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
600/2500 0.24  2. Neutral
300/2500 0.12  3. Opposed to the proposal, GIVEN that
 the employee selected is a female
600/1000 0.60  4. Either a female or opposed to the

proposal 
27The pervious table refers to 2500 employees of
ABC Company, classified by gender and 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
600/2500 0.24  2. Neutral
300/2500 0.12  3. Opposed to the proposal, GIVEN that
 the employee selected is a female
600/1000 0.60  4. Either a female or opposed to the

proposal .. 1000/2500 1000/2500 
600/2500 
1400/2500
0.56
28The pervious table refers to 2500 employees of
ABC Company, classified by gender and 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
600/2500 0.24  2. Neutral
300/2500 0.12  3. Opposed to the proposal, GIVEN that
 the employee selected is a female
600/1000 0.60  4. Either a female or opposed to the

proposal .. 1000/2500 1000/2500 
600/2500 
1400/2500
0.56  5. Are Gender and Opinion (statistically)
independent?
29The pervious table refers to 2500 employees of
ABC Company, classified by gender and 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
600/2500 0.24  2. Neutral
300/2500 0.12  3. Opposed to the proposal, GIVEN that
 the employee selected is a female
600/1000 0.60  4. Either a female or opposed to the

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
of A 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
30Compound ProbabilityAddition Rule
P(A1 or B1 ) P(A1) P(B1)  P(A1 and B1)
Event
Total
B1
B2
Event
P(A1 and B1)
P(A1 and B2)
P(A1)
A1
P(A2 and B2)
P(A2 and B1)
A2
P(A2)
Total
1
P(B2)
P(B1)
For Mutually Exclusive Events P(A or B) P(A)
P(B)
31 Computing Conditional Probability
The Probability of Event A given that Event B has
occurred P(A ?B)
e.g. P(Red Card given that it is an Ace)
32Conditional Probability Using Contingency Table
Conditional Event Draw 1 Card. Note Kind
Color
Color
Type
Total
Black
Red
Revised Sample Space
2
2
4
Ace
24
24
48
NonAce
26
26
52
Total
33 Conditional Probability and Statistical
Independence
Conditional Probability
P(A?B)
Multiplication Rule
P(A and B) P(A ?B) P(B)
P(B ?A) P(A)
34 Conditional Probability and Statistical
Independence (continued)
Events are Independent
P(A ? B) P(A)
Or, P(B ? A) P(B)
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.
35Bayes Theorem
P(Bi ?A)
Adding up the parts of A in all the Bs
Same Event
36 A manufacturer of VCRs purchases a
particular microchip, called the LS24, from
three suppliers Hall Electronics, Spec Sales,
and Crown Components. Thirty percent of the LS24
chips are purchased from Hall, 20 from Spec, and
the remaining 50 from Crown. The manufacturer
has extensive past records for the three
suppliers and knows that there is a 3 chance
that the chips from Hall are defective, a 5
chance that chips from Spec are defective and a
4 chance that chips from Crown are defective.
When LS24 chips arrive at the manufacturer, they
are placed directly into a bin and not inspected
or otherwise identified as to supplier. A worker
selects a chip at random. What
is the probability that the chip is
defective? A worker selects a
chip at random for installation into a VCR and
finds it is defective. What is the probability
that the chip was supplied by Spec Sales?
37Bayes Theorem Contingency Table
What are the chances of repaying a loan,
given a college education?
Loan Status
Prob.
Education
Repay
Default
.2
.05
.25
College
?
?
?
No College
?
?
1
Prob.
P(College and Repay)
?
P(Repay College)
P(College and Repay) P(College and Default)
.20
.80
.25
38Discrete Random Variable
 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)
39Discrete Random Variable
 Discrete Random Variable
 Obtained by Counting (0, 1, 2, 3, etc.)
 Usually finite by number of different values
 e.g.
 Toss a coin 5 times. Count the number of
tails. (0, 1, 2, 3, 4, or 5 times)
40Discrete Probability Distribution Example
Event Toss 2 Coins. Count Tails.
 Probability distribution
 Values probability
 0 1/4 .25
 1 2/4 .50
 2 1/4 .25
T
T
T
T
41Discrete 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 ? p(xi) ? 1
 ? P(xi) 1
42Discrete Random Variable Summary Measures
 Expected value (The mean)
 Weighted average of the probability
distribution  ? E(X) ?xi p(xi)

 E.G. Toss 2 coins, count tails, compute expected
value  ?? 0 ? .25 1 ?.50 2 ? .25 1
Number of Tails
43Discrete Random Variable Summary Measures
 Variance
 Weighted average squared deviation about mean
 ?? E (xi  ? )2? (xi  ? )2p(xi)

