Transcript Lesson3 - Ka

Lesson 3:
A Survey of Probability
Concepts
Ka-fu Wong © 2004
ECON1003: Analysis of Economic Data
Lesson3-1
Outline
Definitions
Learning Exercise #1
Definitions
Learning Exercise #2 and #3
Basic rules of Probability
Independence
Bayes’ Theorem
Probabilities for throwing n dice
Some principles of counting
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ECON1003: Analysis of Economic Data
Lesson3-2
Definition: Probability
A probability is a measure of the likelihood that an event in
the future will happen.

It can only assume a value between 0 and 1.

A value near zero means the event is not likely to happen. A
value near one means it is likely.

There are three definitions of probability: classical,
empirical, and subjective.
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ECON1003: Analysis of Economic Data
Definition: Classical probability
 The classical definition applies when there are n equally
likely outcomes.
 For example, in the tossing of a single perfectly cubical die,
made of completely homogeneous material, the equally
likely events are the appearance of any of the specific
number of dots (from 1 to 6) on its upper face.
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Lesson3-4
Definition: Empirical or statistical
probability
 The empirical or statistical definition applies when the
number of times the event happens is divided by the
number of observations, based on data.
Throughout her teaching career Professor Jones has awarded
186 A’s out of 1,200 students. What is the probability that a
student in her section this semester will receive an A?
186
P(A) 
 0.155
1200
This number may be interpreted as “unconditional probability”.
In most cases, we are interested in the probability of earning
an A for a selected student who study 10 hours or more per
week. We call this “conditional probability”.
P(A | study 10 or more hours per week)
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Definition: Subjective probability
 Subjective probability is based on whatever information is
available, based on subjective feelings.
 Estimating the probability mortgage rates for home
loans will top 8 percent this year.
 Estimating the probability that HK’s economic growth
will be 3% this year.
 Estimating the probability that HK will solve its deficit
problem in 2006.
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Lesson3-6
Example 1
(to be used to illustrate the definitions)
 A fair die is rolled once.
 Peter is concerned with whether the resulted number is
even, i.e., 2, 4, 6.
 Paul is concerned with whether the resulted number is less
than or equal to 3, i.e., 1, 2, 3.
 Mary is concerned with whether the resulted number is 6.
 Sonia is concerned with whether the resulted number is
odd, i.e., 1, 3, 5.
 A fair die is rolled twice.
 John is concerned with whether the resulted number of
first roll is even, i.e., 2, 4, 6.
 Sarah is concerned with whether the resulted number of
second roll is even, i.e., 2, 4, 6.
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Definitions: Experiment and outcome
 An experiment is the observation of some activity or the
act of taking some measurement.
 The experiment is rolling the one die in the first
example, and rolling one die twice in the second
example.
 An outcome is the particular result of an experiment.
 The possible outcomes are the numbers 1, 2, 3, 4, 5, and
6 in the first example.
 The possible outcomes are number pairs (1,1), (1,2), …,
(6,6), in the second example.
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Lesson3-8
Definition: Event
 An event is the collection of one or more outcomes of an
experiment.
 For Peter: the occurrence of an even number, i.e., 2, 4, 6.
 For Paul: the occurrence of a number less than or equal
to 3, i.e., 1, 2, 3.
 For Mary: the occurrence of a number 6.
 For Sonia: the occurrence of an odd number, i.e., 1, 3, 5.
 For John: the occurrence of (2,1), (2,2), (2,3),…, (2,6),
(4,1),…,(4,6), (6,1),…,(6,6)
[John does not care about the result of the second roll].
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ECON1003: Analysis of Economic Data
Lesson3-9
Learning exercise 1:
University Demographics
 Current enrollments by college and by sex appear in the
following table.
College
Ag-For
Arts-Sci
Bus-Econ
Female
500
1500
400
Male
900
1200
1400
2700
Totals
Educ
1000
Engr
Law
Undecl
Totals
200
100
800
4500
500
500 1300
200
900
5500
900
1500 1500
300
1700
10000
 If we select a student at random, what is the probability that the
student is :
 A female or male, i.e., P(Female or Male).
