Download Chap. 3: Probability

© 2011 Pearson Education, Inc
Statistics for Business and
Economics
Chapter 3
Probability
© 2011 Pearson Education, Inc
Contents
1.
2.
3.
4.
5.
6.
7.
8.
Events, Sample Spaces, and Probability
Unions and Intersections
Complementary Events
The Additive Rule and Mutually Exclusive
Events
Conditional Probability
The Multiplicative Rule and Independent
Events
Random Sampling
Baye’s Rule
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Learning Objectives
1. Develop probability as a measure of
uncertainty
2. Introduce basic rules for finding
probabilities
3. Use probability as a measure of reliability
for an inference
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Thinking Challenge
• What’s the probability
of getting a head on
the toss of a single fair
coin? Use a scale from
0 (no way) to 1 (sure
thing).
• So toss a coin twice.
Do it! Did you get one
head & one tail?
What’s it all mean?
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Many Repetitions!*
Total Heads
Number of Tosses
1.00
0.75
0.50
0.25
0.00
0
25
50
75
Number of Tosses
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100
125
3.1
Events, Sample Spaces,
and Probability
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Experiments & Sample Spaces
1. Experiment
• Process of observation that leads to a single
outcome that cannot be predicted with certainty
2. Sample point
• Most basic outcome of an
experiment
Sample Space
Depends on
Experimenter!
3. Sample space (S)
• Collection of all possible outcomes
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Sample Space Properties
1. Mutually Exclusive
•
Experiment: Observe Gender
2 outcomes can not
occur at the same
time
— Male & Female in
same person
2. Collectively Exhaustive
•
One outcome in
sample space must
occur.
— Male or Female
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© 1984-1994 T/Maker Co.
Visualizing
Sample Space
1.
Listing
S = {Head, Tail}
2.
Venn Diagram
H
T
S
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Sample Space Examples
•
•
•
•
•
•
•
Experiment
Sample Space
Toss a Coin, Note Face
Toss 2 Coins, Note Faces
Select 1 Card, Note Kind
Select 1 Card, Note Color
Play a Football Game
Inspect a Part, Note Quality
Observe Gender
{Head, Tail}
{HH, HT, TH, TT}
{2♥, 2♠, ..., A♦} (52)
{Red, Black}
{Win, Lose, Tie}
{Defective, Good}
{Male, Female}
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Events
1. Specific collection of sample points
2. Simple Event
• Contains only one sample point
3. Compound Event
• Contains two or more sample points
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Venn Diagram
Experiment: Toss 2 Coins. Note Faces.
Sample Space
S = {HH, HT, TH, TT}
TH
Outcome
HH
Compound
Event: At
least one
Tail
HT
TT
S
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Event Examples
Experiment: Toss 2 Coins. Note Faces.
Sample Space: HH, HT, TH, TT
•
•
•
•
Event
1 Head & 1 Tail
Head on 1st Coin
At Least 1 Head
Heads on Both
Outcomes in Event
HT, TH
HH, HT
HH, HT, TH
HH
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Probabilities
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What is Probability?
1. Numerical measure of the
likelihood that event will
cccur
• P(Event)
• P(A)
• Prob(A)
1
Certain
.5
2. Lies between 0 & 1
3. Sum of sample points is 1
0
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Impossible
Probability Rules
for Sample Points
Let pi represent the probability of sample point i.
1. All sample point probabilities must lie between 0
and 1 (i.e., 0 ≤ pi ≤ 1).
2. The probabilities of all sample points within a
sample space must sum to 1 (i.e.,  pi = 1).
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Equally Likely Probability
P(Event) = X / T
• X = Number of outcomes in the
event
• T = Total number of sample points
in Sample Space
• Each of T sample points is equally
likely
— P(sample point) = 1/T
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© 1984-1994 T/Maker Co.
Steps for Calculating Probability
1. Define the experiment; describe the process used to
make an observation and the type of observation
that will be recorded
2. List the sample points
3. Assign probabilities to the sample points
4. Determine the collection of sample points contained
in the event of interest
5. Sum the sample points probabilities to get the event
probability
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Combinations Rule
A sample of n elements is to be drawn from a set of N
elements. The, the number of different samples possible
 N
is denoted by   and is equal to
n 
 N
N!
 n   n!N  n !
where the factorial symbol (!) means that
n!  n n  1n  2 L 32 1
For example, 5!  5  4  3 2 1
0! is defined to be 1.
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3.2
Unions and Intersections
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Compound Events
Compound events:
Composition of two or more other events.
Can be formed in two different ways.
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Unions & Intersections
1. Union
•
•
•
Outcomes in either events A or B or both
‘OR’ statement
Denoted by  symbol (i.e., A  B)
2. Intersection
•
•
•
Outcomes in both events A and B
‘AND’ statement
Denoted by  symbol (i.e., A  B)
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Event Union:
Venn Diagram
Experiment: Draw 1 Card. Note Kind, Color &
Suit.
Sample
Space:
2, 2, 2,
..., A
Ace
Black
S
Event Ace:
A, A, A, A
Event Ace  Black:
A, ..., A, 2, ..., K
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Event
Black:
2,
2, ...,
A
Event Union:
Two–Way Table
Experiment: Draw 1 Card. Note Kind, Color &
Suit.
Color
Simple
Sample Space
Type
(S):
Ace
2, 2, 2,
..., A
Non-Ace
Total
Event
Ace  Black:
A,..., A, 2, ..., K
Total
Ace & Ace & Ace
Red
Black
Non & Non & NonRed
Black Ace
Red
Black
S
Red
Black
Simple Event Black:
2, ..., A
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Event
Ace:
A,
A,
A,
A
Event Intersection:
Venn Diagram
Experiment: Draw 1 Card. Note Kind, Color &
Suit.
Sample
Space:
2, 2, 2,
..., A
Ace
Black
S
Event Ace:
A, A, A, A
Event Ace  Black:
A, A
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Event
Black:
2,...,A
Event Intersection:
Two–Way Table
Experiment: Draw 1 Card. Note Kind, Color &
Suit.
Color
Sample Space
Type
(S):
Ace
2, 2, 2,
..., A
Non-Ace
Event
Ace  Black:
A, A
Total
Total
Ace & Ace & Ace
Red
Black
Non & Non & NonRed
Black Ace
Red
Black
S
Red
Black
Simple
Event
Ace:
A, A,
A, A
Simple Event Black: 2, ..., A
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Compound Event Probability
1. Numerical measure of likelihood that
compound event will occur
2. Can often use two–way table
• Two variables only
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Event Probability Using
Two–Way Table
Event
Event
B1
B2
Total
A1
P(A 1  B1) P(A 1  B2) P(A 1)
A2
P(A 2  B1) P(A 2  B2) P(A 2)
Total
Joint Probability
P(B 1)
P(B 2)
1
Marginal (Simple) Probability
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Two–Way Table Example
Experiment: Draw 1 Card. Note Kind & Color.
Color
Type
Red
Black
Total
Ace
2/52
2/52
4/52
Non-Ace
24/52
24/52
48/52
Total
26/52
26/52
52/52
P(Red)
P(Ace  Red)
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P(Ace)
Thinking Challenge
What’s the Probability?
1. P(A) =
2. P(D) =
Event
C
D
4
2
3. P(C  B) =
Event
A
4. P(A  D) =
B
1
3
4
5. P(B  D) =
Total
5
5
10
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Total
6
Solution*
The Probabilities Are:
1. P(A) = 6/10
2. P(D) = 5/10
Event
C
D
4
2
3. P(C  B) = 1/10
Event
A
4. P(A  D) = 9/10
B
1
3
4
5. P(B  D) = 3/10 Total
5
5
10
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Total
6
3.3
Complementary Events
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Complementary Events
Complement of Event A
• The event that A does not occur
• All events not in A
• Denote complement of A by AC
AC
A
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S
Rule of Complements
The sum of the probabilities of complementary events
equals 1:
P(A) + P(AC) = 1
AC
A
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S
Complement of Event
Example
Experiment: Draw 1 Card. Note Color.
Black
Sample
Space:
2, 2, 2,
..., A
Event Black:
2, 2, ..., A
S
Complement of Event Black,
BlackC: 2, 2, ..., A, A
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3.4
The Additive Rule and
Mutually Exclusive Events
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Mutually Exclusive Events
Mutually Exclusive Events
• Events do not occur
simultaneously
• A  B does not contain
any sample points