 E.G. Toss 2 coins, count tails, compute
variance  ?? (0  1)2(.25) (1  1)2(.50)
(2  1)2(.25) .50
44Important Discrete Probability Distribution Models
Discrete Probability Distributions
Binomial
Poisson
45Binomial 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 
46Binomial 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 Another
47Binomial Probability Distribution Function
n
!
X
X
?
n
P(X)
)
?
1
p
p
(
?
X !
(
)
!
?
n
X
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 each success
n sample size
Tails in 2 Tosses of Coin X
P(X) 0 1/4 .25
1 2/4 .50 2
1/4 .25
48Binomial Distribution Characteristics
n 5 p 0.1
P(X)
Mean
.6
E
X
?
?
(
)
?
np
.4
.2
e.g. ? 5 (.1) .5
0
X
0
1
2
3
4
5
Standard Deviation
n 5 p 0.5
P(X)
)
?
p
?
?
np
(
1
.6
.4
.2
e.g. ? 5(.5)(1  .5) 1.118
X
0
0
1
2
3
4
5
49Poisson Distribution
 Poisson process
 Discrete events in an interval
 The probability of one success in
an 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
50Poisson Distribution Function
X
??
?
e
P
X
(
)
?
X
!
P(X ) probability of X successes given
? ? expected (mean) number of
successes e 2.71828 (base of natural
logs) X number of successes per unit
e.g. Find the probability of 4 customers arriving
in 3 minutes when the mean is 3.6.
3.6
4
e
3.6
P(X)
.1912
4!
51Poisson Distribution Characteristics
Mean
?? 0.5
P(X)
.6
?
?
E
X
?
?
(
)
.4
N
.2
?
?
0
X
X
P
X
(
)
i
i
0
1
2
3
4
5
i
?
1
?? 6
P(X)
.6
Standard Deviation
.4
.2
?
?
?
0
X
0
2
4
6
8
10
52Covariance
 X discrete random variable X
 Xi value of the ith outcome of X
 P(xiyi) probability of occurrence of the ith
outcome of X and ith outcome of Y  Y discrete random variable Y
 Yi value of the ith outcome of Y
 I 1, 2, , N
53Computing the Mean for Investment Returns
Return per 1,000 for two types of investments
Investment
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
E(X) ?X (100)(.2) (100)(.5) (250)(.3)
105 E(Y) ?Y (200)(.2) (50)(.5)
(350)(.3) 90
54Computing the Variance for Investment Returns
Investment
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
Var(X) (.2)(100 105)2 (.5)(100 
105)2 (.3)(250  105)2 14,725,
?X 121.35 Var(Y)
(.2)(200  90)2 (.5)(50  90)2 (.3)(350 
90)2 37,900, ?Y
194.68
55Computing the Covariance for Investment Returns
Investment
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
?XY (.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.
56The Normal Distribution
 Bell Shaped
 Symmetrical
 Mean, Median and
 Mode are Equal
 Middle Spread
 Equals 1.33 ??
 Random Variable has
 Infinite Range
f(X)
X
?
Mean Median Mode
57The Mathematical Model
f(X) frequency of random variable
X ? 3.14159 e 2.71828 ? population
standard deviation X value of random variable
(? lt X lt ?) ? population mean
58Many Normal Distributions
There are an Infinite Number
Varying the Parameters ? and ?, we obtain
Different Normal Distributions.
59Normal Distribution Finding Probabilities
Probability is the area under thecurve!
?
)
?
P
c
X
d
(
?
?
f(X)
X
d
c
60Which Table?
Each distribution has its own table?
Infinitely Many Normal Distributions Means
Infinitely Many Tables to Look Up!
61Solution (I) The Standardized Normal Distribution
Standardized Normal Distribution Table (Portion)
? 0 and ?? 1
Z
Z
.0478
.02
Z
.00
.01
Shaded Area Exaggerated
0.0
.0000
.0040
.0080
.0398
.0438
.0478
0.1
0.2
.0793
.0832
.0871
Z 0.12
0.3
.0179
.0217
.0255
Probabilities
Only One Table is Needed
62Solution (II) The Cumulative Standardized Normal
Distribution
Cumulative Standardized Normal Distribution Table
(Portion)
.5478
.02
Z
.00
.01
0.0
.5000
.5040
.5080
Shaded Area Exaggerated
.5398
.5438
.5478
0.1
0.2
.5793
.5832
.5871
Z 0.12
0.3
.5179
.5217
.5255
Probabilities
Only One Table is Needed
63Standardizing Example
Normal Distribution
Standardized Normal Distribution
10
?
1
?
Z
?
6.2
?
0
.12
5
Z
X
Shaded Area Exaggerated
64ExampleP(2.9 lt X lt 7.1) .1664
Normal Distribution
Standardized Normal Distribution
?
?
1
10
Z
.1664
.0832
.0832
Z
0
.21
.21
5
2.9
7.1
X
Shaded Area Exaggerated
65Example P(X ? 8) .3821
.
Normal Distribution
Standardized Normal Distribution
?
1
?
10
.5000
.3821
.1179
?
0
.30
Z
X
?
5
8
Shaded Area Exaggerated
66Finding Z Values for Known Probabilities
What Is Z Given Probability 0.1217?
Standardized Normal Probability Table (Portion)
.01
?
1
Z
.00
0.2
.1217
0.0
.0040
.0000
.0080
0.1
.0398
.0438
.0478
0.2
.0793
.0832
.0871
?
0
.31
Z
.1179
.1255
.1217
0.3
Shaded Area Exaggerated
67Recovering X Values for Known Probabilities
Normal Distribution
Standardized Normal Distribution
?
1
?
10
.1217
.1217
?
Z
?
0
.31
X
?
5
X
8.1
?????? Z?? 5 (0.31)(10)
Shaded Area Exaggerated
68Assessing 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
69Assessing Normality
Normal Probability Plot for Normal Distribution
90
X
60
Z
30
2
1
0
1
2
Look for Straight Line!
70Normal Probability Plots
LeftSkewed
RightSkewed
90
90
60
X
60
X
Z
Z
30
30
2
1
0
1
2
2
1
0
1
2
Rectangular
UShaped
90
90
60
X
60
X
Z
Z
30
30
2
1
0
1
2
2
1
0
1
2
71Chapter 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
72 Summary
 Discussed Binomial and Poisson Distributions
 Addressed Covariance and its Applications in
Finance  Covered Normal Distribution
 Discussed Assessing the Normality Assumption
