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CHAPTER 2 - PROBABILITY CONCEPTS
•Types of Probability
•Fundamentals of Probability
•Statistical Independence and Dependence
•Bayes Theorem
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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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Types of Probability
Objective probability:
P ( event ) =
number of times event occurs
total number of trials or outcomes
Can be determined by experiment or observation:
– Probability of heads on coin flip
– Probably of spades on drawing card from deck
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Types of Probability (cont.)
• Subjective probability is an estimate based on personal belief, experience, or
knowledge of a situation:
•It is often the only means available for making probabilistic estimates.
• Frequently used in making business decisions.
• Different people often arrive at different subjective probabilities.
 Judgment of expert
 Opinion polls
 Delphi method
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Fundamentals of Probability
Outcomes and Events
• An experiment is an activity that results in one of several possible
outcomes which are termed events.
• The probability of an event is always greater than or equal to zero
and less than or equal to one: 0  P(event)  1
• The probabilities of all the events included in an experiment
must sum to one.
• The events in an experiment are mutually exclusive if only one
can occur at a time.
• The probabilities of mutually exclusive events sum to one.
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Fundamentals of Probability
Distributions
• A frequency distribution is an organization of numerical data
about the events in an experiment.
• A list of corresponding probabilities for each event is referred to
as a probability distribution.
• If two or more events cannot occur at the same time they are
termed mutually exclusive.
• A set of events is collectively exhaustive when it includes all the
events that can occur in an experiment.
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Fundamentals of Probability
A Frequency Distribution Example
Example:
Business School at the State University has analyzed the records of the 3000 students
who have taken Management Science course during the past four years. The Dean
wants to determine the number of students who made each grade.
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Fundamentals of Probability
Mutually Exclusive Events and Marginal Probabilities
• A marginal probability is the probability of a single event occuring, denoted P(A).
• For mutually exclusive events, the probability that one or the other of several events
will occur is found by summing the individual probabilities of the events:
P(A or B) = P(A) + P(B)
A Venn diagram is used to show mutually exclusive events.
Venn diagram for mutually exclusive events
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Fundamentals of Probability
Non-Mutually Exclusive Events and Joint Probabilities
• Probability that non-mutually exclusive events A and B or both will occur expressed as:
P(A or B) = P(A) + P(B) - P(AB)
• A joint probability, P(AB), is the probability that two or more events that are not mutually
exclusive can occur simultaneously.
Venn diagram for non–mutually exclusive events and the joint event
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Statistical Independence and Dependence
Independent Events
• A succession of events that do not affect each other are independent.
• The probability of independent events occurring in a succession is computed by
multiplying the probabilities of each event.
• A conditional probability is the probability that an event will occur given that
another event has already occurred, denoted as P(AB). If events A and B are
independent, then:
1. P(AB) = P(A)  P(B)
2. P(AB) = P(A)
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Statistical Independence and Dependence
Independent Events - Probability Trees
- For coin tossed three consecutive times;
Probability tree for
coin-tossing example
- Probability of getting head on first toss, tail on second, tail on third is .125:
P(HTT) = P(H)  P(T)  P(T) = (.5)(.5)(.5) = .125
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Statistical Independence and Dependence
Dependent Events
•If the occurrence of one event affects the probability of the occurrence of another
event, the events are dependent.
•Coin toss to select bucket, draw for blue ball.
•If tail occurs, 1/6 chance of drawing blue ball from bucket 2; if head results, no
possibility of drawing blue ball from bucket 1.
•Probability of event “drawing a blue ball” dependent on event “flipping a coin”.
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Statistical Independence and Dependence
Dependent Events - Unconditional/Conditional Probabilities
- Unconditional: P(H) = .5; P(T) = .5, must sum to one.
- Conditional: P(RH) =.33, P(WH) = .67, P(RT) = .83, P(WT) = .17
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Statistical Independence and Dependence
Mathematical Formulation of Conditional Probabilities
- Given two dependent events A and B:
P(AB) = P(AB)/ P(B)
This formula can be rearranged to obtain the following formula:
P(AB) = P(AB) • P(B)
- With data from previous example:
P(RH) = P(RH)  P(H) = (.33)(.5) = .165
P(WH) = P(WH)  P(H) = (.67)(.5) = .335
P(RT) = P(RT)  P(T) = (.83)(.5) = .415
P(WT) = P(WT)  P(T) = (.17)(.5) = .085
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Statistical Independence and Dependence
Summary of Example Problem probabilities
Probability tree with
marginal, conditional,
and joint probabilities
Joint Probability Table
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Statistical Independence and Dependence
Bayesian Analysis
Bayes’ theorem can be used to calculate revised or posterior
probabilities
Prior
Probabilities
Bayes’
Process
Posterior
Probabilities
New
Information
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Statistical Independence and Dependence
Bayesian Analysis
•In Bayesian analysis, additional information is used to alter the marginal probability of
the occurance of an event.
•A posterior probability is the altered marginal probability of an event based on
additional information.
•Bayes’s Rule for two events, A and B, and third event, C, conditionally dependent on A
and B :
P( A C ) =
P(C A) P( A)
P(C A) P( A)  P(C B) P( B)
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Statistical Independence and Dependence
Bayesian Analysis - Example
- Machine setup; if correct 10% chance of defective part; if incorrect, 40%.
- 50% chance setup will be correct or incorrect.
- What is probability that machine setup is incorrect if sample part is defective?
