Chapter 4
Probability
Chap 4-1
Chapter Goals
After completing this chapter, you should be
able to:
n Explain basic probability concepts and definitions
n Use a Venn diagram or tree diagram to illustrate
simple probabilities
n Apply common rules of probability
n Compute conditional probabilities
n Determine whether events are statistically
independent
n Use Bayes’ Theorem for conditional probabilities
Chap 4-2
Important Terms
n
n
n
n
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 4-3
Important Terms
(continued)
n
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
AÇ B
B
Chap 4-4
Important Terms
(continued)
n
A and B are Mutually Exclusive Events if they
have no basic outcomes in common
n
i.e., the set A ∩ B is empty
S
A
B
Chap 4-5
Important Terms
(continued)
n
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 4-6
Important Terms
(continued)
n
Events E1, E2, … Ek are Collectively Exhaustive
events if E1 U E2 U . . . U Ek = S
n
n
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 4-7
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 4-8
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 Ç B = [4, 6]
Unions:
A Ç B = [5]
A È B = [2, 4, 5, 6]
A È A = [1, 2, 3, 4, 5, 6] = S
Chap 4-9
Examples
(continued)
S = [1, 2, 3, 4, 5, 6]
n
B = [4, 5, 6]
Mutually exclusive:
n
A and B are not mutually exclusive
n
n
A = [2, 4, 6]
The outcomes 4 and 6 are common to both
Collectively exhaustive:
n
A and B are not collectively exhaustive
n
A U B does not contain 1 or 3
Chap 4-10
Probability
n
Probability – the chance that
an uncertain event will occur
(always between 0 and 1)
0 ≤ P(A) ≤ 1 For any event A
1
Certain
.5
0
Impossible
Chap 4-11
Assessing Probability
n
There are three approaches to assessing the
probability of an uncertain event:
1. classical probability
probability of event A =
n
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 4-12
Counting sample Points
n
n
Multiplication Rule (with replacement): if an
operation can be performed in n1 ways, and if for each
of these, a second operation can be performed in n2
ways, the the two operations can be performed in n1n2
ways.
Without replacement (ordered): If only k positions are
to be filled with objects selected from n different objects,
k £ n, the number of possible ordered arrangements is
called Permutation.
n!
P =
( n - k )!
k
n
Chap 4-13
Counting sample Points
n
Without replacement (unordered): The number of
ways in which k objects can be selected without
replacement from n objects, when the order of selection
is disregarded, is
n!
C =
k !( n - k ) !
k
n
a combination of the n objects taken k at a time.
Chap 4-14
Counting the Possible Outcomes
n
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
n
where
n
n
n! = n(n-1)(n-2)…(1)
0! = 1 by definition
Chap 4-15
Assessing Probability
Three approaches (continued)
2. relative frequency probability
probability of event A =
n
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 4-16
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) = å P(Oi )
A
(the notation means that the summation is over all the basic outcomes in A)
3. P(S) = 1
Chap 4-17
Probability Rules
n
The Complement rule:
P(A) = 1- P(A)
n
i.e., P(A) + P(A) = 1
The Addition rule:
n
The probability of the union of two events is
P(A È B) = P(A) + P(B) - P(A Ç B)
Chap 4-18
A Probability Table
Probabilities and joint probabilities for two events A
and B are summarized in this table:
B
B
A
P(A Ç B)
P(A Ç B )
P(A)
A
P(A Ç B)
P(A Ç B )
P(A)
P(B)
P( B )
P(S) = 1.0
Chap 4-19
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 4-20
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 4-21
Addition Rule Example
n
A survey of 1000 people determines that 80%
like walking and 60% like biking, and all like at
least one of the two activities. What is the
probability that a randomly chosen person in
this survey likes biking but not walking?
Chap 4-22
Conditional Probability
n
A conditional probability is the probability of one
event, given that another event has occurred:
P(A Ç B)
P(A | B) =
P(B)
The conditional
probability of A given
that B has occurred
P(A Ç B)
P(B | A) =
P(A)
The conditional
probability of B given
that A has occurred
Chap 4-23
Conditional Probability Example
n
n
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)
Statistics for Business and Economics, 6e © 2007 Pearson Education, Inc.
Chap 4-24
Conditional Probability Example
(continued)
n
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
P(CD Ç AC) .2
P(CD | AC) =
= = .2857
P(AC)
.7
Chap 4-25
Conditional Probability Example
(continued)
n
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 Ç AC) .2
P(CD | AC) =
= = .2857
P(AC)
.7
Chap 4-26
Multiplication Rule
n
Multiplication rule for two events A and B:
P(A Ç B) = P(A | B) P(B)
n
also
P(A Ç B) = P(B | A) P(A)
Chap 4-27
Multiplication Rule Example
P(Red ∩ Ace) = P(Red| Ace)P(Ace)
æ 2 öæ 4 ö 2
= ç ÷ç ÷ =
è 4 øè 52 ø 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 4-28
Statistical Independence
n
Two events are statistically independent
if and only if:
P(A Ç B) = P(A) P(B)
n
n
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 4-29
Statistical Independence Example
n
n
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?