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Mutually Exclusive
Events Example
Experiment: Draw 1 Card. Note Kind & Suit.
Sample
Space:
2, 2,
2, ..., A


Event Spade:
2, 3, 4, ..., A
S
Outcomes
in Event
Heart:
2, 3, 4
, ..., A
Events  and are Mutually Exclusive
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Additive Rule
1. Used to get compound probabilities for
union of events
2. P(A OR B) = P(A  B)
= P(A) + P(B) – P(A  B)
3. For mutually exclusive events:
P(A OR B) = P(A  B) = P(A) + P(B)
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Additive Rule Example
Experiment: Draw 1 Card. Note Kind & Color.
Color
Type
Ace
Red
Black
2
2
Total
4
Non-Ace
24
24
48
Total
26
26
52
P(Ace  Black) = P(Ace) + P(Black) – P(Ace  Black)
4
26
2 28
=
+
–
=
52
52 52 52
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Thinking Challenge
Using the additive rule, what is the probability?
1. P(A  D) =
2. P(B  C) =
Event
A
Event
C
D
4
2
Total
6
B
1
3
4
Total
5
5
10
© 2011 Pearson Education, Inc
Solution*
Using the additive rule, the probabilities are:
1. P(A  D) = P(A) + P(D) – P(A  D)
6
5
2
9
=
+
–
=
10 10 10 10
2. P(B  C) = P(B) + P(C) – P(B  C)
4
5
1
8
=
+
–
=
10 10 10 10
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3.5
Conditional Probability
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Conditional Probability
1. Event probability given that another event
occurred
2. Revise original sample space to account for
new information
• Eliminates certain outcomes
3. P(A | B) = P(A and B) = P(A  B
P(B)
P(B)
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Conditional Probability Using
Venn Diagram
Ace
Black
S
Event (Ace  Black)
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Black ‘Happens’:
Eliminates All
Other Outcomes
Black
(S)
Conditional Probability Using
Two–Way Table
Experiment: Draw 1 Card. Note Kind & Color.
Color
Type
Red
Black
Ace
2
2
Total
4
Non-Ace
24
24
48
Total
26
26
52
Revised
Sample
Space
P(Ace  Black) 2 / 52
2
P(Ace | Black) =