 Not from Agricultural and Forestry, i.e., P(not-Ag-For)
 A female given that the student is known to be from BusEcon,
i.e., P(Female |BusEcon).
 A female and from BusEcon, i.e., P(Female and BusEcon).
 From BusEcon, i.e., P(BusEcon).
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Lesson3-10
Learning exercise 1:
University Demographics
College
Ag-For
Arts-Sci
Bus-Econ
Educ
Engr
Female
500
1500
400
1000
200
100
800
4500
Male
900
1200
500
500 1300
200
900
5500
1400
2700
900
1500 1500
300
1700
10000
Totals
Law
Undecl
Totals
P(Female or Male) =(4500 + 5500)/10000 = 1
P(not-Ag-For) =(10000 – 1400) /10000 = 0.86
P(Female | Bus Econ) = 400 /900 = 0.44
P(Female and BusEcon) = 400 /10000 = 0.04
P(BusEcon) = 900 /10000 = 0.09
P(Female and BusEcon) = P(BusEcon) P(Female | Bus Econ)
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Definition: Mutually Exclusive events
 Events are mutually exclusive if the occurrence of any one
event means that none of the others can occur at the same
time.
 Peter’s event and Paul’s event are not mutually
exclusive – both contains 2.
 Peter’s event and Mary’s event are not mutually
exclusive – both contains 6.
 Paul’s event and Mary’s event are mutually exclusive –
no common numbers.
 Peter’s event and Sonia’s event are mutually exclusive –
no common numbers.
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Definition: Exhaustive events
 Events are collectively exhaustive if at least one of the
events must occur when an experiment is conducted.
 Peter’s event (even numbers) and Sonia’s event (odd
numbers) are collectively exhaustive.
 Peter’s event (even numbers) and Mary’s event
(number 6) are not collectively exhaustive.
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Lesson3-13
Conditional Probability
 A conditional probability is the probability of a particular
event occurring, given that another event has occurred.
 The probability of the event A given that the event B has
occurred is written P(A|B).
 P(female | BusEcon)
 P(girl | test says girl)
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Lesson3-14
Learning exercise 2:
Predicting Sex of Babies
 Many couples take advantage of ultrasound exams to determine
the sex of their baby before it is born. Some couples prefer not to
know beforehand. In any case, ultrasound examination is not
always accurate. About 1 in 5 predictions are wrong.
 In one medical group, the proportion of girls correctly
identified is 9 out of 10, i.e., applying the test to 100 baby girls,
90 of the tests will indicate girls.
and
 the number of boys correctly identified is 3 out of 4.
i.e., applying the test to 100 baby boys, 75 of the tests will
indicate boys.
 The proportion of girls born is 48 out of 100.
 What is the probability that a baby predicted to be a girl actually
turns out to be a girl?
Formally, find P(girl | test says girl).
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Lesson3-15
Learning exercise 2:
Predicting Sex of Babies
 P(girl | test says girl)
 In one medical group, the proportion of girls correctly
identified is 9 out of 10 and
 the number of boys correctly identified is 3 out of 4.
 The proportion of girls born is 48 out of 100.
 Think about the next 1000 births handled by this medical group.
 480 = 1000*0.48 are girls
 520 = 1000*0.52 are boys
 Of the girls, 432 (=480*0.9) tests indicate that they are girls.
 Of the boys, 130 (=520*0.25) tests indicate that they are
girls.
 In total, 562 (=432+130) tests indicate girls. Out of these
562 babies, 432 are girls.
 Thus P(girl | test says girl ) = 432/562 = 0.769
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Learning exercise 2:
Predicting Sex of Babies
With the information given, we can fill in the following table
in sequence from (1) to (9), with the initial assumption of
1000 babies in total.
For the question at hand, i.e., P(girl | test says girl ), we
only need to fill in the cells from (1) to (6).
Test says girl
Test says boy
Totals
Girl
(4) 432
(7) 48
(2) 480
Boy
(5) 130
(8) 390
(3) 520
Totals
(6) 562
(9) 438
(1) 1000
P(girl | test says girl ) = 432/562 = 0.769
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Lesson3-17
Learning exercise 2:
Predicting Sex of Babies
1000*P(girls)
 480 = 1000*0.48 are girls
 520 = 1000*0.52 are boys
1000*P(boys)
 Of the girls, 432 (=480*0.9) tests indicate that they are girls.
