Probability

1. Chap 4-1
Chapter 4
Using Probability and
Probability Distributions

2. Chap 4-2
Chapter Goals
After completing this chapter, you should be
able to:
 Explain three approaches to assessing
probabilities
 Apply common rules of probability
 Use Bayes’ Theorem for conditional probabilities
 Distinguish between discrete and continuous
probability distributions
 Compute the expected value and standard
deviation for a discrete probability distribution

3. Chap 4-3
Important Terms
 Probability – the chance that an uncertain event
will occur (always between 0 and 1)
 Experiment – a process of obtaining outcomes
for uncertain events
 Elementary Event – the most basic outcome
possible from a simple experiment
 Sample Space – the collection of all possible
elementary outcomes

4. Chap 4-4
Examples for Experiment
 Recording opinions of the customers
 Tossing a coin
 Choosing a chairperson
 Conduct a sales call
 Play a football game
 Roll a dice
 Invest money on business
 Getting a contract
 Getting returns from investment

5. Chap 4-5
Events
 Events
Outcomes of an experiment are called
events
(or )
Results of an experiment are called events

6. Chap 4-6
Examples for Event
 From the example of experiment choosing a
chairperson of 10 persons each person can be
denoted by an event.
 In tossing of a coin experiment head is an event
and tail is another event
 In a departmental store purchasing by a
customer is an event and not purchasing by the
customer is also another event
 From the example of getting returns from the
investment the each possible outcome can be
denoted by event. After investment outcome

7. Chap 4-7
Sample Space
The Sample Space is the collection of all
possible outcomes
e.g. All 6 faces of a die:
e.g. All 52 cards of a bridge deck:

8. Chap 4-8
Probability Concepts
 Mutually Exclusive Events
 If E1 occurs, then E2 cannot occur
 E1 and E2 have no common elements
Black
Cards
Red
Cards
A card cannot be
Black and Red at
the same time.
E1
E2

9. Chap 4-9
 Independent and Dependent Events
 Independent: Occurrence of one does not
influence the probability of
occurrence of the other
 Dependent: Occurrence of one affects the
probability of the other
Probability Concepts

10. Chap 4-10
 Independent Events
E1 = heads on one flip of fair coin
E2 = heads on second flip of same coin
Result of second flip does not depend on the result of
the first flip.
 Dependent Events
E1 = rain forecasted on the news
E2 = take umbrella to work
Probability of the second event is affected by the
occurrence of the first event
Independent vs. Dependent Events

11. Chap 4-11
Assigning Probability
 Classical Probability Assessment
 Relative Frequency of Occurrence
 Subjective Probability Assessment
P(Ei) =
Number of ways Ei can occur
Total number of elementary events
Relative Freq. of Ei =
Number of times Ei occurs
N
An opinion or judgment by a decision maker about
the likelihood of an event

12. A Counting Rule for
Multiple-Step Experiments
 If an experiment consists of a sequence of k
steps in which there are n1 possible results for
the first step, n2 possible results for the
second step, and so on, then the total number
of experimental outcomes is given by (n1)(n2)
. . . (nk).
 A helpful graphical representation of a
multiple-step experiment is a tree diagram.

13. 13Slide
Example: Bradley Investments
Bradley has invested in two stocks, Markley Oil and
Collins Mining. Bradley has determined that the
possible outcomes of these investments three months
from now are as follows.
Investment Gain or Loss
in 3 Months (in $000)
Markley Oil Collins Mining
10 8
5 -2
0
-20

14. 14Slide
Example: Bradley Investments
 A Counting Rule for Multiple-Step Experiments
Bradley Investments can be viewed as a two-step
experiment; it involves two stocks, each with a set of
experimental outcomes.
Markley Oil: n1 = 4
Collins Mining: n2 = 2
Total Number of
Experimental Outcomes: n1n2 = (4)(2) = 8

