1. Basic Probability And Probability Distributions
Business Statistics
A First Course
(3rd Edition)
Chapter 4
Basic Probability And Probability Distributions

2. Chapter Topics Sample Spaces and Events, Simple
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 Covariance and its Applications in Finance
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 Black 2 24 26
Ace Not Ace Total
Black
Red
Total

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 Club & diamond on 1 card draw
Complement of event
For event A,
All events not In A:
Null Event 

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 Total Ace Red 2 24 26
Not an
Ace
Total
Ace
Red 2 24 26
Black 2 24 26
Total 4 48 52
Sample Space

12. 2500 Employees of Company ABC
Contingency Table
2500 Employees of Company ABC
Agree
Neutral
Opposed | Total
MALE
200 |
900
300 |
FEMALE
100
2500 |
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. 1
Certain .5
Impossible

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
Event B1 B2
Total A1
P(A1 and B1)
P(A1 and B2)
P(A1) A2
P(A2 and B1)
P(A2 and B2)
P(A2)
Total
P(B1)
P(B2) 1
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. 2500 Employees of Company ABC
Contingency Table
2500 Employees of Company ABC
Agree
Neutral
Opposed | Total
MALE
200 |
900
300 |
FEMALE
100
2500 |
Total
1200
300
Sample Space

20. 1. A female opposed to the proposal
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. 1. A female opposed to the proposal 600/2500 = 0.24
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 /2500 = 0.24

22. 1. A female opposed to the proposal 600/2500 = 0.24 2. Neutral
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 /2500 = 0.24
2. Neutral

23. 1. A female opposed to the proposal 600/2500 = 0.24
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 /2500 = 0.24
2. Neutral /2500 = 0.12

24. 1. A female opposed to the proposal 600/2500 = 0.24
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 /2500 = 0.24
2. Neutral /2500 = 0.12
3. Opposed to the proposal, GIVEN that
the employee selected is a female

25. 1. A female opposed to the proposal 600/2500 = 0.24
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 /2500 = 0.24
2. Neutral /2500 = 0.12
3. Opposed to the proposal, GIVEN that
the employee selected is a female /1000 = 0.60

26. 1. A female opposed to the proposal 600/2500 = 0.24
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 /2500 = 0.24
2. Neutral /2500 = 0.12
3. Opposed to the proposal, GIVEN that
the employee selected is a female /1000 = 0.60
4. Either a female or opposed to the
proposal

27. 1. A female opposed to the proposal 600/2500 = 0.24
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 /2500 = 0.24
2. Neutral /2500 = 0.12
3. Opposed to the proposal, GIVEN that
the employee selected is a female /1000 = 0.60
4. Either a female or opposed to the
proposal ……… / / /2500 =
1400/ = 0.56

28. 1. A female opposed to the proposal 600/2500 = 0.24
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 /2500 = 0.24
2. Neutral /2500 = 0.12
3. Opposed to the proposal, GIVEN that
the employee selected is a female /1000 = 0.60
4. Either a female or opposed to the
proposal ……… / / /2500 =
1400/ = 0.56
5. Are Gender and Opinion (statistically) independent?

29. 1. A female opposed to the proposal 600/2500 = 0.24
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 /2500 = 0.24
2. Neutral /2500 = 0.12
3. Opposed to the proposal, GIVEN that
the employee selected is a female /1000 = 0.60
4. Either a female or opposed to the
proposal ……… / / /2500 =
1400/ = 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 Probability Addition Rule
P(A1 or B1 ) = P(A1) +P(B1) - P(A1 and B1)
Event
Event B1 B2
Total A1
P(A1 and B1)
P(A1 and B2)
P(A1) A2
P(A2 and B1)
P(A2 and B2)
P(A2)
Total
P(B1)
P(B2) 1
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
Red
Black
Total
Revised Sample Space
Ace 2 2 4
Non-Ace 24 24 48
Total 26 26 52

33. Conditional Probability and Statistical Independence
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 Adding up the parts of A in all the B’s P(Bi A) =
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
Education
Repay
Default
Prob.
College .2
.05
.25 ? ? ?
No College ? ?
Prob. 1
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
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
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! X n  X
P(X)  p ( 1  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)
/4 = .25
/4 = .50
/4 = .25

