Basic Probability And Probability Distributions

1
Business Statistics
Basic Probability And Probability Distributions
2
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

3
Chapter Topics

• Binomial and Poisson Distributions
• Covariance and its Applications in Finance
• The Normal Distribution
• Assessing the Normality Assumption

4
Sample Spaces
Collection of all Possible Outcomes
e.g. All 6 faces of a die e.g. All 52 cards
of a bridge deck
5
Events

• 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.

6
Visualizing Events

• Contingency Tables
• Tree Diagrams

Ace Not Ace Total
Black 2 24 26
Red 2 24 26
Total 4 48 52

7
Simple Events
The Event of a Happy Face
There are 5 happy faces in this collection of 18
objects
8
Joint Events
The Event of a Happy Face AND Light Colored
3 Happy Faces which are light in color
9
Special Events
Null Event

• Null event
• Club diamond on 1 card draw
• Complement of event
• For event A,
• All events not In A

10
Dependent 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
11
Contingency 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
12
Contingency 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
13
Tree Diagram
Event Possibilities
Ace
Red Cards
Not an Ace
Full Deck of Cards
Ace
Black Cards
Not an Ace
14
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
1
.5
Impossible
0
15
Computing 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)

17
Joint 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)
19
Contingency 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
20
The 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

21
The 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

22
The 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

23
The 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

24
The 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
  

25
The 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

26
The 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

27
The 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

28
The 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?

29
The 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

30
Compound 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)
32
Conditional 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
Non-Ace
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.
35
Bayes 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 LS-24, from
three suppliers Hall Electronics, Spec Sales,
and Crown Components. Thirty percent of the LS-24
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 LS-24 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?

   
37
Bayes 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
38
Discrete 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)

39
Discrete 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)

40
Discrete 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
41
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 ? p(xi) ? 1
• ? P(xi) 1

42
Discrete 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
43
Discrete 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

44
Important Discrete Probability Distribution Models
Discrete Probability Distributions
Binomial
Poisson
45
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

46
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 Another

47
Binomial 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
48
Binomial 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
49
Poisson 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

50
Poisson 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!
51
Poisson 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
52
Covariance

• 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

53
Computing 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
54
Computing 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
55
Computing 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.
56
The 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
57
The 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
58
Many Normal Distributions
There are an Infinite Number
Varying the Parameters ? and ?, we obtain
Different Normal Distributions.
59
Normal Distribution Finding Probabilities

Probability is the area under thecurve!
?
)
?
P
c
X
d
(
?
?
f(X)
X
d
c
60
Which Table?
Each distribution has its own table?
Infinitely Many Normal Distributions Means
Infinitely Many Tables to Look Up!
61
Solution (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
62
Solution (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
63
Standardizing Example
Normal Distribution
Standardized Normal Distribution
10
?
1
?
Z
?
6.2
?
0
.12
5
Z
X
Shaded Area Exaggerated
64
ExampleP(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
65
Example P(X ? 8) .3821
.
Normal Distribution
Standardized Normal Distribution
?
1
?
10
.5000
.3821
.1179
?
0
.30
Z
X
?
5
8
Shaded Area Exaggerated
66
Finding 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
67
Recovering 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
68
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

69
Assessing Normality
Normal Probability Plot for Normal Distribution
90
X
60
Z
30
-2
-1
0
1
2
Look for Straight Line!
70
Normal Probability Plots
Left-Skewed
Right-Skewed
90
90
60
X
60
X
Z
Z
30
30
-2
-1
0
1
2
-2
-1
0
1
2
Rectangular
U-Shaped
90
90
60
X
60
X
Z
Z
30
30
-2
-1
0
1
2
-2
-1
0
1
2
71
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

72
Summary

• Discussed Binomial and Poisson Distributions
• Addressed Covariance and its Applications in
  Finance
• Covered Normal Distribution
• Discussed Assessing the Normality Assumption