15. 15Slide
Example: Bradley Investments
 Tree Diagram
Markley Oil Collins Mining Experimental
(Stage 1) (Stage 2) Outcomes
Gain 5
Gain 8
Gain 8
Gain 10
Gain 8
Gain 8
Lose 20
Lose 2
Lose 2
Lose 2
Lose 2
Even
(10, 8) Gain $18,000
(10, -2) Gain $8,000
(5, 8) Gain $13,000
(5, -2) Gain $3,000
(0, 8) Gain $8,000
(0, -2) Lose $2,000
(-20, 8) Lose $12,000
(-20, -2) Lose $22,000

16. Chap 4-16
Visualizing Events
 A automobile consultant records fuel type and
vehicle type for a sample of vehicles
2 Fuel types: Gasoline, Diesel
3 Vehicle types: Truck, Car, SUV
6 possible elementary events:
e1 Gasoline, Truck
e2 Gasoline, Car
e3 Gasoline, SUV
e4 Diesel, Truck
e5 Diesel, Car
e6 Diesel, SUV
Car
Car
e1
e2
e3
e4
e5
e6

17. Chap 4-17
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 52 Cards
Sample
Space
Sample
Space2
24
2
24

18. 18Slide
Another useful counting rule enables us to count the
number of experimental outcomes when n objects are to
be selected from a set of N objects.
 Number of combinations of N objects taken n at a
time
where N! = N(N - 1)(N - 2) . . . (2)(1)
n! = n(n - 1)( n - 2) . . . (2)(1)
0! = 1
Counting Rule for Combinations
C
N
n
N
n N n
n
N






 

!
!( )!

19. 19Slide
Counting Rule for Permutations
A third useful counting rule enables us to count the
number of experimental outcomes when n objects are to
be selected from a set of N objects where the order of
selection is important.
 Number of permutations of N objects taken n at a
time
P n
N
n
N
N n
n
N






 

!
!
( )!

20. 20Slide
Classical Method
If an experiment has n possible outcomes, this method
would assign a probability of 1/n to each outcome.
 Example
Experiment: Rolling a die
Sample Space: S = {1, 2, 3, 4, 5, 6}
Probabilities: Each sample point has a 1/6 chance
of occurring.

21. 21Slide
Example: Lucas Tool Rental
 Relative Frequency Method
Lucas would like to assign probabilities to the
number of floor polishers it rents per day. Office
records show the following frequencies of daily rentals
for the last 40 days.
Number of Number
Polishers Rented of Days
0 4
1 6
2 18
3 10
4 2

22. 22Slide
 Relative Frequency Method
The probability assignments are given by dividing
the number-of-days frequencies by the total frequency
(total number of days).
Number of Number
Polishers Rented of Days Probability
0 4 .10 = 4/40
1 6 .15 = 6/40
2 18 .45 etc.
3 10 .25
4 2 .05
40 1.00
Example: Lucas Tool Rental

23. 23Slide
Subjective Method
 When economic conditions and a company’s
circumstances change rapidly it might be
inappropriate to assign probabilities based solely on
historical data.
 We can use any data available as well as our
experience and intuition, but ultimately a probability
value should express our degree of belief that the
experimental outcome will occur.
 The best probability estimates often are obtained by
combining the estimates from the classical or relative
frequency approach with the subjective estimates.

24. 24Slide
Example: Bradley Investments
Applying the subjective method, an analyst
made the following probability assignments.
Exper. Outcome Net Gain/Loss Probability
( 10, 8) $18,000 Gain .20
( 10, -2) $8,000 Gain .08
( 5, 8) $13,000 Gain .16
( 5, -2) $3,000 Gain .26
( 0, 8) $8,000 Gain .10
( 0, -2) $2,000 Loss .12
(-20, 8) $12,000 Loss .02
(-20, -2) $22,000 Loss .06

25. 25Slide
Example: Bradley Investments
 Events and Their Probabilities
Event M = Markley Oil Profitable
M = {(10, 8), (10, -2), (5, 8), (5, -2)}
P(M) = P(10, 8) + P(10, -2) + P(5, 8) + P(5, -2)
= .2 + .08 + .16 + .26
= .70
Event C = Collins Mining Profitable
P(C) = .48 (found using the same logic)

26. 26Slide
 The union of events A and B is the event containing
all sample points that are in A or B or both.
 The union is denoted by A B
 The union of A and B is illustrated below.
Sample Space S
Event A Event B
Union of Two Events