48. Binomial Distribution Characteristics
n = 5 p = 0.1
Mean
P(X) .6   E ( X )  np .4 .2
e.g.  = 5 (.1) = .5 X 1 2 3 4 5
Standard Deviation
n = 5 p = 0.5   )
P(X) np ( 1  p .6 .4
e.g.  = 5(.5)(1 - .5) = 1.118 .2 X 1 2 3 4 5

49. Poisson Distribution P X x ( | !   e 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 P X x ( | !   e -

50. Poisson Distribution Function X e  P ( X )  X !
P(X ) = probability of X successes given 
 = expected (mean) number of ‘successes’
e = (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)   E ( X )   .6 .4 N .2   X P ( X ) X i i 1 2 3 4 5 i  1
= 6
P(X)
Standard Deviation .6 .4    .2 X 2 4 6 8 10

52. Covariance Xi = value of the ith outcome of X
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 $1,000 for two types of investments
Investment
P(XiYi) Economic condition Dow Jones fund X Growth Stock Y
Recession -$ $200
Stable Economy
Expanding Economy
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
P(XiYi) Economic condition Dow Jones fund X Growth Stock Y
Recession -$ $200
Stable Economy
Expanding Economy
Var(X) = = (.2)( )2 + (.5)( )2 + (.3)( )2
= 14,725, X =
Var(Y) = = (.2)( )2 + (.5)( )2 + (.3)( )2
= 37,900, Y =

55. Computing the Covariance for Investment Returns
P(XiYi) Economic condition Dow Jones fund X Growth Stock Y
Recession -$ $200
Stable Economy
Expanding Economy
XY = (.2)( )( ) + (.5)( )( )
+ (.3)( )( ) = 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
 = ; e =
 = population standard deviation
X = value of random variable (- < X < )
 = 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 the curve! ? )  P ( c  X  d
f(X) X c d

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
Table (Portion)  = 0 and  = 1 Z Z
.0478
.02 Z
.00
.01
0.0
.0000
.0040
.0080
Shaded Area Exaggerated
0.1
.0398
.0438
.0478
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
0.1
.5398
.5438
.5478
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 
= 5
6.2 X 
= 0
.12 Z
Shaded Area Exaggerated

64. Example: P(2.9 < X < 7.1) = .1664
Normal Distribution
Standardized Normal Distribution 
= 10 
= 1 Z
.1664
.0832
.0832
2.9 5
7.1 X
-.21
.21 Z
Shaded Area Exaggerated

65. Example: P(X  8) = .3821  = 10  = 1  = 5 8 X  = 0 .30 Z .3821 .
Normal Distribution
Standardized Normal Distribution 
= 10 
= 1
.5000
.3821
.1179 
= 5 8 X 
= 0
.30 Z
Shaded Area Exaggerated

66. Finding Z Values for Known Probabilities
What Is Z Given Probability = ?
Standardized Normal Probability Table (Portion) 
= 1
.01 Z
.00
0.2
.1217
0.0
.0000
.0040
.0080
0.1
.0398
.0438
.0478 
= 0
.31 Z
0.2
.0793
.0832
.0871
0.3
.1179
.1217
.1255
Shaded Area Exaggerated

67. Recovering X Values for Known Probabilities
Normal Distribution
Standardized Normal Distribution 
= 10 
= 1
.1217
.1217 ? 
= 5 X 
= 0
.31 Z X
 Z= 5 + (0.31)(10) =
8.1
Shaded Area Exaggerated

68. Compare Data Characteristics to Properties of Normal Distribution
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. Normal Probability Plot for Normal Distribution
Assessing Normality
Normal Probability Plot for Normal Distribution 90 X 60 30 Z -2 -1 1 2
Look for Straight Line!

70. Normal Probability Plots
Left-Skewed
Right-Skewed 90 90 X 60 X 60 30 Z 30 Z -2 -1 1 2 -2 -1 1 2
Rectangular
U-Shaped 90 90 X 60 X 60 30 Z 30 Z -2 -1 1 2 -2 -1 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. Chapter Summary Discussed Binomial and Poisson Distributions
Addressed Covariance and its Applications in Finance
Covered Normal Distribution
Discussed Assessing the Normality Assumption