1000*P(girls)*P(test says girls|girls)
 Of the boys, 130 (=520*0.25) tests indicate that they are girls.
1000*P(boys)*P(test says girls | boys)
 In total, 562 tests indicate girls.
1000*[P(girls)*P(test says girls|girls) + P(boys)*P(test says girls|boys)]
 Out of these 562 babies, 432 are girls.
 Thus P(girls | test syas girls ) = 432/562 = 0.769
1000*P(girls)*P(test says girls|girls)
1000*[P(girls)*P(test says girls|girls) + P(boys)*P(test says girls|boys)]
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ECON1003: Analysis of Economic Data
Lesson3-18
Learning exercise 3:
Putting in Extra Trunk Lines
 Given recent flooding (or other condition more appropriate
to your area) between Town A and Town B, the local
telephone company is assessing the value of adding an
independent trunk line between the two towns. The second
line will fail independently of the first because it will
depend on different equipment and routing (we assume a
regional disaster is highly unlikely).
 Under current conditions, the present line works 98 out of
100 times someone wishes to make a call. If the second
line performs as well, what is the chance that a caller will
be able to get through? Formally,
 P( Line 1 works ) = 98/100
 P( Line 2 works ) = 98/100
 Find P( Line 1 or Line 2 works ).
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Lesson3-19
Learning exercise 3:
Putting in Extra Trunk Lines
 P( Line 1 works ) = 98/100
 P( Line 2 works ) = 98/100
 Find P( Line 1 or Line 2 works ).
P( Line 1 or Line 2 works )
= 1 – P(Both line1 and Line 2 fail)
= 1 – P(Line 1 fails)*P(line 2 fails)
= 1 – 0.02*0.02
= 0.9996.
Line 2 works
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Line 2 fails
Line 1 works
0.98*0.98
0.98*0.02
Line 1 fails
0.02*0.98
0.02*0.02
ECON1003: Analysis of Economic Data
Lesson3-20
Learning exercise 3:
Putting in Extra Trunk Lines
 P( individual line works ) = 50/100
 Find the number of lines we must install so that P(at least
one of the lines works ) is larger than 0.9.
Suppose we install k independent lines
P( at least one of the k lines works )
= 1 – P(all the k lines fail)
= 1 – P(Line 1 fails)*P(line 2 fails)*… *P(line k fails)
= 1 – 0.5k
We want to find the smallest integer k such that
1 – 0.5k >0.9 or 0.1> 0.5k
Log 0.1< k log(0.5) or k>log 0.1 / log(0.5) = 3.32
Hence k=4.
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Lesson3-21
Basic Rules of Probability
If two events A and B are mutually exclusive, the
special rule of addition states that the probability of
A or B occurring equals the sum of their respective
probabilities:
P(A or B) = P(A) + P(B)
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EXAMPLE 3
 New England Commuter Airways recently supplied the
following information on their commuter flights from
Boston to New York:
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Arrival
Frequency
Early
100
On Time
800
Late
75
Canceled
25
Total
1000
ECON1003: Analysis of Economic Data
Lesson3-23
EXAMPLE 3
Arrival
Early
Frequency
100
On Time
800
Late
75
Canceled
25
Total
1000
 If A is the event that a
flight arrives early, then
P(A) = 100/1000 = .10.
 If B is the event that a
flight arrives late, then
P(B) = 75/1000 = .075.
 The probability that a flight is either early or late is:
P(A or B) = P(A) + P(B) = .10 + .075 =.175.
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Lesson3-24
The Complement Rule
 The complement rule is used to determine the probability
of an event occurring by subtracting the probability of the
event not occurring from 1.
 If P(A) is the probability of event A and P(~A) is the
complement of A,
P(A) + P(~A) = 1 or P(A) = 1 - P(~A).
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Lesson3-25
The Complement Rule continued
 A Venn diagram illustrating the complement rule would
appear as:
A
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~A
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Lesson3-26
EXAMPLE 4
 Recall EXAMPLE 3. Use the complement rule to find the
probability of an early (A) or a late (B) flight
Arrival
Early
Frequency
100
On Time
800
Late
75
Canceled
25
Total
1000
 P(A or B) = 1 - P(C or D)
 If C is the event that a flight
arrives on time, then
P(C) = 800/1000 = .8.
