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Business Statistics, Can. ed.
By Black, Chakrapani & Castillo
Discrete Distributions
Chapter 4
Probability
Prepared by Dr. Clarence S. Bayne
JMSB, Concordia University
Business Statistics, Can. Ed. © 2010 John Wiley & Sons Canada, Ltd.
Learning Objectives
• Understanding what is probability
• Comprehend the different ways of assigning probability.
• Understand and apply marginal, union, joint, and conditional probabilities.
• Select the appropriate laws of probability to use in solving problems.
• Solve problems using the laws of probability including the laws of addition, multiplication and conditional probability
• Revise probabilities using Bayes’ rule.
Business Statistics, Can. Ed. © 2010 John Wiley & Sons Canada, Ltd.
Methods of Assigning Probabilities
There are three methods for assigning
probabilities:
• The classical method (mathematical rules and laws)
• Relative frequency of occurrence (based on historical data: empirical )
• Subjective probability (based on personal intuition or reasoning)
Business Statistics, Can. Ed. © 2010 John Wiley & Sons Canada, Ltd.
Classical Probability
• Number of outcomes leading to the event divided by the total number of outcomes possible
• Each outcome is equally likely
P (E ) 
Where
:
N  total
n
• Determined a priori ‐‐ before performing the experiment
• Applicable to games of chance
• Objective ‐‐ everyone correctly using the method assigns an identical probability
Business Statistics, Can. Ed. © 2010 John Wiley & Sons Canada, Ltd.
e
n
e
N
number
 number
of outcomes
of outcomes
in E
Relative Frequency Probability
• Based on historical data
• Computed after performing the experiment
n
P (E ) 
Where
N
:
N  total
• Number of times an event occurred divided by the number of trials
• Objective ‐‐ everyone correctly using the method assigns an identical probability
Business Statistics, Can. Ed. © 2010 John Wiley & Sons Canada, Ltd.
n
e
e
number
 number
producing
of trials
of outcomes
E
Subjective Probability
• Subjective probability comes from a person’s intuition or reasoning
• However different individuals may (correctly) assign different numeric probabilities to the same event
• Expresses an individual’s degree of belief
• Useful for unique (single‐trial) experiments
– New product introduction
– Initial public offering of common stock
– Site selection decisions
– Sporting events
Business Statistics, Can. Ed. © 2010 John Wiley & Sons Canada, Ltd.
Fundamental Concepts and Laws of
Probability Theory
•
•
•
•
•
•
•
•
•
Experiment
Event
Elementary Events
Sample Space
Unions and Intersections
Mutually Exclusive Events
Independent Events
Collectively Exhaustive Events
Complementary Events
Business Statistics, Can. Ed. © 2010 John Wiley & Sons Canada, Ltd.
Experiment
• Experiment: a process that produces outcomes
– More than one possible outcome
– Only one outcome per trial
• Trial: one repetition of the process
• Elementary Event: cannot be decomposed or broken down into other events
• Event: an outcome of an experiment
– May be an elementary event, or
– May be an aggregate of elementary events
– Usually represented by an uppercase letter, e.g., A, E1
Business Statistics, Can. Ed. © 2010 John Wiley & Sons Canada, Ltd.
An Example Experiment
Experiment: randomly select, without replacement, two families from the residents of Tiny Town
 Elementary Event: the sample includes families A and C
 Event: each family in the sample has children in the household
 Event: the sample families own a total of four automobiles
Tiny Town Population
Family
Children in
Household
Number of
Automobiles
A
B
C
D
Yes
Yes
No
Yes
3
2
1
2
Business Statistics, Can. Ed. © 2010 John Wiley & Sons Canada, Ltd.
Sample Space
• The set of all elementary events for an experiment
• Methods for describing a sample space
– roster or listing
– tree diagram
– set builder notation
– Venn diagram
Business Statistics, Can. Ed. © 2010 John Wiley & Sons Canada, Ltd.
Sample Space: Roster Example
• Experiment: randomly select, without replacement, two families from the residents of Tiny Town
• Each ordered pair in the sample space is an elementary event, for example ‐‐ (D,C)
Family
A
B
C
D
Children in
Household
Number of
Automobiles
Yes
Yes
No
Yes
3
2
1
2
Business Statistics, Can. Ed. © 2010 John Wiley & Sons Canada, Ltd.
Listing of Sample Space
(A,B), (A,C), (A,D),
(B,A), (B,C), (B,D),
(C,A), (C,B), (C,D),
(D,A), (D,B), (D,C)
Sample Space: Tree Diagram for
Random Sample of Two Families
A
B
C
D
Business Statistics, Can. Ed. © 2010 John Wiley & Sons Canada, Ltd.
B
C
D
A
C
D
A
B
D
A
B
C
Sample Space: Set Notation for
Random Sample of Two Families
• S = {(x,y) | x is the family selected on the first draw, and y is the family selected on the second draw}
• Concise description of large sample spaces
Business Statistics, Can. Ed. © 2010 John Wiley & Sons Canada, Ltd.
Sample Space
• Useful for discussion of general principles and concepts
Listing of Sample Space
(A,B), (A,C), (A,D),
(B,A), (B,C), (B,D),
(C,A), (C,B), (C,D),
(D,A), (D,B), (D,C)
Business Statistics, Can. Ed. © 2010 John Wiley & Sons Canada, Ltd.
Venn Diagram
Union of Sets
• The union of two sets contains an instance of each element of the two sets.
X  1,4 ,7 ,9
Y  2 ,3,4 ,5,6
X
Y
X  Y  1,2 ,3,4 ,5,6,7 ,9
C   IBM , DEC , Apple
F   Apple, Grape, Lime
C  F   IBM , DEC , Apple, Grape, Lime
Business Statistics, Can. Ed. © 2010 John Wiley & Sons Canada, Ltd.
XY
Intersection of Sets
• The intersection of two sets contains only those element common to the two sets.
X

