1. Probability Concepts and Applications
Chapter 2
Probability Concepts and Applications
To accompany Quantitative Analysis for Management, 8e
by Render/Stair/Hanna
2-1

2. Learning Objectives Students will be able to:
Understand the basic foundations of probability analysis
Understand the difference between mutually exclusive and collectively exhaustive events
Describe statistically dependent and independent events
Use Bayes’ theorem to establish posterior probabilities
To accompany Quantitative Analysis for Management, 8e
by Render/Stair/Hanna
2-2

3. Learning Objectives - continued
Describe and provide examples of both discrete and continuous random variables
Explain the difference between discrete and continuous probability distributions
Calculate expected values and variances
Use the Normal table
To accompany Quantitative Analysis for Management, 8e
by Render/Stair/Hanna
2-3

4. Chapter Outline 2.1 Introduction 2.2 Fundamental Concepts
2.3 Mutually Exclusive and Collectively Exhaustive Events
2.4 Statistically Independent Events
2.5 Statistically Dependent Events
2.6 Revising Probabilities with Bayes’ Theorem
To accompany Quantitative Analysis for Management, 8e
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2-4

5. Chapter Outline - continued
2.7 Further Probability Revisions
2.8 Random Variables
2.9 Probability Distributions
2.10 The Binomial Distribution
2.11 The Normal Distribution
2.12 The Exponential Distribution
2.13 The Poisson Distribution
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2-5

6. Introduction Life is uncertain! We must deal with risk!
A probability is a numerical statement about the likelihood that an event will occur
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2-6

7. Basic Statements About Probability
The probability, P, of any event or state of nature occurring is greater than or equal to 0 and less than or equal to 1.
That is: 0  P(event)  1
2. The sum of the simple probabilities for all possible outcomes of an activity must equal 1.
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2-7

8. Example 2.1
Demand for white latex paint at Diversey Paint and Supply has always been 0, 1, 2, 3, or 4 gallons per day. (There are no other possible outcomes; when one outcome occurs, no other can.) Over the past 200 days, the frequencies of demand are represented in the following table:
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2-8

9. Quantity Demanded (Gallons)
Example continued
Frequencies of Demand
Quantity Demanded (Gallons) 1 2 3 4
Number of Days 40 80 50 20 10
Total 200
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2-9

10. Probabilities of Demand
Example continued
Probabilities of Demand
Probability
(40/200) = 0.20
(80/200) = 0.40
(50/200) = 0.25
(20/200) = 0.10
(10/200) = 0.05
Total
Prob = 1.00
Quant Freq.
Demand (days)
Total days = 200
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2-10

11. Types of Probability Objective probability:
Determined by experiment or observation:
Probability of heads on coin flip
Probably of spades on drawing card from deck
occurrences or
outcomes of
number
Total
occurs
event
times
Number ) ( = P
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2-11

12. Types of Probability Subjective probability: Based upon judgement
Determined by:
judgement of expert
opinion polls
Delphi method
etc.
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2-12

13. Mutually Exclusive Events
Events are said to be mutually exclusive if only one of the events can occur on any one trial
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14. Collectively Exhaustive Events
Events are said to be collectively exhaustive if the list of outcomes includes every possible outcome: heads and tails as possible outcomes of coin flip
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2-14

15. Example 2 Rolling a die has six possible outcomes Outcome of Roll 1 23 4 5 6
Probability
1/6
Total = 1
Rolling a die has six possible outcomes
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2-15

16. Example 2a Outcome of Roll = 5 Die 1 Die 2 1 4 2 3 3 2 4 1 Probability
Probability
1/36
Rolling two dice results in a total of five spots showing. There are a total of 36 possible outcomes.
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2-16

17. Example 3 Draw Mutually Collectively Exclusive Exhaustive Yes No
Draw a space and a club
Draw a face card and a number card
Draw an ace and a 3
Draw a club and a nonclub
Draw a 5 and a diamond
Draw a red card and a diamond
Yes No
Yes Yes
No No
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2-17

18. Probability : Mutually Exclusive
P(event A or event B) =
P(event A) + P(event B)
or:
P(A or B) = P(A) + P(B)
i.e.,
P(spade or club) = P(spade) + P(club)
= 13/ /52
= 26/52 = 1/2 = 50%
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2-18

19. Probability: Not Mutually Exclusive
P(event A or event B) =
P(event A) + P(event B) -
P(event A and event B both occurring) or
P(A or B) = P(A)+P(B) - P(A and B)
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20. P(A and B) (Venn Diagram)
P(B)
P(A and B)
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21. P(A or B) + - = P(A) P(B) P(A and B) P(A or B) 2-21
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2-21