 If D is the event that a flight
is canceled, then
P(D) = 25/1000 = .025.
P(A or B) = 1 - P(C or D) = 1 - [.8 +.025] =.175
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Lesson3-27
EXAMPLE 4 continued
 P(A or B) = 1 - P(C or D)
= 1 - [.8 +.025] =.175
C
.8
D
.025
~(C or D) = (A or B)
.175
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Lesson3-28
The General Rule of Addition
 If A and B are two events that are not mutually exclusive,
then P(A or B) is given by the following formula:
P(A or B) = P(A) + P(B) - P(A and B)
B
A and B
A
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Lesson3-29
EXAMPLE 5
 In a sample of 500 students, 320 said they had a stereo,
175 said they had a TV, and 100 said they had both:
Both
100
Stereo
320 220
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TV
175 75
ECON1003: Analysis of Economic Data
Lesson3-30
EXAMPLE 5 continued
 In a sample of 500 students, 320 said they had a stereo, 175
said they had a TV, and 100 said they had both.
 If a student is selected at random, what is the probability
that the student has only a stereo, only a TV, and both a
stereo and TV?
P(student has a stereo) = 320/500 = .64.
P(student has a TV) = 175/500 = .35.
P(student has both a stereo and TV) = 100/500 = .20.
P(student as only a stereo) = 220/500 = .44.
P(student has only a TV) = 75/500 = .15.
P(student as only a stereo) = P(student has a stereo)
- P(student has both a stereo and TV)
P(student has only a TV) = P(student has a TV)
- P(student has both a stereo and TV)
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Lesson3-31
EXAMPLE 5 continued
 In a sample of 500 students, 320 said they had a stereo,
175 said they had a TV, and 100 said they had both.
P(S) = 320/500 = .64.
P(T) = 175/500 = .35.
P(S and T) = 100/500 = .20.
 If a student is selected at random, what is the
probability that the student has either a stereo or a TV
in his or her room?
P(either S or T)
= P(only S) + P (only T)
= [P(S) - P(S and T)] +[P(T) - P(S and T)]
= .64 - .20 +.35 - .20 = .59.
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Lesson3-32
Joint Probability
 A joint probability measures the likelihood that two or
more events will happen concurrently.
 An example would be the event that a student has both a
stereo and TV in his or her dorm room.
 In an experiment of throwing only one die, let x denote the
outcome,
P(x=5, and x=6) =0
because the two outcomes are mutually exclusive.
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Lesson3-33
Special Rule of Multiplication
 The special rule of multiplication requires that two events
A and B are independent.
 Two events A and B are independent if the occurrence of
one has no effect on the probability of the occurrence of
the other.
 This rule is written:
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P(A and B) = P(A)P(B)
ECON1003: Analysis of Economic Data
Lesson3-34
EXAMPLE 6
 Chris owns two stocks, IBM and General Electric (GE). The
probability that IBM stock will increase in value next year
is .5 and the probability that GE stock will increase in
value next year is .7. Assume the two stocks are
independent. What is the probability that both stocks will
increase in value next year?
P(IBM and GE) = (.5)(.7) = .35.
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Lesson3-35
EXAMPLE 6 continued
 The probability that IBM stock will increase in value
next year is .5 and the probability that GE stock will
increase in value next year is .7.
 What is the probability that at least one of these
stocks increase in value during the next year? (This
means that either one can increase or both.)
Approach 1:
P(at least one)
= P(GE only) + P(IBM only) + P(both)
= (.5)(.3) + (.5)(.7) +(.7)(.5)
= .85.
Approach 2:
P(at least one)
= P(GE) + P(IBM) – P(both)
= 0.5 + 0.7 – 0.35
= .85.
Approach 3:
P(at least one) = 1- P(both GE and IBM do not increase)
= 1 – 0.15 = .85.
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Lesson3-36
General Multiplication Rule
 The general rule of multiplication is used to find the joint
probability that two events will occur.
