Transcript No Slide Title

Lesson 3:
A Survey of Probability
Concepts
Ka-fu Wong © 2007
ECON1003: Analysis of Economic Data
Lesson3-1
Outline
Learning Exercises
Definitions
Basic rules of Probability
Independence
Tree Diagram
Bayes’ Theorem
Ka-fu Wong © 2007
ECON1003: Analysis of Economic Data
Lesson3-2
Learning exercise 1:
University Demographics
 Current enrollments by college and by sex appear in the
following table.
College
Ag-For
Arts-Sci
Bus-Econ
Educ
Engr
Law
Undecl
Totals
Female
500
1500
400
1000
200
100
800
4500
Male
900
1200
500
500
1300
200
900
5500
Totals
1400
2700
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).
Ka-fu Wong © 2007
ECON1003: Analysis of Economic Data
Lesson3-3
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 | BusEcon) = 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 | BusEcon)
Ka-fu Wong © 2007
ECON1003: Analysis of Economic Data
Lesson3-4
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).
Ka-fu Wong © 2007
ECON1003: Analysis of Economic Data
Lesson3-5
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
Ka-fu Wong © 2007
ECON1003: Analysis of Economic Data
Lesson3-6
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
Ka-fu Wong © 2007
ECON1003: Analysis of Economic Data
Lesson3-7
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)]
Ka-fu Wong © 2007
ECON1003: Analysis of Economic Data
Lesson3-8
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.
Ka-fu Wong © 2007
ECON1003: Analysis of Economic Data
Definitions: Experiment and outcome
 A random 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.
 Sample Space – the collection of all possible outcomes of a
random experiment.
Ka-fu Wong © 2007
ECON1003: Analysis of Economic Data
Lesson3-10
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].
Ka-fu Wong © 2007
ECON1003: Analysis of Economic Data
Lesson3-11
Definition: Intersection of Events
 Intersection of Events – If A and B are two events in a sample
space S, then the intersection, A ∩ B, is the set of all outcomes in S
that belong to both A and B
S
A




The intersection
The intersection
The intersection
The intersection
Ka-fu Wong © 2007
AB
B
of Peter’s event and Paul’s event contains 2.
of Peter’s event and Mary’s event contains 6.
of Paul’s event and Mary’s event contains nothing.
of Peter’s event and Sonia’s event contains nothing.
ECON1003: Analysis of Economic Data
Lesson3-12
Definition: Mutually Exclusive events
 A and B are Mutually Exclusive Events if they have no basic
outcomes in common
 i.e., the set A ∩ B is empty
S
A
Ka-fu Wong © 2007
B
ECON1003: Analysis of Economic Data
Lesson3-13
Example 2: Intersections and Mutually
Exclusive events
 Peter’s event and Paul’s event are not mutually exclusive – both
contains 2.
 The intersection of Peter’s event and Paul’s event contains 2.
 Peter’s event and Mary’s event are not mutually exclusive – both
contains 6.
 The intersection of Peter’s event and Mary’s event contains 6.
 Paul’s event and Mary’s event are mutually exclusive – no
common numbers.
 The intersection of Paul’s event and Mary’s event contains nothing.
 Peter’s event and Sonia’s event are mutually exclusive – no
common numbers.
 The intersection of Peter’s event and Sonia’s event contains nothing.
Ka-fu Wong © 2007
ECON1003: Analysis of Economic Data
Lesson3-14
Definition: Union
 Union of Events – If A and B are two events in a sample
space S, then the union, A U B, is the set of all outcomes in
S that belong to either
A or B
S
A
Ka-fu Wong © 2007
B
The entire shaded
area represents
AUB
ECON1003: Analysis of Economic Data
Lesson3-15
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.
 Events E1, E2, … Ek are Collectively Exhaustive events if
E1 U E2 U . . . U Ek = S
 i.e., the events completely cover the sample space
Ka-fu Wong © 2007
ECON1003: Analysis of Economic Data
Lesson3-16
Complement
 The Complement of an event A is the set of all basic
outcomes in the sample space that do not belong to A. The
complement is denoted A
S
A
Ka-fu Wong © 2007
A
ECON1003: Analysis of Economic Data
Lesson3-17
Example 3
Let the Sample Space be the collection of all possible
outcomes of rolling one die:
S = [1, 2, 3, 4, 5, 6]
Let A be the event “Number rolled is even”
Let B be the event “Number rolled is at least 4”
Then the two events contains
A = [2, 4, 6]
Ka-fu Wong © 2007
B = [4, 5, 6]
ECON1003: Analysis of Economic Data
Lesson3-18
Example 3
S = [1, 2, 3, 4, 5, 6]
A = [2, 4, 6]
B = [4, 5, 6]
Complements:
B  [1, 2, 3]
A  [1,3, 5]
Intersections:
A  B  [4, 6]
Unions:
A  B  [5]
A  B  [2, 4, 5, 6]
A  A  [1, 2, 3, 4, 5, 6]  S
Ka-fu Wong © 2007
ECON1003: Analysis of Economic Data
Lesson3-19
Example 3
S = [1, 2, 3, 4, 5, 6]
A = [2, 4, 6]
B = [4, 5, 6]
 Mutually exclusive:
 Are A and B mutually exclusive?
No. The outcomes 4 and 6 are common to both
 Collectively exhaustive:
 Are A and B collectively exhaustive?
No. A U B does not contain 1 and 3
Ka-fu Wong © 2007
ECON1003: Analysis of Economic Data
Lesson3-20
Probability
 Probability – the chance that an
uncertain event will occur (always
between 0 and 1)
0 ≤ P(A) ≤ 1 For any event A
1
.5
0
Ka-fu Wong © 2007
ECON1003: Analysis of Economic Data
Certain
Impossible
Lesson3-21
Assessing Probability
 There are three approaches to assessing the probability of an
uncertain event:
1. classical probability
probability of event A 
NA
number of outcomes that satisfythe event

