Transcript P(A and B)

Chapter 2
Probability
Concepts and
Applications
To accompany Quantitative Analysis
for Management, 8e
by Render/Stair/Hanna
2-1
© 2003 by Prentice Hall, Inc.,
Upper Saddle River, NJ 07458
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
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2-2
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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
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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
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2-4
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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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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
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Basic Statements About
Probability
1. 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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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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Example 2.1 - continued
Frequencies of Demand
Quantity
Number of Days
Demanded
(Gallons)
0
40
1
80
2
50
3
20
4
10
Total 200
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Example 2.1 - continued
Probabilities of Demand
Quant.
Freq.
Demand
(days)
Probability
0
40
(40/200) = 0.20
1
80
(80/200) = 0.40
2
50
(50/200) = 0.25
3
20
(20/200) = 0.10
4
10
(10/200) = 0.05
Total
Prob
= 1.00
Total days = 200
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2-10
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Types of Probability
Objective probability:
P ( event ) =
Number of times event occurs
Total number of outcomes or occurrences
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
Subjective probability:
Based upon judgement
Determined by:
• judgement of expert
• opinion polls
• Delphi method
• etc.
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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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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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Example 2
Rolling a
die has six
possible
outcomes
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Outcome Probability
of Roll
1
1/6
2
1/6
3
1/6
4
1/6
5
1/6
6
1/6
Total = 1
2-15
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Example 2a
Outcome
Probability
Rolling two
of Roll = 5
dice results in Die 1 Die 2
a total of five
1
4
spots
2
3
showing.
3
2
There are a
total of 36
4
1
possible
outcomes.
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2-16
1/36
1/36
1/36
1/36
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Example 3
Draw
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
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Mutually Collectively
Exclusive Exhaustive
Yes
No
Yes
Yes
Yes
No
Yes
Yes
No
No
No
No
2-17
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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 + 13/52
= 26/52 = 1/2 = 50%
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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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P(A and B)
(Venn Diagram)
P(A)
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P(B)
2-20
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P(A or B)
-
+
P(A)
P(B)
=
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P(A and B)
P(A or B)
2-21
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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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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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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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Probability(A|B)
Independent Events
P(B)
P(B|A)
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P(A|B)
2-25
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Statistically
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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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)
2-26
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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
(conditional probability)
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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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Probability(A|B)
/
P(A)
P(AB)
P(B)
P(A|B) = P(AB)/P(B)
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Probability(B|A)
/
P(B)
P(AB)
P(A)
P(B|A) = P(AB)/P(A)
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Statistically Dependent
Events
Then:
Assume that we
• P(WL) = 4/10 = 0.40
have an urn
• P(WN) = 2/10 = 0.20
containing 10 balls
of the following
• P(W) = 6/10 = 0.60
descriptions:
• P(YL) = 3/10 = 0.3
•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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• P(YN) = 1/10 = 0.1
• P(Y) = 4/10 = 0.4
2-31
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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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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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Joint Probabilities, Dependent
Events - continued
Let M represent
the event of the
Then:
stock market
P(MT) =P(T|M)P(M)
reaching the
= (0.70)(0.40)
10,500 point
= 0.28
level, and T
represent the
event that
Tubeless goes
up.
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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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General Form of
Bayes’ Theorem
P( AB)
P( A | B) =
P( B)
or
P( B | A) P( A)
P( A | B) =
P( B | A) P( A)  P( B | A ) P( A )
where A = complement of the event A
For example, if the event A is " fair" die,
then the event A is " unfair" die.
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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 0.166. 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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Posterior Probabilities
Continued
• We know that:
P(fair) = 0.50
P(loaded) = 0.50
• And:
P(3|fair) = 0.166
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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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) = 0.083 + 0.0300 = 0.383.