27. Business Statistics: A Decision-Making Approach, 6e © 2005 Prentice-Hall, Inc. Chap 4-27
 A credit card customer at BigBazar can use Visa (V),
MasterCard (M), or American Express (A). The
merchandise may be books (B), Electronic Media (E), or
other (O).
 a) Define the Experiment
 b) Enumerate the elementary events in the sample
space related to the experiment
 c) would each elementary event be equally likely

28. Chap 4-28
 A survey asked tax accounting firms their
business from ( S= sole proprietorship, P
=Partnership, C= corporation) and type of risk
insurance they carry ( L = liability only, T =
property loss only, B = both liability and
property).
 1. Enumerate the Elementary events in the
sample space
 2. Would these elementary events in the
sample space be equally likely ? Explain

29. Chap 4-29
Rules of Probability
Rules for
Possible Values
and Sum
Individual Values Sum of All Values
0 ≤ P(ei) ≤ 1
For any event ei
1)P(e
k
1i
i 
where:
k = Number of elementary events
in the sample space
ei = ith elementary event

30. Chap 4-30
Addition Rule for Elementary Events
 The probability of an event Ei is equal to the
sum of the probabilities of the elementary
events forming Ei.
 That is, if:
Ei = {e1, e2, e3}
then:
P(Ei) = P(e1) + P(e2) + P(e3)

31. Chap 4-31
Complement Rule
 The complement of an event E is the collection of
all possible elementary events not contained in
event E. The complement of event E is
represented by E.
 Complement Rule:
P(E)1)EP(  E
E
1)EP(P(E) Or,

32. Chap 4-32
Addition Rule for Two Events
P(E1 or E2) = P(E1) + P(E2) - P(E1 and E2)
E1 E2
P(E1 or E2) = P(E1) + P(E2) - P(E1 and E2)
Don’t count common
elements twice!
■ Addition Rule:
E1 E2+ =

33. Chap 4-33
Addition Rule Example
P(Red or Ace) = P(Red) +P(Ace) - P(Red and Ace)
= 26/52 + 4/52 - 2/52 = 28/52
Don’t count
the two red
aces twice!
Black
Color
Type Red Total
Ace 2 2 4
Non-Ace 24 24 48
Total 26 26 52

34. Chap 4-34
Addition Rule for
Mutually Exclusive Events
 If E1 and E2 are mutually exclusive, then
P(E1 and E2) = 0
So
P(E1 or E2) = P(E1) + P(E2) - P(E1 and E2)
= P(E1) + P(E2)
E1 E2

35. The employees of a certain company have
elected 5 of their number to represent them on the
employee-management productivity council. Profiles of
the 5 are as follows:
Gender Age
Male 30
Male 32
Female 45
Female 20
Male 40
This group decides to elect a spokesperson by
drawing a name from a chit. What is the probability
the spokesperson will be either female or over 35?

36. An inspector of the Alaska pipeline has the task of
comparing the reliability of two pumping stations.
Each station is susceptible to two kinds of failure:
pump failure and leakage. When either (or both)
occur, the station must be shut down. The data at
hand indicate that the following probabilities prevail:
Station P (Pump failure) P (Leakage) P (Both)
1 0.07 0.10 0
2 0.09 0.12 0.06
Which station has the higher probability of being
shut down?

37. Chap 4-37
Conditional Probability
 Conditional probability for any
two events E1 , E2:
)P(E
)EandP(E
)E|P(E
2
21
21 
0)P(Ewhere 2 

38. Chap 4-38
 What is the probability that a car has a CD
player, given that it has AC ?
i.e., we want to find P(CD | AC)
Conditional Probability Example
 Of the cars on a used car lot, 70% have air
conditioning (AC) and 40% have a CD player
(CD). 20% of the cars have both.

