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© 2011 Pearson Education, Inc
© 2011 Pearson Education, Inc
Statistics for Business and Economics
Chapter 3
Probability
© 2011 Pearson Education, Inc
Contents1. Events, Sample Spaces, and Probability2. Unions and Intersections3. Complementary Events4. The Additive Rule and Mutually Exclusive
Events5. Conditional Probability6. The Multiplicative Rule and Independent
Events7. Random Sampling8. 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!*
Number of Tosses
Total Heads Number of Tosses
0.00
0.25
0.50
0.75
1.00
0 25 50 75 100 125
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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
3. Sample space (S) • Collection of all possible outcomes
Sample Space Depends on Experimenter!
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Sample Space Properties
Experiment: Observe Gender
© 1984-1994 T/Maker Co.
1. Mutually Exclusive
• 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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Visualizing Sample Space
1. Listing
S = {Head, Tail}
2. Venn Diagram
HT
S
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Sample Space Examples
• Toss a Coin, Note Face {Head, Tail}• Toss 2 Coins, Note Faces {HH, HT, TH, TT}• Select 1 Card, Note Kind {2♥, 2♠, ..., A♦} (52)• Select 1 Card, Note Color {Red, Black} • Play a Football Game {Win, Lose, Tie}• Inspect a Part, Note Quality {Defective, Good}• Observe Gender {Male, Female}
Experiment Sample Space
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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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S
HH
TT
THHT
Sample Space S = {HH, HT, TH, TT}
Venn Diagram
Outcome
Experiment: Toss 2 Coins. Note Faces.
Compound Event: At least one Tail
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Event Examples
• 1 Head & 1 Tail HT, TH
• Head on 1st Coin HH, HT
• At Least 1 Head HH, HT, TH
• Heads on Both HH
Experiment: Toss 2 Coins. Note Faces.
Sample Space: HH, HT, TH, TT
Event Outcomes in Event
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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)
2. Lies between 0 & 1
3. Sum of sample points is 1
11
.5 .5
00
CertainCertain
ImpossibleImpossible
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Probability Rulesfor 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
© 1984-1994 T/Maker Co.
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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 RuleA sample of n elements is to be drawn from a set of N elements. The, the number of different samples possible
is denoted byN
n
⎛⎝⎜
⎞⎠⎟
and is equal to
N
n
⎛⎝⎜
⎞⎠⎟=
N!n! N−n( )!
where the factorial symbol (!) means that
n!=n n−1( ) n−2( )L 3( ) 2( ) 1( )
5!=5⋅4⋅3⋅2⋅1For example, 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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S
BlackAce
Event Union: Venn Diagram
Event Ace Black:
A, ..., A, 2, ..., K
Event Black:
2,
2,...,
A
Sample Space:
2,2,
2, ..., A
Event Ace:
A, A, A, A
Experiment: Draw 1 Card. Note Kind, Color & Suit.
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EventAce Black:
A,..., A, 2, ..., K
Event Union: Two–Way Table
Sample Space (S):
2, 2,
2, ..., A
Simple Event Ace:
A,
A,
A,
A
Simple Event Black:
2, ..., A
Experiment: Draw 1 Card. Note Kind, Color & Suit. Color
Type Red Black TotalAce Ace &
RedAce &Black
Ace
Non &Red
Non &Black
Non-Ace
Total Red Black S
Non-Ace
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S
BlackAce
Event Intersection: Venn Diagram
Event Ace Black:
A, A
Event Black:
2,...,A
Sample Space:
2, 2,
2, ..., A
Experiment: Draw 1 Card. Note Kind, Color & Suit.
Event Ace:
A, A, A, A
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Sample Space (S):
2, 2,
2, ..., A
Event Intersection: Two–Way Table
Experiment: Draw 1 Card. Note Kind, Color & Suit.
Event
Ace Black:
A, A
Simple Event Ace:
A, A,
A, A
Simple Event Black: 2, ..., A
ColorType Red Black Total
Ace Ace &Red
Ace &Black
Ace
Non &Red
Non &Black
Non-Ace
Total Red Black S
Non-Ace
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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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EventEvent B1 B2 Total
A1 P(A 1 B1) P(A1 B2) P(A1)
A2 P(A 2 B1) P(A2 B2) P(A2)
P(B1) P(B2) 1
Event Probability Using Two–Way Table
Joint Probability Marginal (Simple) Probability
Total
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ColorType 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
Two–Way Table Example
Experiment: Draw 1 Card. Note Kind & Color.
P(Ace)
P(Ace Red)P(Red)
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1. P(A) =
2. P(D) =
3. P(C B) =
4. P(A D) =
5. P(B D) =
Thinking Challenge
EventEvent C D Total
A 4 2 6
B 1 3 4
Total 5 5 10
What’s the Probability?