- Solution: P(C) = .50, P(IC) = .50, P(D|C) = .10, P(D|IC) = .40
where C = correct, IC = incorrect, D = defective
- Posterior probabilities:
P(IC D) =
P(D IC)P(IC)
P(D IC)P(IC)  P(D C)P(C)
(.40)(.50)
(.40)(.50)  (.10)(.50)
= .80
=
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Bayesian Analysis - Example
Example:
A particular type of printer ribbon is produced by only
two companies A and B. Company A produces 65% of
the ribbons while company B produces 35%. 8% of the
ribbons produced by company A are defective while 12%
of the ribbons produced by company B are defective
A customer purchases a new ribbon.
a) What is the probability that company A produced the
ribbon?
b) It is known that the ribbon is produced by company B.
What is the probability that the ribbon is defective?
c) The ribbon is tested, and it is defective. What is the
probability that company A produced the ribbon?
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Bayesian Analysis – Example (cont.)
Solution: P(A) = 0.65, P(B) = 0.35, P(D|A) = 0.08, P(D|B) = 0.12
where A = ribbon produced by company A,
B = ribbon produced by company B
D = defective ribbon
a)
b)
P(A) = 0.65
P(D|B) = P(DB)/ P(B) = 0.12
c)
P(A)  P(D/A)
(0.65)(0.08)
=
P(A|D) =
P(A)  P(D/A) + P(B)  P(D/B) (0.65)(0.08)  (0.35)(0.12)
= 0.052 / 0.094 = 0.553
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Bayesian Analysis – Example (cont.)
Event
(Ei)
Prior
Probability
P(Ei)
Conditional
Probability
P(D | Ei)
Joint
Probability
P(Ei  D)
Posterior /
Revised
Probability
P(Ei | D)
A
0.65
0.08
0.052
0.052/0.094 =
0.553
B
0.35
0.12
0.042
0.042/0.094 =
0.447
P(D) = 0.94
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Expected Value
Random Variables
- When the values of variables occur in no particular order or sequence,
the variables are referred to as random variables.
- Random variables are represented symbolically by a letter x, y, z, etc.
- Although exact values of random variables are not known prior to
events, it is possible to assign a probability to the occurrence of possible
values.
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Expected Value
Example (1 of 4)
- Machines break down 0, 1, 2, 3, or 4 times per month.
- Relative frequency of breakdowns , or a probability distribution:
Random Variable x
(number of breakdowns)
0
1
2
3
4
P(x)
.10
.20
.30
.25
.15
1.00
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Expected Value
Example (2 of 4)
- The expected value of a random variable is computed by multiplying each
possible value of the variable by its probability and summing these products.
- The expected value is the weighted average, or mean, of the probability
distribution of the random variable.
- Expected value of number of breakdowns per month:
E(x) =  x P(x)
E(x) = (0)(.10) + (1)(.20) + (2)(.30) + (3)(.25) + (4)(.15)
= 0 + .20 + .60 + .75 + .60
= 2.15 breakdowns
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Expected Value
Example (3 of 4)
- Variance is a measure of the dispersion of random variable values about the mean.
- Variance computed as follows:
1. Square each value, x
2. Multiply square of each value by the probability of each value and sum it.
3. Subtract square of mean from the value obtained in step 2
- General formula:
Variance, 2 = [xi - E(xi)] 2 P(xi)
=  x2 P(x) – [E(x)]2
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Expected Value
Example (4 of 4)
- Standard deviation computed by taking the square root of the variance.
- For example data:
xi
P(xi)
xi P(xi)
xi2 P(xi)
0
1
2
3
4
.10
.20
.30
.25
.15
1.00
0
0.20
0.60
0.75
0.60
2.15
0
0.20
1.20
2.25
2.40
6.04
0.20
Variance, 2 =  x2 P(x) – [E(x)]2
2 = 6.04 – (2.15)2 = 1.4175 breakdowns per month
standard deviation =  = 1.4175 = 1.19 breakdowns per month
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The Normal Distribution
Continuous Random Variables
• A continuous random variable can take on an infinite number of values within
some interval.
• Continuous random variables have values that are not specifically countable and
are often fractional.
• Cannot assign a unique probability to each value of a continuous random
variable.
•In a continuous probability distribution the probability refers to a value of the
random variable being within some range.
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The Normal Distribution
Definition
• The normal distribution is a continuous probability distribution that is symmetrical on both
sides of the mean.
• The center of a normal distribution is its mean .
• The area under the normal curve represents probability, and total area under the curve sums to
one.
The normal curve
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The Normal Distribution Example
- Mean weekly carpet sales of 4,200 yards, with standard deviation of 1,400 yards.
- What is probability of sales exceeding 6,000 yards ?
 = 4,200 yd;  = 1,400 yd; probability that number of yards of carpet will be equal to or greater
than 6,000 expressed as: P(x  6,000).
The normal distribution
for carpet demand
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The Normal Distribution Example
Z = (x - )/  = (6,000 - 4,200)/1,400 = 1.29 standard deviations
P(x  6,000) = .5000 - .4015 = .0985
Determination of the Z value
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The Normal Distribution Example
- Determine probability that demand will be 5,000 yards or less.
Z = (x - )/  = (5,000 - 4,200)/1,400 = .57 standard deviations
P(x 5,000) = .5000 + .2157 = .7157
Normal distribution
for P(x  5,000 yards)
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The Normal Distribution Example
- Determine probability that demand will be between 3,000 yards and 5,000 yards.
Z = (3,000 - 4,200)/1,400 = -1,200/1,400 = -.86
P(3,000  x  5,000) = .2157 + .3051= .5208
Normal distribution with
P(3,000  x  5,000)
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