Y 
1 , 4 , 7 , 9 
2 , 3 , 4 , 5 , 6 
X
Y
X  Y  4 
C  IBM , DEC , Apple 
F  Apple , Grape , Lime 
C  F  Apple 
Business Statistics, Can. Ed. © 2010 John Wiley & Sons Canada, Ltd.
XY
Mutually Exclusive Events
• Events with no common outcomes
• Occurrence of one event precludes the occurrence of the other event
C   IBM , DEC , Apple
P( X Y)  0
X
Y
X  1,7 ,9 
F   Grape , Lime
Y  2 ,3,4 ,5,6
CF 
X Y 

Business Statistics, Can. Ed. © 2010 John Wiley & Sons Canada, Ltd.

Independent Events
• Occurrence of one event does not affect the occurrence or nonoccurrence of the other event
• The conditional probability of X given Y is equal to the marginal probability of X.
• The conditional probability of Y given X is equal to the marginal probability of Y.
P( X| Y )  P( X) and P(Y| X)  P(Y )
Business Statistics, Can. Ed. © 2010 John Wiley & Sons Canada, Ltd.
Collectively Exhaustive Events
• Contains all elementary events for an experiment
E1
E2
E3
Sample Space with three
collectively exhaustive events
Business Statistics, Can. Ed. © 2010 John Wiley & Sons Canada, Ltd.
Complementary Events
• All elementary events not in the event ‘A’
are in its complementary event.
Sample
Space
A
A
P(SampleSpace)  1
P( A)  1  P( A)
Business Statistics, Can. Ed. © 2010 John Wiley & Sons Canada, Ltd.
Counting the Possibilities
• mn Rule
• Sampling from a Population with Replacement
• Combinations: Sampling from a Population without Replacement
Business Statistics, Can. Ed. © 2010 John Wiley & Sons Canada, Ltd.
mn Rule
• If an operation can be done in m ways and a second operation can be done in n ways, then there are mn ways for the two operations to occur in order.
• A cafeteria offers 5 salads, 4 meats, 8 vegetables, 3 breads, 4 desserts, and 3 drinks. A meal consists of one serving of each of the items. How many meals are available? • (Ans: 548343 = 5,760 meals.)
Business Statistics, Can. Ed. © 2010 John Wiley & Sons Canada, Ltd.
Sampling from a Population with
Replacement
• A tray contains 1,000 individual tax returns. If 3 returns are randomly selected with replacement from the tray, how many possible samples are there?
• (N)n = (1,000)3 = 1,000,000,000
Business Statistics, Can. Ed. © 2010 John Wiley & Sons Canada, Ltd.
Combinations: Sampling from a
Population without Replacement
• This counting method uses combinations
• Selecting n items from a population of N without replacement
Business Statistics, Can. Ed. © 2010 John Wiley & Sons Canada, Ltd.
Combinations: Sampling from a
Population without Replacement
• For example, suppose a small law firm has 16 employees and three are to be selected randomly to represent the company at the annual meeting of the Bar Association.
• How many different combinations of lawyers could be sent to the meeting?
• Answer: NCn = 16C3 = 16!/(3!13!) = 560.
Business Statistics, Can. Ed. © 2010 John Wiley & Sons Canada, Ltd.
Four Types of Probability
• Marginal Probability
• Union Probability
• Joint Probability
• Conditional Probability
Business Statistics, Can. Ed. © 2010 John Wiley & Sons Canada, Ltd.
Four Types of Probability
Marginal
Union
Joint
Conditional
P( X )
P( X  Y )
P( X  Y )
P( X| Y )
The probability
of X occurring
X
The probability
of X or Y
occurring
The probability
of X and Y
occurring
X Y
The probability
of X occurring
given that Y
has occurred
X Y
Y
Business Statistics, Can. Ed. © 2010 John Wiley & Sons Canada, Ltd.
General Law of Addition
P( X Y)  P( X)  P(Y)  P( X Y)
X
Y
Business Statistics, Can. Ed. © 2010 John Wiley & Sons Canada, Ltd.
General Law of Addition -- Example
P( N  S)  P( N )  P(S)  P( N  S)
.70
P ( N )  .7 0
S
N
.56
.67
P ( S )  .6 7
P ( N  S )  .5 6
P ( N  S )  .7 0  .6 7  .5 6
 0 .8 1
Business Statistics, Can. Ed. © 2010 John Wiley & Sons Canada, Ltd.
Office Design Problem
Probability Matrix
Noise
Reduction
Yes
No
Total
Increase
Storage Space
Yes
No
.14
.56
.19
.11
.33
.67
Business Statistics, Can. Ed. © 2010 John Wiley & Sons Canada, Ltd.
Total
.70
.30
1.00
Office Design Problem
Probability Matrix
Noise
Reduction
Yes
No
Total
Increase
Storage Space
Yes
No
.14
.56
.19
.11
.33
.67
Total
.70
.30
1.00
P( N  S )  P( N )  P(S )  P( N  S )
 .70  .67  .56
 .81
Business Statistics, Can. Ed. © 2010 John Wiley & Sons Canada, Ltd.
Venn Diagram of the X or Y
but not Both Case
X
Y
Business Statistics, Can. Ed. © 2010 John Wiley & Sons Canada, Ltd.
The Neither/Nor Region
X
Y
P( X  Y )  1  P( X  Y )
Business Statistics, Can. Ed. © 2010 John Wiley & Sons Canada, Ltd.
The Neither/Nor Region
N
S
P( N  S )  1  P( N  S )
 1.81
.19
Business Statistics, Can. Ed. © 2010 John Wiley & Sons Canada, Ltd.
Special Law of Addition
If X and Y are mutually exclusive,
P( X  Y )  P( X )  P(Y )
Y
X
Business Statistics, Can. Ed. © 2010 John Wiley & Sons Canada, Ltd.
Special Law of Multiplication
for Independent Events
• General Law
P( X  Y )  P( X)  P(Y| X)  P(Y )  P( X| Y )
• Special Law
If events X and Y are independent,
P( X )  P( X | Y ), and P(Y )  P(Y | X ).
Consequently,
P( X  Y )  P( X )  P( Y )
Business Statistics, Can. Ed. © 2010 John Wiley & Sons Canada, Ltd.
Law of Conditional Probability
• The conditional probability of X given Y is the joint probability of X and Y divided by the marginal probability of Y.
P( X  Y ) P(Y| X)  P( X)
P( X| Y ) 