N
total number of outcomes in the sample space
(Assumes all outcomes in the sample space are equally likely to occur.)
Ka-fu Wong © 2007
ECON1003: Analysis of Economic Data
Lesson3-22
Counting the Possible Outcomes
 Use the Combinations formula to determine the
number of combinations of n things taken k at a time
C(n,k) 
n!
k!(n  k)!
 where
 n! = n(n-1)(n-2)…(1)
 0! = 1 by definition
Ka-fu Wong © 2007
ECON1003: Analysis of Economic Data
Lesson3-23
Example 4: Combination
C(n,k) 
n!
k!(n  k)!
 Suppose there are 2 stocks in our portfolio. We would like to
select 1 stocks and sell them. What are all the possible
combinations.
 [1], [2]
 C(2,1) = 2!/(1!(2-1)!)=2
Ka-fu Wong © 2007
ECON1003: Analysis of Economic Data
Lesson3-24
Example 4: Combination
n!
C(n,k) 
k!(n  k)!
 Suppose there are 3 stocks in our portfolio. We would like to
select 1 stocks and sell them. What are all the possible
combinations.
 [1], [2], [3]
 C(3,1) = 3!/(1!(3-1)!)=3*2*1/[1*(2*1)]=3
 Suppose there are 3 stocks in our portfolio. We would like to
select 2 stocks and sell them. What are all the possible
combinations.
 [1,2], [1,3], [2,3]
 C(3,2) = 3!/(2!(3-2)!)=3*2*1/[(2*1)*1]=3
Ka-fu Wong © 2007
ECON1003: Analysis of Economic Data
Lesson3-25
Example 5: Combination
n!
C(n,k) 
k!(n  k)!
 Suppose there are 4 stocks in our portfolio. We would like to
select 1 stocks and sell them. What are all the possible
combinations.
 [1], [2], [3], [4]
 C(4,1) = 4!/(1!(4-1)!)=4*3*2*1/[1*(3*2*1)]=4
 Suppose there are 4 stocks in our portfolio. We would like to
select 2 stocks and sell them. What are all the possible
combinations.
 [1,2], [1,3], [1,4], [2,3], [2,4], [3,4]
 C(4,2) = 4!/(2!(4-2)!)=4*3*2*1/[(2*1)*(2*1)]=6
Ka-fu Wong © 2007
ECON1003: Analysis of Economic Data
Lesson3-26
Assessing Probability
Three approaches (continued)
2. relative frequency probability
probability of event A 
nA
number of events in the population that satisfyevent A

