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EF 507
QUANTITATIVE METHODS FOR
ECONOMICS AND FINANCE
FALL 2008
Chapter 4
Probability
Chap 4-1
Important Terms




Random Experiment – a process leading to an
uncertain outcome
Basic Outcome – a possible outcome of a
random experiment
Sample Space – the collection of all possible
outcomes of a random experiment
Event – any subset of basic outcomes from the
sample space
Chap 4-2
Important Terms
(continued)

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
AB
B
Chap 4-3
Important Terms
(continued)

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
B
Chap 4-4
Important Terms
(continued)

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
B
The entire shaded
area represents
AUB
Chap 4-5
Important Terms
(continued)

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
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
A
Chap 4-6
Examples
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
A = [2, 4, 6]
and
B = [4, 5, 6]
Chap 4-7
Examples
(continued)
S = [1, 2, 3, 4, 5, 6]
A = [2, 4, 6]
B = [4, 5, 6]
Complements:
A  [1, 3, 5]
B  [1, 2, 3]
Intersections:
A  B  [4, 6]
Unions:
A  B  [5]
A  B  [2, 4, 5, 6]
A  A  [1, 2, 3, 4, 5, 6]  S
Chap 4-8
Examples
(continued)
S = [1, 2, 3, 4, 5, 6]

B = [4, 5, 6]
Mutually exclusive:

A and B are not mutually exclusive


A = [2, 4, 6]
The outcomes 4 and 6 are common to both
Collectively exhaustive:

A and B are not collectively exhaustive

A U B does not contain 1 or 3
Chap 4-9
Probability

Probability – the chance that
an uncertain event will occur
(always between 0 and 1)
0 ≤ P(A) ≤ 1 For any event A
1
Certain
0.5
0
Impossible
Chap 4-10
Assessing Probability

There are three approaches to assessing the
probability of an uncertain event:
1. classical probability (sample)
probability of event A 

NA
number of outcomes that satisfy the event

N
total number of outcomes in the sample space
Assumes all outcomes in the sample space are equally likely to
occur
Chap 4-11
Counting the Possible Outcomes

Use the Combinations formula to determine the
number of combinations of n things taken k at a
time
n!
C 
k! (n  k)!
n
k

where


n! = n(n-1)(n-2)…(1)
0! = 1 by definition
Chap 4-12
Assessing Probability
Three approaches (continued)
2. relative frequency probability (population)
probability of event A 

nA
number of events in the population that satisfy event 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
Chap 4-13
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. Then
P(A)   P(Oi )
A
(the notation means that the summation is over all the basic outcomes in A)
3. P(S) = 1
Chap 4-14
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)
Chap 4-15
A Probability Table
Probabilities and joint probabilities for two events A
and B are summarized in this table:
B
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
Chap 4-16
Addition Rule Example
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
Chap 4-17
Addition Rule Example
(continued)
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
Don’t count
the two red
aces twice!
Chap 4-18
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
Chap 4-19
Conditional Probability Example


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)
Chap 4-20
Conditional Probability Example
(continued)

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
0.2
0.5
0.7
No AC
0.2
0.1
0.3
Total
0.4
0.6
1.0
P(CD  AC) 0.2
P(CD|AC) 

 0.2857
P(AC)
0.7
Chap 4-21
Conditional Probability Example
(continued)

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
0.2
0.5
0.7
No AC
0.2
0.1
0.3
Total
0.4
0.6
1.0
P(CD  AC) 0.2
P(CD|AC) 

 0.2857
P(AC)
0.7
Chap 4-22
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)
Chap 4-23
Multiplication Rule Example
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
Color
Red
Black
Total
Ace
2
2
4
Non-Ace
24
24
48
Total
26
26
52
Chap 4-24
Statistical Independence

Two events are statistically independent if
and only if (works in both directions):
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
P(A | B)  P(A)
if P(B)>0
P(B | A)  P(B)
if P(A)>0
Chap 4-25
Statistical Independence Example