22. Statistical Dependence
Events are either
statistically independent (the occurrence of one event has no effect on the probability of occurrence of the other) or
statistically dependent (the occurrence of one event gives information about the occurrence of the other)
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2-22

23. Which Are Independent? (a) Your education (b) Your income level
(a) Draw a Jack of Hearts from a full 52 card deck
(b) Draw a Jack of Clubs from a full 52 card deck
(a) Chicago Cubs win the National League pennant
(b) Chicago Cubs win the World Series
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2-23

24. Probabilities - Independent Events
Marginal probability: the probability of an event occurring:
[P(A)]
Joint probability: the probability of multiple, independent events, occurring at the same time
P(AB) = P(A)*P(B)
Conditional probability (for independent events):
the probability of event B given that event A has occurred P(B|A) = P(B)
or, the probability of event A given that event B has occurred P(A|B) = P(A)
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2-24

25. Probability(A|B) Independent Events
P(B)
P(A)
P(A|B)
P(B|A)
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2-25

26. Statistically Independent Events
1. P(black ball drawn on first draw)
P(B) = 0.30 (marginal probability)
2. P(two green balls drawn)
P(GG) = P(G)*P(G) = 0.70*0.70 = 0.49 (joint probability for two independent events)
A bucket contains 3 black balls, and 7 green balls. We draw a ball from the bucket, replace it, and draw a second ball
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2-26

27. Statistically Independent Events - continued
1. P(black ball drawn on second draw, first draw was green)
P(B|G) = P(B) = 0.30
(conditional probability)
2. P(green ball drawn on second draw, first draw was green)
P(G|G) = 0.70
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2-27

28. Probabilities - Dependent Events
Marginal probability: probability of an event occurring P(A)
Conditional probability (for dependent events):
the probability of event B given that event A has occurred P(B|A) = P(AB)/P(A)
the probability of event A given that event B has occurred P(A|B) = P(AB)/P(B)
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2-28

29. Probability(A|B) / P(A|B) = P(AB)/P(B) P(AB) P(B) P(A) 2-29
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2-29

30. Probability(B|A) / P(B|A) = P(AB)/P(A) P(AB) P(B) P(A) 2-30
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2-30

31. Statistically Dependent Events
Then:
P(WL) = 4/10 = 0.40
P(WN) = 2/10 = 0.20
P(W) = 6/10 = 0.60
P(YL) = 3/10 = 0.3
P(YN) = 1/10 = 0.1
P(Y) = 4/10 = 0.4
Assume that we have an urn containing 10 balls of the following descriptions:
4 are white (W) and lettered (L)
2 are white (W) and numbered N
3 are yellow (Y) and lettered (L)
1 is yellow (Y) and numbered (N)
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2-31

32. Statistically Dependent Events - Continued
Then:
P(L|Y) = P(YL)/P(Y)
= 0.3/0.4 = 0.75
P(Y|L) = P(YL)/P(L)
= 0.3/0.7 = 0.43
P(W|L) = P(WL)/P(L)
= 0.4/0.7 = 0.57
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2-32

33. Joint Probabilities, Dependent Events
Your stockbroker informs you that if the stock market reaches the 10,500 point level by January, there is a 70% probability the Tubeless Electronics will go up in value. Your own feeling is that there is only a 40% chance of the market reaching 10,500 by January.
What is the probability that both the stock market will reach 10,500 points, and the price of Tubeless will go up in value?
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2-33

34. Joint Probabilities, Dependent Events - continued
Let M represent the event of the stock market reaching the 10,500 point level, and T represent the event that Tubeless goes up.
Then:
P(MT) =P(T|M)P(M)
= (0.70)(0.40)
= 0.28
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2-34

35. Revising Probabilities: Bayes’ Theorem
Bayes’ theorem can be used to calculate revised or posterior probabilities
Prior Probabilities
Bayes’ Process
Posterior Probabilities
New Information
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2-35

36. General Form of Bayes’ Theorem
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2-36

37. Posterior Probabilities
A cup contains two dice identical in appearance. One, however, is fair (unbiased), the other is loaded (biased). The probability of rolling a 3 on the fair die is 1/6 or The probability of tossing the same number on the loaded die is 0.60.
We have no idea which die is which, but we select one by chance, and toss it. The result is a 3.
What is the probability that the die rolled was fair?
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2-37