 It states that for two events A and B, the joint probability
that both events will happen is found by multiplying the
probability that event A will happen by the conditional
probability of B given that A has occurred.
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Lesson3-37
General Multiplication Rule
 The joint probability, P(A and B) is given by the following
formula:
P(A and B) = P(A)P(B|A)
or
P(A and B) = P(B)P(A|B)
 Examples:
 P(test says girl and girl)
= P(girls) * P(test says girls | girls)
 P(test says boy and boy)
= P(boys) * P(test says boys | boys)
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Lesson3-38
Independence -- an illustration
 Consider whether the decision of a young man going to party
depends on whether his girlfriend goes to the same party.
 Assume the probability of the young man going to party is 0.7
(i.e., he goes to 70 out of 100 parties on average).
 If he tends to go to whichever party his girlfriend goes, his party
behavior depends on his girlfriend’s. That is, the probability of
going to a party conditional on his girlfriend’s presence is larger
than 0.7 (extreme case being 1.0).
 If he tends to avoid going to whichever party his girlfriend goes,
his party behavior also depends on his girlfriend’s. That is, the
probability of going to a party conditional on his girlfriend’s
presence is less than 0.7 (extreme case being 0.0).
 If in making the party decision, he never considers whether his
girlfriend is going to a party, his party behavior does not depends
on his girlfriend’s. That means, the probability of going to a party
conditional on his girlfriend’s presence is 0.7.
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ECON1003: Analysis of Economic Data
Lesson3-39
Independence -- an illustration
 Define events:
 A: a young man goes to a party
 B: his girlfriend goes to the same party.
 Assume P(A) =0.7
 His party behavior does not depend on his girlfriend’s only if
P(A|B) =P(A) = 0.7. And, event A is said to be independent of
event B.
 P(the young man and his girlfriends shows up in a party) = P(A &
B) = P(B)*P(A|B).
 If he always goes to whichever party his girlfriend goes,
P(A|B) = 1. Hence, P(A & B) = P(B)*P(A|B) = P(B).
 If he always avoid to whichever party his girlfriend goes,
P(A|B) = 0. Hence, P(A & B) = P(B)*P(A|B) = 0.
 If in making the party decision, he never considers whether
his girlfriend is going to a party, P(A|B) = 0.7. Hence, P(A & B)
= P(B)*P(A|B) = P(B)*P(A) = 0.7.
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ECON1003: Analysis of Economic Data
Lesson3-40
Independence
 Event A is independent of B
 P(A|B) = P (A)
 Event B is independent of A
 P(B|A) = P (B)
 P(B|A) = P (B) implies
P(A&B) = P(B|A) * P(A) = P(B) * P(A)
 P(A|B) = P (A) implies
P(A&B) = P(A|B) * P(B) = P(A) * P(B)
 Thus, if P(A&B) = P(B) * P(A), we must have either P(A|B)
= P (A) or P(B|A) = P (B).
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ECON1003: Analysis of Economic Data
Lesson3-41
EXAMPLE 7
The Dean of the School of Business at Owens University
collected the following information about undergraduate
students in her college:
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MAJOR
Male
Female
Total
Accounting
170
110
280
Finance
120
100
220
Marketing
160
70
230
Management
150
120
270
Total
600
400
1000
ECON1003: Analysis of Economic Data
EXAMPLE 7 continued
MAJOR
Male
Female
Total
Accounting
170
110
280
Finance
120
100
220
Marketing
160
70
230
Management
150
120
270
Total
600
400
1000
If a student is selected at
random, what is the probability
that the student is a female (F)
accounting major (A)
P(A and F) = 110/1000.
Given that the student is a female, what is the probability
that she is an accounting major?
Approach 1:
P(A|F) = P(A and F)/P(F)
= [110/1000]/[400/1000]
= .275
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Approach 2:
P(A|F)
= 110/400
= .275
ECON1003: Analysis of Economic Data
Lesson3-43
Tree Diagrams
 A tree diagram is useful for portraying conditional and
joint probabilities. It is particularly useful for analyzing
business decisions involving several stages.
 EXAMPLE 8: In a bag containing 7 red chips and 5 blue
chips you select 2 chips one after the other without
replacement. Construct a tree diagram showing this
information.