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
0.2
0.5
0.7
No AC
0.2
0.1
0.3
Total
0.4
0.6
1.0
Are the events AC and CD statistically independent?
Chap 4-26
Statistical Independence Example
(continued)
CD
No CD
Total
AC
0.2
0.5
0.7
No AC
0.2
0.1
0.3
Total
0.4
0.6
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
Chap 4-27
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)
Chap 4-28
Joint and Marginal Probabilities

The probability of a joint event, A ∩ B:
P(A  B) 

number of outcomes satisfying A 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
Chap 4-29
Marginal Probability Example
P(Ace)
 P(Ace  Red)  P(Ace  Black) 
Type
2
2
4


52 52 52
Color
Red
Black
Total
Ace
2
2
4
Non-Ace
24
24
48
Total
26
26
52
Chap 4-30
Using a Tree Diagram
Given AC or
no AC:
.2
.7
.5
.7
All
Cars
.2
.3
.1
.3
P(AC ∩ CD) = 0.2
P(AC ∩ CD) = 0.5
P(AC ∩ CD) = 0.2
P(AC ∩ CD) =0.1
Chap 4-31
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)
Chap 4-32
Odds: Example

Calculate the probability of winning (=A) 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
Chap 4-33
Overinvolvement Ratio

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(A 1 | B1 )
P(A 1 | B 2 )

An overinvolvement ratio greater than 1 implies that
event A1 increases the conditional odds ration in favor
of B1:
P(B1 | A1 ) P(B1 )

P(B 2 | A1 ) P(B 2 )
Chap 4-34
Bayes’ Theorem
Refers to situations where the outcome of an experiment depends on what happens
in various intermediate stages. The probability that event A was reached via the
r-th branch of the tree diagram, given that it was reached via one of its k
branches, is the ratio of the probability associated with the r-th branch to the
sum of the probabilities associated with all k branches of the tree.
P(A/B1)
B1
A
P(B1) P(A/B1)
P(B1)
P(B2) P(A/B2)
B2
A
P(A/B2)
P(B2)
P(BK)
BK
P(A/BK)
A
P(BK) P(A/BK)
P(B r |A) 
P(A|B r )P(B r )
k
 P(A|B )P(B )
i
i
i1
Chap 4-35
Bayes’ Theorem
P(E i | A) 


P(A | E i )P(E i )
P(A)
P(A | E i )P(E i )
P(A | E 1 )P(E 1 )  P(A | E 2 )P(E 2 )    P(A | E k )P(E k )
where:
Ei = ith event of k mutually exclusive and collectively
exhaustive events
A = new event that might impact P(Ei)
Chap 4-36
Bayes’ Theorem Example

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?
Chap 4-37
Bayes’ Theorem Example
(continued)

Let S = successful well
U = unsuccessful well

P(S) = 0.4 , P(U) =0.6

Define the detailed test event as D

Conditional probabilities:
P(D|S) = 0.6

(prior probabilities)
P(D|U) = 0.2
Goal is to find P(S|D)
Chap 4-38
Bayes’ Theorem Example
(continued)
Apply Bayes’ Theorem:
P(D|S)P(S)
P(S|D) 
P(D|S)P(S)  P(D|U)P(U)
(0.6)(0.4)

(0.6)(0.4)  (0.2)(0.6)
0.24

 0.667
0.24  0.12
So the revised probability of success (from the original estimate of 0.4),
given that this well has been scheduled for a detailed test, is 0.667
Chap 4-39
Bayes’ Theorem
S
P(D/S)
P(D/S) P(S)
P(S)
successful well
P(S|D) 
P(S/D)
P(D|S)P(S)
P(D|S)P(S)  P(D|U)P(U)
(0.6)(0.4)
(0.6)(0.4)  (0.2)(0.6)
0.24

 0.667
0.24  0.12

P(D/U) P(U)
P(U)
unsuccessful well
U
P(D/U)
Chap 4-40