n
total number of events in the population
the limit of the proportion of times that an event A occurs
in a large number of trials, n
3. subjective probability
an individual opinion or belief about the probability of occurrence
Ka-fu Wong © 2007
ECON1003: Analysis of Economic Data
Lesson3-27
Probability Postulates
1. If A is any event in the sample space S, then
0  P(A)  1
2. Let A be an event in S, and let Oi denote the basic outcomes (mutually
exclusive). Then
P(A)  P(Oi )
A
(the notation means that the summation is over all the basic
outcomes in A)
3.
P(S) = 1
Ka-fu Wong © 2007
ECON1003: Analysis of Economic Data
Lesson3-28
Probability Rules
 The Complement rule:
P( A)  1 P(A)
i.e., P(A)  P( A)  1
 The Addition rule:
 The probability of the union of two events is
P(A  B)  P(A)  P(B)  P(A  B)
Ka-fu Wong © 2007
ECON1003: Analysis of Economic Data
Lesson3-29
A Probability Table
Probabilities and joint probabilities for two events A
and B are summarized in this table:
B
Ka-fu Wong © 2007
B
A
P(A  B)
P(A  B)
P(A)
A
P( A  B)
P( A  B)
P( A)
P(B)
P( B )
P(S)  1.0
ECON1003: Analysis of Economic Data
Lesson3-30
Example 5: Addition Rule
P(A  B)  P(A)  P(B)  P(A  B)
Consider a standard deck of 52 cards, with four suits:
♥♣♦♠
Let event A = card is an Ace
Let event B = card is from a red suit
Ka-fu Wong © 2007
ECON1003: Analysis of Economic Data
Lesson3-31
Example 5: Addition Rule
P(A  B)  P(A)  P(B)  P(A  B)
P(Red U Ace) = P(Red) + P(Ace) - P(Red ∩ Ace)
= 26/52 + 4/52 - 2/52 = 28/52
Type
Color
Red
Black
Total
Ace
2
2
4
Non-Ace
24
24
48
Total
26
26
52
Ka-fu Wong © 2007
ECON1003: Analysis of Economic Data
Don’t count
the two red
aces twice!
Lesson3-32
Conditional Probability
 A conditional probability is the probability of one event,
given that another event has occurred:
P(A  B)
P(A | B) 
P(B)
The conditional
probability of A given
that B has occurred
P(A  B)
P(B | A) 
P(A)
The conditional
probability of B given
that A has occurred
Ka-fu Wong © 2007
ECON1003: Analysis of Economic Data
Lesson3-33
Example 6: Conditional Probability
P(A | B) 
P(A  B)
P(B)
 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.
 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)
Ka-fu Wong © 2007
ECON1003: Analysis of Economic Data
Lesson3-34
Example 6: Conditional Probability
P(A | B) 
P(A  B)
P(B)
 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.
Ka-fu Wong © 2007
CD
No CD
Total
AC
.2
.5
.7
No AC
.2
.1
.3
Total
.4
.6
1.0
ECON1003: Analysis of Economic Data
Lesson3-35
Example 6: Conditional Probability
P(A | B) 
P(A  B)
P(B)
 Given AC, we only consider the top row (70% of the cars). Of
these, 20% have a CD player. 20% of 70% is 28.57%.
CD
No CD
Total
AC
.2
.5
.7
No AC
.2
.1
.3
Total
.4
.6
1.0
P(CD  AC) .2
P(CD | AC) 
  .2857
P(AC)
.7
Ka-fu Wong © 2007
ECON1003: Analysis of Economic Data
Lesson3-36
Multiplication Rule
 Multiplication rule for two events A and B:
P(A  B)  P(A | B) P(B)
 also
P(A  B)  P(B | A)P(A)
Examples:
1. P(test says girl and girl) = P(girls) * P(test says girls | girls)
2. P(test says boy and boy) = P(boys) * P(test says boys | boys)
Ka-fu Wong © 2007
ECON1003: Analysis of Economic Data
Lesson3-37
Example 7: Multiplication Rule P(A  B) P(A | B)P(B)
P(Red ∩ Ace) = P(Red| Ace)P(Ace)
 2  4  2
    