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Lesson3-44
EXAMPLE 8 continued
6/11
7/12
5/12
R2
R1
5/11
B2
7/11
R2
B1
4/11
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B2
Lesson3-45
EXAMPLE 8 continued
The tree diagram is very illustrative about the relation
between joint probability and conditional probability
Let A (B) be the event of a red chip in the first (second) draw.
6/11
P(B|A) = 6/11
R2
P(A) = 7/12
7/12
5/12
R1
5/11
B2
7/11
R2
B1
4/11
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P(A and B)
= P(A)*P(B|A)
= 6/11 * 7/12
B2
ECON1003: Analysis of Economic Data
Lesson3-46
Bayes’ Theorem
 Bayes’ Theorem is a method for revising a probability
given additional information.
 If A1 and A2 are mutually exclusive and exhaustive, and A1
and A2 together covers event B, we have :
P(A1 )P(B | A1 )
P(A1 | B) 
P(A1 )P(B | A1 )  P(A 2 )P(B | A 2 )
Bayes Theorem is an essential tool to understand options
and real options in finance.
Ka-fu Wong © 2004
ECON1003: Analysis of Economic Data
Lesson3-47
Bayes’ Theorem
 A1 and A2 are mutually exclusive means
P (A1 or A2 ) = P(A1) + P(A2 )
P (A1 and A2 ) = 0
 A1 and A2 are exhaustive means
P (A1 or A2 ) = P(A1) + P(A2 ) = 1
 A1 and A2 together covers event B means
P(B) = P(A1 & B) + P(A2 & B)
(see the Venn diagram illustration)
Ka-fu Wong © 2004
ECON1003: Analysis of Economic Data
Lesson3-48
Venn Diagram illustration of Bayes’
theorem
A1
B
A1 & B
Ka-fu Wong © 2004
A2
A2 & B
ECON1003: Analysis of Economic Data
Lesson3-49
Bayes’ Theorem
P(A1 | B) 
P(A1 )P(B | A1 )
P(A1 )P(B | A1 )  P(A 2 )P(B | A 2 )
 Bayes’ Theorem can be derived based on simple manipulation of
the general multiplication rule.
P(A1|B)
= P(A1 & B) /P(B)
= [P(A1) P(B|A1)] / P(B)
= [P(A1) P(B|A1)] / [P(A1 & B) + P(A2 & B)]
= [P(A1) P(B|A1) ]/ [P(A1) P(B|A1) + P(A2) P(B|A2)]
Ka-fu Wong © 2004
ECON1003: Analysis of Economic Data
Lesson3-50
EXAMPLE 9
 Duff Cola Company recently received several complaints
that their bottles are under-filled. A complaint was
received today but the production manager is unable to
identify which of the two Springfield plants (A or B) filled
this bottle. The following table summarizes the Duff
production experience.
% of Total Production
% of under-filled bottles
A
55
3
B
45
4
 What is the probability that the under-filled bottle came
from plant A?
Ka-fu Wong © 2004
ECON1003: Analysis of Economic Data
Lesson3-51
Example 9 continued
% of Total Production
% of under-filled bottles
A
55
3
B
45
4
 What is the probability that the under-filled bottle came
from plant A?
P(A)P(U | A)
P(A/U) 
P(A)P(U | A)  P(B)P(U | B)
.55(.03)

 .4783
.55(.03)  .45(.04)
The likelihood the bottle was filled in Plant A is reduced from
.55 to .4783.
Without the information about U, the manager will say the
under-filled bottle is likely from plant A. With the additional
information about U, the manager will say the under-filled
bottle is likely from plant B.
Ka-fu Wong © 2004
ECON1003: Analysis of Economic Data
Lesson3-52
Probabilities for throwing n Dice
 Suppose we roll n regular balanced six-sided dice. If the
event X is the sum of the n values which appear, what are
the probabilities associated for each value of X, for the
possible values X = n, ... , 6n?
 In the case n =1, these probabilities are all 1/6.
