Chapter 4
Probability
McGraw-Hill/Irwin
Copyright © 2009 by The McGraw-Hill Companies, Inc. All Rights Reserved.
Probability
4.1 The Concept of Probability
4.2 Sample Spaces and Events
4.3 Some Elementary Probability Rules
4.4 Conditional Probability and
Independence
4-2
The Concept of Probability
• An experiment is any process of
observation with an uncertain outcome
• The possible outcomes for an
experiment are called the experimental
outcomes
• Probability is a measure of the chance
that an experimental outcome will occur
when an experiment is carried out
4-3
Probability
If E is an experimental outcome, then
P(E) denotes the probability that E will
occur and:
Conditions
1. 0  P(E)  1 such that:
•
If E can never occur, then P(E) = 0
•
If E is certain to occur, then P(E) = 1
2. The probabilities of all the experimental
outcomes must sum to 1
4-4
Assigning Probabilities to Experimental
Outcomes
• Classical Method
– For equally likely outcomes
• Long-run relative frequency
– In the long run
• Subjective
– Assessment based on experience,
expertise or intuition
4-5
Classical Method
• All the experimental outcomes are
equally likely to occur
• Example: tossing a “fair” coin
– Two outcomes: head (H) and tail (T)
– If the coin is fair, then H and T are equally
likely to occur any time the coin is tossed
– So P(H) = 0.5, P(T) = 0.5
• 0 < P(H) < 1, 0 < P(T) < 1
• P(H) + P(T) = 1
4-6
Long-Run Relative Frequency Method
• Let E be an outcome of an experiment
• If it is performed many times, P(E) is the
relative frequency of E
– P(E) is the percentage of times E occurs in
many repetitions of the experiment
• Use sampled or historical data
• Example: Of 1,000 randomly selected
consumers, 140 preferred brand X
• The probability of randomly picking a person
who prefers brand X is 140/1000 = 0.14 or 14%
4-7
Subjective Probability
• Using experience, intuitive judgment, or
expertise to assess a probability
• May or may not have relative frequency
interpretation
4-8
Sample Spaces and Events
• A sample space of an experiment is the
set of all possible experimental
outcomes
– Heads and tails when flipping a coin
• The experimental outcomes in the
sample space are often called sample
space outcomes
4-9
Example 4.2: Genders of Two Children
• Let: B be the outcome that child is boy
G be the outcome that child is girl
• Sample space S = {BB, BG, GB, GG}
• If B and G are equally likely , then
P(B) = P(G) = ½ and
• P(BB) = P(BG) = P(GB) = P(GG) = ¼
4-10
A Tree Diagram of the Genders of Two
Children
4-11
Events
• An event is a set (or collection) of
sample space outcomes
• The probability of an event is the sum of
the probabilities of the sample space
outcomes that correspond to the event
4-12
Example 4.4: Gender of Two Children
• Experimental Outcomes:
BB, BG, GB, GG
• All outcomes equally likely:
P(BB) = … = P(GG) = ¼
• P(one boy and one girl) =
P(BG) + P(GB) = ¼ + ¼ = ½
• P(at least one girl) =
P(BG) + P(GB) + P(GG) = ¼+¼+¼ = ¾
4-13
Probabilities: Equally Likely Outcomes
• If the sample space outcomes (or
experimental outcomes) are all equally
likely, then the probability that an event
will occur is equal to the ratio:
– The number of ways the event can occur
– Over the total number of outcomes
Number of sample space outcomes that correspond to the event
Total number of sample space outcomes
4-14
Example 4.7: AccuRatings Case
• Of 5528 residents sampled,
445 prefer KPWR
• Estimated Share P(KPWR) =
445 / 5528 = 0.0805
• So the probability that any
resident chosen at random
prefers KPWR is 0.0805
• Assuming 8,300,000 Los Angeles
residents aged 12 or older:
• # Listeners = Population x Share
so 8,300,000 x 0.0805 = 668,144
4-15
Example: AccuRatings Case
Continued
4-16
Some Elementary Probability Rules
1. Complement
2. Union
3. Intersection
4. Addition
5. Conditional probability
6. Multiplication
4-17
Complement
• The complement (Ā) of an event A is the
set of all sample space outcomes not in
A
• P(Ā) = 1 – P(A)
4-18
Union and Intersection
• The union of A and B are elementary
events that belong to either A or B or
both
– Written as A  B
• The intersection of A and B are
elementary events that belong to both A
and B
– Written as A ∩ B
4-19