Chap 4-30
Statistical Independence Example
(continued)
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(CD) = 0.4
P(AC)P(CD) = (0.7)(0.4) = 0.28
P(AC ∩ CD) = 0.2 ≠ P(AC)P(CD) = 0.28
So the two events are not statistically independent
Chap 4-31
Statistical Independence Example
(continued)
n
Suppose that a fair six-sided die is tossed. We define
the following events:
A = “the number of tossed is even”
B = “the number of tossed is £ 3”
C = “the number of tossed is a 1 or 2”
D = “the number of tossed doesn’t start with the letter ‘f’
or ‘t’”
a) Find P(B|A), P(B|C) and P(A|C).
b) Check the independent of events A and B, B and C, A
and C; A and D.
Chap 4-32
Bivariate Probabilities
Outcomes for bivariate events:
B1
B2
...
Bk
A1
P(A1ÇB1)
P(A1ÇB2)
...
P(A1ÇBk)
A2
P(A2ÇB1)
P(A2ÇB2)
...
P(A2ÇBk)
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
Ah
P(AhÇB1)
P(AhÇB2)
...
P(AhÇBk)
Chap 4-33
Joint and
Marginal Probabilities
n
The probability of a joint event, A ∩ B:
P(A Ç B) =
n
number of outcomes satisfying A and B
total number of elementary outcomes
Computing a marginal probability:
P(A) = P(A Ç B1 ) + P(A Ç B 2 ) +  + P(A Ç Bk )
n
Where B1, B2, …, Bk are k mutually exclusive and collectively
exhaustive events
Chap 4-34
Marginal Probability Example
P(Ace)
2
2
4
= P(Ace Ç Red) + P(Ace Ç Black) =
+
=
52 52 52
Type
Color
Red
Black
Total
Ace
2
2
4
Non-Ace
24
24
48
Total
26
26
52
Statistics for Business and Economics, 6e © 2007 Pearson Education, Inc.
Chap 4-35
Using a Tree Diagram
Given AC or
no AC:
P(A
H
All
Cars
A
as
.7
=
)
C
C
Do
e
h a v s n ot
eA
P(A
C
C)=
.3
.2
.7
D
C
H as
P(AC ∩ CD) = .2
D oe
s
h a ve n o t . 5
CD
P(AC ∩ CD) = .5
.7
.2
.3
D
C
H as
D oe
s
h a ve n o t . 1
CD
.3
P(AC ∩ CD) = .2
P(AC ∩ CD) = .1
Chap 4-36
The law of total Probability
n
Two - event case:
P( A) = P( B) P( A | B) + P( B ) P( A | B )
n
More than two events
P( A) = P( B1 ) P( A | B1 ) + P( B2 ) P( A | B2 ) +…+ P( Bn ) P( A | Bn )
=
n
å P( B ) P( A | B )
i =1
i
i
where there are n events in the partition: Bi, i =
1,…, n.
Chap 4-37
Example
n
An analyst believes the stock market has a
0.75 probability of going up in the next year if
the economy should do well, and a 0.30
probability of going up if the economy should
not do well during the year. The analyst futher
believes there is a 0.80 probability that the
economy will do well in the coming year. What
is the probability that the stock market will go
up next year (using the analyst’s assessments)?
Chap 4-38
Bayes’ Theorem
P(E i | A) =
=
n
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 4-39
Bayes’ Theorem Example
n
n
n
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 4-40
Bayes’ Theorem Example
(continued)
n
Let S = successful well
U = unsuccessful well
n
P(S) = .4 , P(U) = .6
n
Define the detailed test event as D
n
Conditional probabilities:
P(D|S) = .6
n
(prior probabilities)
P(D|U) = .2
Goal is to find P(S|D)
Chap 4-41
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)
(.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
Chap 4-42
Example
n
A test for a disease correctly diagnoses a
diseased person as having the disease with
probability 0.85. The test incorrectly diagnoses
someone without the disease as having the
disease with probability 0.01. If 1% of the
people in a population have the disease, what
is the chance that a person from this population
who tests positive for the disease actually has
the disease?
Chap 4-43
Chapter Summary
n
Defined basic probability concepts
n
n
Sample spaces and events, intersection and union of events,
mutually exclusive and collectively exhaustive events,
complements
Examined basic probability rules
n
Complement rule, addition rule, multiplication rule
n
Defined conditional, joint, and marginal probabilities
n
Reviewed odds and the overinvolvement ratio
n
Defined statistical independence
n
Discussed Bayes’ theorem
Chap 4-44