 4  52  52
number of cards that are red and ace
2


total number of cards
52
Type
Ka-fu Wong © 2007
Color
Red
Black
Total
Ace
2
2
4
Non-Ace
24
24
48
Total
26
26
52
ECON1003: Analysis of Economic Data
Lesson3-38
Example 8: Independence
Should I go to a party without my girlfriend?
 The probability of the my going to party is 0.7 (i.e., I go to
70 out of 100 parties on average).
 If I tend to go to whichever party my girlfriend (Venus)
goes, my party behavior depends on Venus’s. That is, my
probability of going to a party conditional on Venus’s
presence is larger than 0.7 (extreme case being 1.0).
 If I tends to avoid going to whichever party Venus goes,
my party behavior also depends on Venus. That is, my
probability of going to a party conditional on Venus’s
presence is less than 0.7 (extreme case being 0.0).
 If in making the party decision, I never consider whether
Venus is going to a party, my party behavior does not
depends on Venus’s. That means, the probability of going
to a party conditional on Venus’s presence is 0.7.
Ka-fu Wong © 2007
ECON1003: Analysis of Economic Data
Lesson3-39
Example 8: Independence
Should I go to a party without my girlfriend?
 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) = P(B)*0.7.
Ka-fu Wong © 2007
ECON1003: Analysis of Economic Data
Lesson3-40
Statistical Independence
 Two events are statistically independent if and only if:
P(A  B)  P(A)P(B)
 Events A and B are independent when the probability of
one event is not affected by the other event
 If A and B are independent, then
Ka-fu Wong © 2007
P(A | B)  P(A)
if P(B)>0
P(B | A)  P(B)
if P(A)>0
ECON1003: Analysis of Economic Data
Lesson3-41
Example 9: Statistical Independence
 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.
CD
No CD
Total
AC
.2
.5
.7
No AC
.2
.1
.3
Total
.4
.6
1.0
 Are the events AC and CD statistically independent?
Ka-fu Wong © 2007
ECON1003: Analysis of Economic Data
Lesson3-42
Example 9: Statistical Independence
CD
No CD
Total
AC
.2
.5
.7
No AC
.2
.1
.3
Total
.4
.6
1.0
P(AC ∩ CD) = 0.2
P(AC) = 0.7
P(AC)P(CD) = (0.7)(0.4) = 0.28
P(CD) = 0.4
P(AC ∩ CD) = 0.2 ≠ P(AC)P(CD) = 0.28
So the two events are not statistically independent
Ka-fu Wong © 2007
ECON1003: Analysis of Economic Data
Lesson3-43
Bivariate Probabilities
Outcomes for bivariate events:
B1
B2
...
Bk
A1
P(A1B1)
P(A1B2)
...
P(A1Bk)
A2
P(A2B1)
P(A2B2)
...
P(A2Bk)
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
Ah
P(AhB1)
P(AhB2)
...
P(AhBk)
Ka-fu Wong © 2007
ECON1003: Analysis of Economic Data
Lesson3-44
Joint and Marginal Probabilities
 The probability of a joint event, A ∩ B:
P(A  B) 
number of outcomes satisfyingA and B
total number of elementary outcomes
 Computing a marginal probability:
P(A) P(A  B1 )  P(A  B2 )  ...  P(A  Bk )
Where B1, B2, …, Bk are k mutually exclusive and
collectively exhaustive events
Ka-fu Wong © 2007
ECON1003: Analysis of Economic Data
Lesson3-45
Example 10: Marginal Probability
P(Ace)
 P(Ace  Red)  P(Ace  Black) 
Type
Ka-fu Wong © 2007
2
2
4