Ka-fu Wong © 2004
ECON1003: Analysis of Economic Data
Lesson3-53
Probabilities for throwing n Dice
For two dice , it is easiest to
consider a table of possible
outcomes:
Next, we can consider the
sums associated with these
outcomes:
(1,1) (1,2) (1,3) (1,4) (1,5) (1,6)
2
3
4
5
6
7
(2,1) (2,2) (2,3) (2,4) (2,5) (2,6)
3
4
5
6
7
8
(3,1) (3,2) (3,3) (3,4) (3,5) (3,6)
4
5
6
7
8
9
(4,1) (4,2) (4,3) (4,4) (4,5) (4,6)
5
6
7
8
9
10
(5,1) (5,2) (5,3) (5,4) (5,5) (5,6)
6
7
8
9
10
11
(6,1) (6,2) (6,3) (6,4) (6,5) (6,6)
7
8
9
10
11
12
Ka-fu Wong © 2004
ECON1003: Analysis of Economic Data
Lesson3-54
Probabilities for throwing n Dice
Next, we can consider
the sums associated
with these outcomes:
Since there are 36 outcomes, each
equally likely, we see that the
probabilities are:
X
#
P(X)
X
#
P(X)
2
1
1/36
8
5
5/36
3
2
2/36
9
4
4/36
4
3
3/36
10
3
3/36
2
3
4
5
6
7
3
4
5
6
7
8
4
5
6
7
8
9
5
6
7
8
9
10
6
7
8
9
10
11
5
4
4/36
11
2
2/36
7
8
9
10
11
12
6
5
5/36
12
1
1/36
7
6
6/36
Ka-fu Wong © 2004
ECON1003: Analysis of Economic Data
Lesson3-55
Probabilities for throwing n Dice
 If there are more than two dice, it is difficult to make
tables. The number of outcomes will be 6n, so to
calculate the probabilities it is sufficient to count
the number of times each sum occurs among the 6n
possible outcomes. This is easily done by considering
the generating function for the number of times each
sum appears:
f(x) = (x + x2 + x3 + x4 + x 5+x6)n
 In our example, n=2
(x + x2 + x3 + x4 + x 5+x6)2
=x2 + 2x3 +3x4 + 4x5+5x6+6x7+5x8+4x9+3x10+2x11+x12
 The coefficient of x3 is number of times the sum of 3
occurs.
Note: The formula is for reference only. No need to memorize it.
Ka-fu Wong © 2004
ECON1003: Analysis of Economic Data
Lesson3-56
Some Principles of Counting
 The multiplication formula indicates that if there are m
ways of doing one thing and n ways of doing another thing,
there are m x n ways of doing both.
 Example 10: Dr. Delong has 10 shirts and 8 ties. How
many shirt and tie outfits does he have?
(10)(8) = 80
Ka-fu Wong © 2004
ECON1003: Analysis of Economic Data
Lesson3-57
Some Principles of Counting
 A permutation is any arrangement of r objects selected
from n possible objects.
 Note: The order of arrangement is important in
permutations.
n!
n Pr 
(n  r)!
n! = n(n-1) (n-2)… 21
n=1, n!=1
n=2, n!=21=2
n=3, n!=321=6
n=4, n!=4321=24
n=5, n!=54321=120
Ka-fu Wong © 2004
ECON1003: Analysis of Economic Data
Lesson3-58
Some Principles of Counting
 A combination is the number of ways to choose r objects
from a group of n objects without regard to order.
n!
n Cr 
r!(n  r)!
Ka-fu Wong © 2004
ECON1003: Analysis of Economic Data
Lesson3-59
EXAMPLE 11
 There are 12 players on the Carolina Forest High
School basketball team. Coach Thompson must pick
five players among the twelve on the team to
comprise the starting lineup. How many different
groups are possible?
12!
 792
12 C 5 
5!(12  5)!
Ka-fu Wong © 2004
ECON1003: Analysis of Economic Data
Lesson3-60
Example 11 continued
 Suppose that in addition to selecting the group, he must
also rank each of the players in that starting lineup
according to their ability.
12!
 95,040
12 P 5 
(12  5)!
Ka-fu Wong © 2004
ECON1003: Analysis of Economic Data
Lesson3-61
Lesson 3:
A Survey of Probability Concepts
- END -
Ka-fu Wong © 2004
ECON1003: Analysis of Economic Data
Lesson3-62