38. Posterior Probabilities Continued
We know that:
P(fair) = P(loaded) = 0.50
And:
P(3|fair) = P(3|loaded) = 0.60
Then:
P(3 and fair) = P(3|fair)P(fair) = (0.166)(0.50)
= 0.083
P(3 and loaded) = P(3|loaded)P(loaded)
= (0.60)(0.50)
= 0.300
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2-38

39. Posterior Probabilities Continued
A 3 can occur in combination with the state “fair die” or in combination with the state ”loaded die.” The sum of their probabilities gives the unconditional or marginal probability of a 3 on a toss:
P(3) = =
Then, the probability that the die rolled was the fair one is given by:
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2-39

40. Further Probability Revisions
To obtain further information as to whether the die just rolled is fair or loaded, let’s roll it again.
Again we get a 3.
Given that we have now rolled two 3s, what is the probability that the die rolled is fair?
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2-40

41. Further Probability Revisions - continued
P(fair) = 0.50, P(loaded) = 0.50 as before
P(3,3|fair) = (0.166)(0.166) = 0.027
P(3,3|loaded) = (0.60)(0.60) = 0.36
P(3,3 and fair) = P(3,3|fair)P(fair)
= (0.027)(0.05)
= 0.013
P(3,3 and loaded) = P(3,3|loaded)P(loaded)
= (0.36)(0.5)
= 0.18
P(3,3) = = 0.193
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2-41

42. Further Probability Revisions - continued
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2-42

43. Further Probability Revisions - continued
To give the final comparison:
P(fair|3) = 0.22
P(loaded|3) = 0.78
P(fair|3,3) = 0.067
P(loaded|3,3) = 0.933
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2-43

44. Random Variables
Discrete random variable - can assume only a finite or limited set of values- i.e., the number of automobiles sold in a year
Continuous random variable - can assume any one of an infinite set of values - i.e., temperature, product lifetime
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2-44

45. Random Variables (Numeric)
Experiment
Outcome
Random Variable
Range of
Random
Variable
Stock 50
Xmas trees
Number of
trees sold
X = number of
0,1,2,, 50
Inspect 600
items
Number
acceptable
Y = number
0,1,2,…,
600
Send out
5,000 sales
letters
people e
responding
Z = number of
people responding
5,000
Build an
apartment
building
%completed
after 4
months
R = %completed
after 4 months R
100
Test the
lifetime of a
light bulb
(minutes)
Time bulb
lasts -
up to
80,000
minutes
S = time bulb
burns S
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2-45

46. Random Variables (Non-numeric)
Experiment
Outcome
Random
Range of
Variable
Random
Variable
Students
Strongly agree (SA)
X = 5 if SA
1,2,3,4,5
respond to a
Agree (A)
4 if A
questionnaire
Neutral (N)
3 if N
Disagree (D)
2 if D
Strongly Disagree (SD)
1 if SD
One machine is
Defective
Y = 0 if defective
0,1
inspected
Not defective
1 if not defective
Consumers
Good
Z = 3 if good
1,2,3
respond to how
Average
2 if average
they like a
Poor
1 if poor
product
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2-46

47. Probability Distributions
Figure 2.5
Probability Function
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48. Expected Value of a Discrete Probability Distributionå = n i ) X ( P E 1
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49. Variance of a Discrete Probability Distribution
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50. Binomial Distribution
Assumptions:
1. Trials follow Bernoulli process – two possible outcomes
2. Probabilities stay the same from one trial to the next
3. Trials are statistically independent
4. Number of trials is a positive integer
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2-50

51. Binomial Distribution
n = number of trials
r = number of successes
p = probability of success
q = probability of failure
Probability of r successes
in n trials
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52. Binomial Distribution
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53. Binomial Distribution
N = 5, p = 0.50
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54. Probability Distribution Continuous Random Variable
Normal Distribution
Probability density function - f(X)
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2-54

55. Normal Distribution for Different Values of 
=50
=60
=40
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2-55

56. Normal Distribution for Different Values of 
 = 1
=0.1
=0.3
=0.2
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57. Three Common Areas Under the Curve
Three Normal distributions with different areas
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58. Three Common Areas Under the Curve
Three Normal distributions with different areas
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59. The Relationship Between Z and X
=100
=15
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60. Haynes Construction Company Example Fig. 2.12
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61. Haynes Construction Company Example Fig. 2.13
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62. Haynes Construction Company Example Fig. 2.14
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63. The Negative Exponential Distribution
Expected value = 1/
Variance = 1/2
=5
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64. The Poisson Distribution
Expected value = 
Variance = 
=2
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2-64