39. Chap 4-39
Conditional Probability Example
No CDCD Total
AC .2 .5 .7
No AC .2 .1 .3
Total .4 .6 1.0
 Of the cars on a used car lot, 70% have air conditioning
(AC) and 40% have a CD player (CD).
20% of the cars have both.
.2857
.7
.2
P(AC)
AC)andP(CD
AC)|P(CD 
(continued)

40. Chap 4-40
Conditional Probability Example
No CDCD Total
AC .2 .5 .7
No AC .2 .1 .3
Total .4 .6 1.0
 Given AC, we only consider the top row (70% of the cars). Of these,
20% have a CD player. 20% of 70% is about 28.57%.
.2857
.7
.2
P(AC)
AC)andP(CD
AC)|P(CD 
(continued)

41. Chap 4-41
For Independent Events:
 Conditional probability for
independent events E1 , E2:
)P(E)E|P(E 121  0)P(Ewhere 2 
)P(E)E|P(E 212  0)P(Ewhere 1 

42. Chap 4-42
Multiplication Rules
 Multiplication rule for two events E1 and E2:
)E|P(E)P(E)EandP(E 12121 
)P(E)E|P(E 212 Note: If E1 and E2 are independent, then
and the multiplication rule simplifies to
)P(E)P(E)EandP(E 2121 

43. Chap 4-43
Tree Diagram Example
Diesel
P(E2) = 0.2
Gasoline
P(E1) = 0.8
Car: P(E4|E1) = 0.5
P(E1 and E3) = 0.8 x 0.2 = 0.16
P(E1 and E4) = 0.8 x 0.5 = 0.40
P(E1 and E5) = 0.8 x 0.3 = 0.24
P(E2 and E3) = 0.2 x 0.6 = 0.12
P(E2 and E4) = 0.2 x 0.1 = 0.02
P(E3 and E4) = 0.2 x 0.3 = 0.06
Car: P(E4|E2) = 0.1

44. Assessing Uncertainty at the
Bender Company
 Bender Company supplies contractors with material
for the construction of houses
 Currently it has a contract with one of its customers
to fill an order by the end of July
 However there is some uncertainty about whether
this dead line can be met, due to uncertainty about
whether Bender will receive the material it needs
from one of its suppliers by the middle of July. Right
now it is July 1day
 To asses the situation of uncertainty bender used
various probability rules.

45. Problem ?
 Bender will meet its end of July dead line, given
the information the company has at the
beginning of July

46. In this case what are the two major event
Bender should Identify
A = Bender meets its end-of-July deadline
B = Bender receives the material from its
supplier by the middle of July

47.  What are the various possible happenings
 Bender will get its materials on time and meet its end
of July deadline
 Bender will not get its materials on time and meet its
end of July deadline
 Bender will get its material on time and will not meet
its end of July deadline
 Bender will not get its material on time and will not
meet its end of July deadline
47

48.  From the past data what are the chances Bender
should estimated
1. The chances of getting the material on
time from its supplier are 2 out of 3
2. The chances of meeting the end of July
deadline after receiving the material
receiving material is 3 out of 4
3. The chances of meeting the end of July
deadline are 1 out of 5 if the material do not
arrive on time

49. The above probabilities can be
represented by
P(B)= 2/3 and P(B/A) = 3/4
 i. Then the probability of happening both
P(B and A) = P(B) P(A/B) = (3/4) (2/3)=0.5
That is there is only fifty-fifty chances that
Bender will get its materials on time and
meet its end of July.

50. 
B

51. Chap 4-51

52. The bottom line of the case

Business Statistics: A Decision-Making Approach, 6e © 2005 Prentice-Hall, Inc. Chap 4-52

53. Assessing Uncertainty at the
Bender Company
 Objective: To apply several of the essential
probability rules in determining the probability
that Bender will meet its end-of-July deadline,
given the information the company has at the
beginning of July.
 Solution: Use multiplication rule and build a
probability tree. All calculations can be
performed once the probability tree is
available.