Some Elementary Probability Rules
4-20
Mutually Exclusive
• A and B are mutually exclusive if they
have no sample space outcomes in
common
• In other words:
P(A∩B) = 0
4-21
The Addition Rule
• If A and B are mutually exclusive, then
the probability that A or B (the union of A
and B) will occur is
P(AB) = P(A) + P(B)
• If A and B are not mutually exclusive:
P(AB) = P(A) + P(B) – P(A∩B)
where P(A∩B) is the joint probability of
A and B both occurring together
4-22
Example: Newspaper Subscribers #1
• Define events:
– A = event that a randomly selected household
subscribes to the Atlantic Journal
– B = event that a randomly selected household
subscribes to the Beacon News
• Given:
– total number in city, N = 1,000,000
– number subscribing to A, N(A) = 650,000
– number subscribing to B, N(B) = 500,000
– number subscribing to both, N(A∩B) = 250,000
4-23
Example: Newspaper Subscribers #2
• Use the relative frequency method to
assign probabilities
650,000
P  A 
 0.65
1,000,000
500,000
P B  
 0.50
1,000,000
250,000
P A  B  
 0.25
1,000,000
4-24
Example: Newspaper Subscribers #3
• Refer to the contingency table in Table
4.3 for all data
• For example, the chance that a
household does not subscribe to either
newspaper
– Want PA  B  , so from middle row and
middle column of Table 4.3
100,000
PA  B  
 0.10
1,000,000
4-25
Example: Newspaper Subscribers #4
• The chance that a household subscribes to
either newspaper:
P(A  B)=P(A)+P(B)  P(A  B)
 0.65  0.50  0.25
 0.90
• Note that if the joint probability was not
subtracted the would have gotten 1.15, which
is greater than 1, which is absurd
– The subtraction avoids double counting the joint
probability
4-26
Conditional Probability and
Independence
• The probability of an event A, given that
the event B has occurred, is called the
conditional probability of A given B
– Denoted as P(A|B)
• Further, P(A|B) = P(A∩B) / P(B)
– P(B) ≠ 0
4-27
Interpretation
• Restrict sample space to just event B
• The conditional probability P(A|B) is the
chance of event A occurring in this new
sample space
• In other words, if B occurred, then what
is the chance of A occurring
4-28
Example: Newspaper Subscribers
• Of the households that subscribe to the
Atlantic Journal, what is the chance that
they also subscribe to the Beacon
News?
PA  B
• Want P(B|A), where PB | A  
PA 
0.25

0.65
 0.3846
4-29
Independence of Events
• Two events A and B are said to be
independent if and only if:
P(A|B) = P(A)
• This is equivalently to
P(B|A) = P(B)
4-30
Example: Newspaper Subscribers
• Of the Atlantic Journal subscribers, what is
the chance that they also subscribe to the
Beacon News?
– If independent, the P(B|A) = P(A)
• Is P(B|A) = P(A)?
– Know that P(A) = 0.65
– Just calculated that P(B|A) = 0.3846
– 0.65 ≠ 0.3846, so P(B|A) ≠ P(A)
• A is not independent of B
– A and B are said to be dependent
4-31
The Multiplication Rule
• The joint probability that A and B (the
intersection of A and B) will occur is
P(A∩B) = P(A) • P(B|A) = P(B) • P(A|B)
• If A and B are independent, then the
probability that A and B will occur is:
P(A∩B) = P(A) • P(B) = P(B) • P(A)
4-32
Example: Genders of Two Children
• B is the outcome that child is boy
• G is the outcome that child is girl
• Sample space S = {BB, BG, GB, GG}
• If B and G are equally likely, then
P(B) = P(G) = ½ and
• P(BB) = P(BG) = P(GB) = P(GG) = ¼
4-33
Example: Genders of Two Children
Continued
• Of two children, what is the probability of
having a girl first and then a boy second?
• Want P(G first and B second)?
– Want P(G∩B)
• P(G∩B) = P(G)  P(B|G)
• But gender of siblings is independent
– So P(B|G) = P(B)
– Then P(G∩B) = P(G)  P(B) = ½  ½ = ¼
• Consistent with the tree diagram
4-34
Contingency Tables
P(R1 )
P(R1  C1 )
R1
R2
Total
P(R 2  C2 )
C1
.4
.1
.5
C2
.2
.3
.5
Total
.6
.4
1.00
P(C 2 )
4-35
EXAMPLE 4.16 The AccuRatings Case:
Estimating Radio Station Share by Daypart
• 5,528 L.A. residents sampled
• 2,827 of residents sampled listen during
some portion of the 6-10 a.m. daypart
• Of those, 201 prefer KIIS
• KIIS Share for 6-10 a.m. daypart:
P(KIIS|6-10 a.m.) =
P(KIIS  6-10 a.m.) / P(6-10 a.m.) =
(201/5528)  (2827/5528) =
201 / 2827 = 0.0711
4-36
Example 4.16
Continued
4-37