52 52 52
Color
Red
Black
Total
Ace
2
2
4
Non-Ace
24
24
48
Total
26
26
52
ECON1003: Analysis of Economic Data
Lesson3-46
Tree Diagrams
 A tree diagram is useful for portraying conditional and
joint probabilities. It is particularly useful for analyzing
business decisions involving several stages.
Ka-fu Wong © 2007
ECON1003: Analysis of Economic Data
Lesson3-47
EXAMPLE 11: Tree Diagram
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.
6/11
7/12
5/12
R2
R1
5/11
B2
7/11
R2
B1
4/11
Ka-fu Wong © 2007
ECON1003: Analysis of Economic Data
B2
Lesson3-48
EXAMPLE 11: Tree Diagram
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
Ka-fu Wong © 2007
P(A and B)
= P(A)*P(B|A)
= 6/11 * 7/12
B2
ECON1003: Analysis of Economic Data
Lesson3-49
Example 12: Using a Tree Diagram
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.
P(AC ∩ CD) = .2
All
Cars
P(AC ∩ CD) = .5
P(AC ∩ CD) = .2
P(AC ∩ CD) = .1
Ka-fu Wong © 2007
ECON1003: Analysis of Economic Data
Lesson3-50
Odds
 The odds in favor of a particular event are given by the
ratio of the probability of the event divided by the
probability of its complement
 The odds in favor of A are
P(A)
P(A)
odds 

1- P(A) P( A)
Ka-fu Wong © 2007
ECON1003: Analysis of Economic Data
Lesson3-51
Example 13: Odds
 Calculate the probability of winning if the odds of
winning are 3 to 1:
3
P(A)
odds  
1 1- P(A)
 Now multiply both sides by 1 – P(A) and solve for
P(A):
3 x (1- P(A)) = P(A)
3 – 3P(A) = P(A)
3 = 4P(A)
P(A) = 0.75
Ka-fu Wong © 2007
ECON1003: Analysis of Economic Data
Lesson3-52
Overinvolvement Ratio
 Does A1 depends on B1 and B2 differently?
 The probability of event A1 conditional on event B1 divided
by the probability of A1 conditional on activity B2 is defined
as the overinvolvement ratio:
P (A1 | B1 )
P (A1 | B2 )
 An overinvolvement ratio greater than 1 implies that event
A1 increases the conditional odds ratio in favor of B1:
P(B1 | A1 ) P(B1 )

P(B2 | A1 ) P(B2 )
Ka-fu Wong © 2007
ECON1003: Analysis of Economic Data
Lesson3-53
Example 14: Overinvolvement ratio
 If the probability to get lung cancer is 0.5% for smokers and 0.1%
for non-smokers, what is the overinvolvement ratio?
Prob(lung cancer | smokers) = 0.005
Prob(lung cancer | non-smokers) = 0.001
Overinvolvement ratio = P(L|S)/P(L|N) = 5
Ka-fu Wong © 2007
ECON1003: Analysis of Economic Data
Lesson3-54
Bayes’ Theorem
P (Ei | A) 

P (A | E i )P (Ei )
P (A)
P (A | E i )P (Ei )
P (A | E 1 )P (E1 )  P (A | E 2 )P (E2 )  ...  P (A | E k )P (Ek )
 where:
Ei = ith event of k mutually exclusive and collectively
exhaustive events
A = new event that might impact P(Ei)
Ka-fu Wong © 2007
ECON1003: Analysis of Economic Data
Lesson3-55
Bayes’ Theorem
P(A1 | B) 
P(A1 )P(B | A1 )
P(A1 )P(B | A1 )  P(A2 )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 © 2007
ECON1003: Analysis of Economic Data
Lesson3-56
Example 15: Bayes’ Theorem
 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?
Ka-fu Wong © 2007
ECON1003: Analysis of Economic Data
Lesson3-57
Example 15: Bayes’ Theorem
 Let S = successful well
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
 Goal is to find P(S|D)
Ka-fu Wong © 2007
ECON1003: Analysis of Economic Data
Lesson3-58
Example 15: Bayes’ Theorem
Apply Bayes’ Theorem:
P(D | S)P(S)
P(S | D) 
P(D | S)P(S)  P(D | U)P(U)
(.6)(.4)

(.6)(.4)  (.2)(.6)

.24
 .667
.24  .12
So the revised probability of success (from the original
estimate of .4), given that this well has been scheduled for
a detailed test, is .667
Ka-fu Wong © 2007
ECON1003: Analysis of Economic Data
Lesson3-59
Example 16: Bayes’ Theorem
 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 © 2007
ECON1003: Analysis of Economic Data
Lesson3-60
Example 16: Bayes’ Theorem
% 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 © 2007
ECON1003: Analysis of Economic Data
Lesson3-61
Lesson 3:
A Survey of Probability Concepts
- END -
Ka-fu Wong © 2007
ECON1003: Analysis of Economic Data
Lesson3-62