• Then, the probability that the die
rolled was the fair one is given by:
P(Fair | 3) =
P(Fair and 3) 0.083
=
= 0.22
P(3)
0.383
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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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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) = 0.013 + 0.18 = 0.193
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Further Probability
Revisions - continued
P(3,3 and Fair)
P(Fair | 3,3) =
P(3,3)
0.013
=
= 0.067
0.193
P(3,3 and Loaded)
P(Loaded | 3,3) =
P(3,3)
0.18
=
= 0.933
0.193
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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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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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Random Variables
(Numeric)
Experiment
Outcome
Random Variable
Range of
Random
Variable
Stock 50
Xmas trees
Number of
trees sold
X = number of
trees sold
Inspect 600
items
Number
acceptable
Y = number
acceptable
0,1,2,…,
600
Send out
5,000 sales
letters
Number of
peoplee
responding
Z = number of
people responding
0,1,2,…,
5,000
Build an
apartment
building
%completed
after 4
months
R = %completed
after 4 months
0R 100
Test the
lifetime of a
light bulb
(minutes)
Time bulb
lasts - up to
80,000
minutes
S = time bulb
burns
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0,1,2,, 50
0S80,000
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Random
Variables (Non-numeric)
Experiment
Outcome
Random
Variable
Range of
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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Probability Distributions
Table 2.4
Outcome X Number
P(X)
Responding
SA
5
10
0.10
A
4
20
0.20
N
3
30
0.30
D
2
30
0.30
SD
1
10
0.10
D
0.30
0.25
Figure 2.5
Probability
Function
0.20
0.15
0.10
0.05
0.00
1
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2
3
4
2-47
5
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Expected Value of a
Discrete Probability
Distribution
n
E ( X ) =  X i P( X i )
i =1
5
E ( X ) =  X i P( X i )
i =1
= X 1 P( X 1 )  X 2 P( X 2 )  X 3 P( X 3 )
 X 4 P( X 4 )  X 5 P( X 5 )
= (5)(0.1)  (4)(0.2)  (3)(0.3)
 (2)(0.3)  (1)(0.1)
= 2.9
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Variance of a Discrete
Probability Distribution
n
 =  X i  E  X  P X i 
2
2
i =1
 2 = 5  2.92 0.1  4  2.92 0.2
 3  2.9 0.3  (2 - 2.9) 2 (0.3)
2
 (1  2.9) 2 (0.1)
= 0.44 - 0.242  0.003  0.243  0.361
= 1.29
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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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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
n!
r nr
=
pq
r!(n - r)!
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Binomial Distribution
 = np
  = np( 1  p )
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Binomial Distribution
N = 5, p = 0.50
0.35
0.30
P(r)
0.25
0.20
0.15
0.10
0.05
0.00
1
2
3
4
5
6
(r) Number of Successes
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Probability Distribution
Continuous Random
Variable
Normal Distribution
Probability density function - f(X)
5
5.05
5.1
f (X ) =
5.15
5.2
5.3
5.35
1 / 2 ( X   ) 2
1
 2
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5.25
2-54
e
2
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5.4
Normal Distribution for
Different Values of 
=40
=50
=60
40
50
60
0
30
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Upper Saddle River, NJ 07458
70
Normal Distribution for
Different Values of 
=1
=0.1
=0.3
0
0.5
To accompany Quantitative Analysis
for Management, 8e
by Render/Stair/Hanna
=0.2
1
2-56
1.5
© 2003 by Prentice Hall, Inc.,
Upper Saddle River, NJ 07458
2
Three Common Areas
Under the Curve
• Three Normal distributions with
different areas
To accompany Quantitative Analysis
for Management, 8e
by Render/Stair/Hanna
2-57
© 2003 by Prentice Hall, Inc.,
Upper Saddle River, NJ 07458
Three Common Areas
Under the Curve
Three
Normal
distributions
with
different
areas
To accompany Quantitative Analysis
for Management, 8e
by Render/Stair/Hanna
2-58
© 2003 by Prentice Hall, Inc.,
Upper Saddle River, NJ 07458
The Relationship Between
Z and X
=100
=15
x
Z=

55
70
85
100
115
130
145
-3
-2
-1
0
1
2
3
To accompany Quantitative Analysis
for Management, 8e
by Render/Stair/Hanna
2-59
© 2003 by Prentice Hall, Inc.,
Upper Saddle River, NJ 07458
Haynes Construction
Company Example
Fig. 2.12
To accompany Quantitative Analysis
for Management, 8e
by Render/Stair/Hanna
2-60
© 2003 by Prentice Hall, Inc.,
Upper Saddle River, NJ 07458
Haynes Construction
Company Example
Fig. 2.13
To accompany Quantitative Analysis
for Management, 8e
by Render/Stair/Hanna
2-61
© 2003 by Prentice Hall, Inc.,
Upper Saddle River, NJ 07458
Haynes Construction
Company Example
Fig. 2.14
To accompany Quantitative Analysis
for Management, 8e
by Render/Stair/Hanna
2-62
© 2003 by Prentice Hall, Inc.,
Upper Saddle River, NJ 07458
The Negative
Exponential
Distribution
f ( X ) = e
 x Expected value = 1/
Variance = 1/2
6
5
=5
4
3
2
1
0
0
0.2
To accompany Quantitative Analysis
for Management, 8e
by Render/Stair/Hanna
0.4
0.6
2-63
0.8
1
1.2
© 2003 by Prentice Hall, Inc.,
Upper Saddle River, NJ 07458
The Poisson
Distribution
e
x 
P( X ) =
X!
Expected value = 
Variance = 
0.30
=2
0.25
0.20
0.15
0.10
0.05
0.00
1
2
To accompany Quantitative Analysis
for Management, 8e
by Render/Stair/Hanna
3
4
2-64
5
6
7
8
9
© 2003 by Prentice Hall, Inc.,
Upper Saddle River, NJ 07458