54. Chap 4-54
A bank has the following data on the gender and marital status
of 200 customers.
Male Female
Single 20 30
Married 100 50
1. What is the probability of finding a single female
customer?
2. What is the probability of finding a married male
customer?
3. If a customer is female, what is the probability that she
is single?
4. What percentage of customers is male?
5. If a customer is male, what is the probability that he is
married?
6. Are gender and marital status mutually exclusive?
7. Is marital status independent of gender? Explain using

55. Chap 4-55
Bayes’ Theorem
 where:
Ei = ith event of interest of the k possible events
B = new event that might impact P(Ei)
Events E1 to Ek are mutually exclusive and collectively
exhaustive
)E|)P(BP(E)E|)P(BP(E)E|)P(BP(E
)E|)P(BP(E
B)|P(E
kk2211
ii
i




56. Chap 4-56
Bayes’ Theorem Example
 A drilling company has estimated a 40%
chance of striking oil for their new well.
 A detailed test has been scheduled for more
information. Historically, 60% of successful
wells have had detailed tests, and 20% of
unsuccessful wells have had detailed tests.
 Given that this well has been scheduled for a
detailed test, what is the probability
that the well will be successful?

57. Chap 4-57
 Let S = successful well and U = unsuccessful well
 P(S) = .4 , P(U) = .6 (prior probabilities)
 Define the detailed test event as D
 Conditional probabilities:
P(D|S) = .6 P(D|U) = .2
 Revised probabilities
Bayes’ Theorem Example
Event
Prior
Prob.
Conditional
Prob.
Joint
Prob.
Revised
Prob.
S (successful) .4 .6 .4*.6 = .24 .24/.36 = .67
U (unsuccessful) .6 .2 .6*.2 = .12 .12/.36 = .33
Sum = .36
(continued)

58. Chap 4-58
 Given the detailed test, the revised probability
of a successful well has risen to .67 from the
original estimate of .4
Bayes’ Theorem Example
Event
Prior
Prob.
Conditional
Prob.
Joint
Prob.
Revised
Prob.
S (successful) .4 .6 .4*.6 = .24 .24/.36 = .67
U (unsuccessful) .6 .2 .6*.2 = .12 .12/.36 = .33
Sum = .36
(continued)

59.  A consulting firm submitted a bid for a large research
project. The firms management initially felt they had a
50 – 50 chances of getting the project. However, the
agency to which the bid was submitted subsequently
requested additional information on the bid. Past
experience indicates that for 75% of the successful bids
and 40% of the unsuccessful bids the agency requested
additional information
 What is the prior probability of the bid being successful
(that is prior to the request for additional information)
 What is the probability of a request for additional
information given that the bid will ultimately be
successful.
 Compute the probability that the bid will be successful
given a request for additional information

60. Chap 4-60
 A local bank received its credit card policy with the
intention of recalling some of its credit cards. In the past
approximately 5% of the cardholders defaulted, leaving
the bank unable to collect the outstanding balance.
Hence, management established a prior probability of
0.05 that any particular cardholder will default. The bank
also found that the probability of missing a monthly
payment is 0.20 for customers who do not default. Of
course, the probability of missing a monthly payment for
those who default is 1
 Given that a customer missed one or more monthly
payments, compute the posterior probability tha the
customer will default.
 The bank would like to recall its card if the probability
that a customer will default is greater than 0.20. should
the bank recall its card if the customer misses a monthly
payment? why

62. Chap 4-62
Introduction to Probability
Distributions
 Random Variable
 Represents a possible numerical value from
a random event
Random
Variables
Discrete
Random Variable
Continuous
Random Variable

63. Chap 4-63
Discrete Random Variables
 Can only assume a countable number of values
Examples:
 Roll a die twice
Let x be the number of times 4 comes up
(then x could be 0, 1, or 2 times)
 Toss a coin 5 times.
Let x be the number of heads
(then x = 0, 1, 2, 3, 4, or 5)

64. Chap 4-64
Experiment: Toss 2 Coins. Let x = # heads.
T
T
Discrete Probability Distribution
4 possible outcomes
T
T
H
H
H H
Probability Distribution
0 1 2 x
x Value Probability
0 1/4 = .25
1 2/4 = .50
2 1/4 = .25
.50
.25
Probability

65. Chap 4-65
 A list of all possible [ xi , P(xi) ] pairs
xi = Value of Random Variable (Outcome)
P(xi) = Probability Associated with Value
 xi’s are mutually exclusive
(no overlap)
 xi’s are collectively exhaustive
(nothing left out)
 0  P(xi)  1 for each xi
 S P(xi) = 1
Discrete Probability Distribution