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Solution*
The Probabilities Are:
1. P(A) = 6/10
2. P(D) = 5/10
3. P(C B) = 1/10
4. P(A D) = 9/10
5. P(B D) = 3/10
EventEvent C D Total
A 4 2 6
B 1 3 4
Total 5 5 10
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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
S
AC
A
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Rule of Complements
The sum of the probabilities of complementary events equals 1:
P(A) + P(AC) = 1
S
AC
A
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S
Black
Complement of Event Example
Event Black:
2, 2, ..., A
Complement of Event Black,
BlackC: 2, 2, ..., A, A
Sample Space:
2, 2,
2, ..., A
Experiment: Draw 1 Card. Note Color.
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3.4
The Additive Rule and Mutually Exclusive Events
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Mutually Exclusive Events
• Events do not occur simultaneously
• A does not contain any sample points
Mutually Exclusive Events
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S
Mutually Exclusive Events Example
Events and are Mutually Exclusive
Experiment: Draw 1 Card. Note Kind & Suit.
Outcomes in Event Heart:
2, 3, 4,
..., A
Sample Space:
2, 2,
2, ..., A
Event Spade:
2, 3, 4, ..., A
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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.
P(Ace Black) = P(Ace) + P(Black) – P(Ace Black)
ColorType Red Black Total
Ace 2 2 4
Non-Ace 24 24 48
Total 26 26 52
52 52 52 52 4 26 2 28
= + – =
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Thinking Challenge
1. P(A D) =
2. P(B C) =
EventEvent C D Total
A 4 2 6
B 1 3 4
Total 5 5 10
Using the additive rule, what is the probability?
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10 10 10 10 6 5 2 9
Solution*
Using the additive rule, the probabilities are:
P(A D) = P(A) + P(D) – P(A D)1.
2. P(B C) = P(B) + P(C) – P(B C)
10 10 10 10
4 5 1 8
= + – =
= + – =
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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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S
BlackAce
Conditional Probability Using Venn Diagram
Black ‘Happens’: Eliminates All Other Outcomes
Event (Ace Black)
(S)Black
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Conditional Probability Using Two–Way Table
Experiment: Draw 1 Card. Note Kind & Color.
Revised Sample Space
ColorType Red Black Total
Ace 2 2 4
Non-Ace 24 24 48
Total 26 26 52
P(Ace Black) 2 / 52 2P(Ace | Black) =
P(Black) 26 / 52 26
∩= =
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Using the table then the formula, what’s the probability?
Thinking Challenge
1. P(A|D) =
2. P(C|B) =
EventEvent C D Total
A 4 2 6
B 1 3 4
Total 5 5 10
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Solution*
Using the formula, the probabilities are:
P A D( )=P A∩B( )
P D( )=
25
510
=25
P C B( )=P C∩B( )
P B( )=110
410
=14
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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 Red Black Total
Ace 2 2 4
Non-Ace 24 24 48
Total 26 26 52
4 2 2
52 4 52⎛ ⎞⎛ ⎞= =⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠
P(Ace Black) = P(Ace)∙P(Black | Ace)
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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)
Statistical Independence
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Thinking Challenge
1. P(C B) =
2. P(B D) =
3. P(A B) =
EventEvent C D Total
A 4 2 6
B 1 3 4
Total 5 5 10
Using the multiplicative rule, what’s the probability?
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Solution*
Using the multiplicative rule, the probabilities are:
P C ∩B( )=P C( )⋅P B C( )=510
⋅15=
110
P B∩D( )=P B( )⋅P D B( )=410
⋅35=
625
P A∩B( )=P A( )⋅P B A( )=0
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Tree DiagramExperiment: Select 2 pens from 20 pens: 14 blue & 6 red. Don’t replace.
Dependent!
BB
RR
BBRR
BB
RR6/20
5/19
14/19
14/206/19
13/19
P(R R)=(6/20)(5/19) =3/38
P(R B)=(6/20)(14/19) =21/95
P(B R)=(14/20)(6/19) =21/95
P(B B)=(14/20)(13/19) =91/190
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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 thatP(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
Factory Factory IIII
Factory Factory II0 .6
0.02
0.98
0 .4 0.01
0.99
DefectiveDefective
DefectiveDefective
GoodGood
GoodGood
P(I | D) =P( I )P(D |I )
P( I )P(D |I ) + P( II )P(D |II )=
0.6⋅0.020.6⋅0.02 + 0.4⋅0.01
=0.75
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Key Ideas
Probability Rules for k Sample Points,
S1, S2, S3, . . . , Sk
1. 0 ≤ P(Si) ≤ 1
2.P Si( )∑ =1
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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!n! N−n( )!
=N N−1( ) N−2( )L N−n+1( )
n n−1( ) n−2( )L 2( ) 1( )
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Key Ideas
Bayes’s Rule
P(S
i| A) =
P(Si )P(A|Si )P(S1)P(A|S1) + P(S2 )P(A|S2 ) + ...+ P(Sk)P(A|Sk)