P(Black)
26 / 52 26
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Thinking Challenge
Using the table then the formula, what’s the
probability?
1. P(A|D) =
2. P(C|B) =
Event
A
Event
C
D
4
2
Total
6
B
1
3
4
Total
5
5
10
© 2011 Pearson Education, Inc
Solution*
Using the formula, the probabilities are:
P A  B  2 5 2
P A D  


5
P D 
5
10
P C  B  110 1
P C B  


4
P B 
4
10
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3.6
The Multiplicative Rule
and Independent Events
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Multiplicative Rule
1. Used to get compound probabilities for
intersection of events
2. P(A and B) = P(A  B)
= P(A)  P(B|A)
= P(B)  P(A|B)
3. For Independent Events:
P(A and B) = P(A  B) = P(A)  P(B)
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Multiplicative Rule Example
Experiment: Draw 1 Card. Note Kind & Color.
Color
Type
Ace
Red
Black
2
2
Total
4
Non-Ace
24
24
48
Total
26
26
52
P(Ace  Black) = P(Ace)∙P(Black | Ace)
 4  2  2
    
 52  4  52
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Statistical Independence
1. Event occurrence does not affect probability of
another event
• Toss 1 coin twice
2. Causality not implied
3. Tests for independence
• P(A | B) = P(A)
• P(B | A) = P(B)
• P(A  B) = P(A)  P(B)
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Thinking Challenge
Using the multiplicative rule, what’s the
probability?
1. P(C  B) =
Event
C
D
4
2
2. P(B  D) =
Event
A
3. P(A  B) =
B
1
3
4
Total
5
5
10
© 2011 Pearson Education, Inc
Total
6
Solution*
Using the multiplicative rule, the probabilities
are:
5 1 1
P C  B   P C  P B C    
10 5 10
4 3 6
P B  D   P B  P D B    
10 5 25
P A  B   P A  P B A   0
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Tree Diagram
Experiment: Select 2 pens from 20 pens: 14
blue & 6 red. Don’t replace.
Dependent!
5/19
6/20
P(R  R)=(6/20)(5/19) =3/38
B
R
P(R  B)=(6/20)(14/19) =21/95
B
P(B  B)=(14/20)(13/19) =91/190
R
14/19
14/20
R
6/19
P(B  R)=(14/20)(6/19) =21/95
B
13/19
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3.7
Random Sampling
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Importance of Selection
How a sample is selected from a population is
of vital importance in statistical inference
because the probability of an observed sample
will be used to infer the characteristics of the
sampled population.
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Random Sample
If n elements are selected from a population in
such a way that every set of n elements in the
population has an equal probability of being
selected, the n elements are said to be a
random sample.
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Random Number Generators
Most researchers rely on random number
generators to automatically generate the
random sample.
Random number generators are available in
table form, and they are built into most
statistical software packages.
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3.8
Bayes’s Rule
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Bayes’s Rule
Given k mutually exclusive and exhaustive
events B1, B1, . . . Bk , such that
P(B1) + P(B2) + … + P(Bk) = 1,
and an observed event A, then
P(Bi  A)
P(Bi | A) 
P( A)
P(Bi )P( A | Bi )

P(B1 )P( A | B1 )  P(B2 )P( A | B2 )  ...  P(Bk )P( A | Bk )
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Bayes’s Rule Example
A company manufactures MP3 players at two
factories. Factory I produces 60% of the MP3
players and Factory II produces 40%. Two
percent of the MP3 players produced at Factory
I are defective, while 1% of Factory II’s are
defective. An MP3 player is selected at random
and found to be defective. What is the
probability it came from Factory I?
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Bayes’s Rule Example
0 .6
0 .4
Factory
I
Factory
II
0.02
Defective
0.98
Good
0.01
Defective
0.99
Good
P( I | D) 
P(I )P( D | I )
0.6  0.02

 0.75
P(I )P( D | I )  P(II )P(D | II ) 0.6  0.02  0.4  0.01
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Key Ideas
Probability Rules for k Sample Points,
S1, S2, S3, . . . , Sk
1. 0 ≤ P(Si) ≤ 1
2.
 P S   1
i
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Key Ideas
Random Sample
All possible such samples have equal
probability of being selected.
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Key Ideas
Combinations Rule
Counting number of samples of n elements
selected from N elements
 N
N N  1N  2 L N  n  1
N!
 n   n!N  n ! 
n n  1n  2 L 2 1
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Key Ideas
Bayes’s Rule
P(Si | A) 
P(Si )P( A | Si )
P(S1 )P( A | S1 )  P(S2 )P( A | S2 )  ...  P(Sk )P( A | Sk )
© 2011 Pearson Education, Inc