66. Chap 4-66
Discrete Random Variable
Summary Measures
 Expected Value of a discrete distribution
(Weighted Average)
E(x) = Sxi P(xi)
 Example: Toss 2 coins,
x = # of heads,
compute expected value of x:
E(x) = (0 x .25) + (1 x .50) + (2 x .25)
= 1.0
x P(x)
0 .25
1 .50
2 .25

67. Chap 4-67
 Standard Deviation of a discrete distribution
where:
E(x) = Expected value of the random variable
x = Values of the random variable
P(x) = Probability of the random variable having
the value of x
Discrete Random Variable
Summary Measures
P(x)E(x)}{xσ 2
x  
(continued)

68. Chap 4-68
 Example: Toss 2 coins, x = # heads,
compute standard deviation (recall E(x) = 1)
Discrete Random Variable
Summary Measures
P(x)E(x)}{xσ 2
x  
.707.50(.25)1)(2(.50)1)(1(.25)1)(0σ 222
x 
(continued)
Possible number of heads
= 0, 1, or 2

69. Chap 4-69
Two Discrete Random Variables
 Expected value of the sum of two discrete
random variables:
E(x + y) = E(x) + E(y)
= S x P(x) + S y P(y)
(The expected value of the sum of two random
variables is the sum of the two expected
values)

70. Business Statistics: A Decision-Making Approach, 6e © 2005 Prentice-Hall, Inc. Chap 4-70
 Bob Walters who frequently invests in the stock market,
carefully studies any potential investment. He is
currently examining the possibility of investing in the
Trinity Power Company. Through studying past
performance, Walters has broken the potential results of
the investment into five possible outcomes with a
accompanying frequencies. Construct a probability
distribution. The outcomes are annual rates of return on
a single share of stock that currently costs $150. Find
the expected value of the return for investing in a single
share of Trinity Power
Return 0 10 15 25 50
Probty 20 25 30 15 10

71. Chap 4-71
 During one holiday season, the Texas lottery
played a game called the Stocking Stuffer. With
this game, total instant winnings of $34.8 million
were available in 70 millions $1 tickets, with
ticket prizes raging from $1 to $1000. shown
here are the various prizes and the probability
of winning each prize. Use these data to
compute the expected value of the game, the
variance of the game, and the standard
deviation of the game
 Prize 1000 100 20 10 4 2 1 0
Probty 0.000
02
0.000
63
0.004 0.006 0.024
03
0.088
77
0.104
79
0.771
75

72. Chap 4-72
Covariance
 Covariance between two discrete random
variables:
σxy = S [xi – E(x)][yj – E(y)]P(xiyj)
where:
xi = possible values of the x discrete random variable
yj = possible values of the y discrete random variable
P(xi ,yj) = joint probability of the values of xi and yj occurring

73. Chap 4-73
 Covariance between two discrete random
variables:
xy > 0 x and y tend to move in the same direction
xy < 0 x and y tend to move in opposite directions
xy = 0 x and y do not move closely together
Interpreting Covariance

74. Chap 4-74
Correlation Coefficient
 The Correlation Coefficient shows the
strength of the linear association between
two variables
where:
ρ = correlation coefficient (“rho”)
σxy = covariance between x and y
σx = standard deviation of variable x
σy = standard deviation of variable y
yx
yx
σσ
σ
ρ 

75. Chap 4-75
 The Correlation Coefficient always falls
between -1 and +1
 = 0 x and y are not linearly related.
The farther  is from zero, the stronger the linear
relationship:
 = +1 x and y have a perfect positive linear relationship
 = -1 x and y have a perfect negative linear relationship
Interpreting the
Correlation Coefficient

76. Chap 4-76
Chapter Summary
 Described approaches to assessing probabilities
 Developed common rules of probability
 Used Bayes’ Theorem for conditional
probabilities
 Distinguished between discrete and continuous
probability distributions
 Examined discrete probability distributions and
their summary measures