P(Y)
P(Y )
Business Statistics, Can. Ed. © 2010 John Wiley & Sons Canada, Ltd.
Law of Conditional Probability
N
S
.56
.70
P ( N )  .7 0
P ( N  S )  .5 6
P(N  S)
P (S| N ) 
P(N )
.5 6

.7 0
 .8 0
Business Statistics, Can. Ed. © 2010 John Wiley & Sons Canada, Ltd.
Independent Events
• If X and Y are independent events, the occurrence of Y does not affect the probability of X occurring.
• If X and Y are independent events, the occurrence of X does not affect the probability of Y occurring. If X and Y are independent events,
P( X | Y )  P( X ), and
P(Y | X )  P(Y ).
Business Statistics, Can. Ed. © 2010 John Wiley & Sons Canada, Ltd.
Independent Events
Demonstration Problem 4.10
P( A  G) 14

 0.33
42
P(G)
P( A | G)  0.33  P( A)  0.28
P( A | G) 
Business Statistics, Can. Ed. © 2010 John Wiley & Sons Canada, Ltd.
P( A)  0.28
Independent Events
Demonstration Problem 4.11
D
E
A
8
12
20
B
20
30
50
C
6
9
15
34
51
85
Business Statistics, Can. Ed. © 2010 John Wiley & Sons Canada, Ltd.
8
P( A| D) 
.2353
34
20
P( A) 
.2353
85
P( A| D)  P( A)  0.2353
Revision of Probabilities: Bayes’ Rule
• An extension to the conditional law of probabilities
• Enables revision of original probabilities with new information
P ( X i| Y ) 
P ( Y | Xi ) P ( Xi )
P ( Y | X 1) P ( X 1)  P ( Y | X 2 ) P ( X 2 )   P ( Y | X n ) P ( X n )
Business Statistics, Can. Ed. © 2010 John Wiley & Sons Canada, Ltd.
Revision of Probabilities
with Bayes' Rule: Ribbon Problem
P(Pr airie)  0.65
P( Badlands)  0.35
P(d | Pr airie)  0.08
P(d | Badlands)  0.12
P(d | Prairie)  P(Pr airie)
P(d | Pr airie)  P(Pr airie)  P(d | Badlands)  P( Badlands)
(0.08)(0.65)

 0.553
(0.08)(0.65)  (0.12)(0.35)
P(d | Badlands)  P( Badlands)
P( Badlands | d ) 
P(d | Pr airie)  P(Pr airie)  P(d | Badlands)  P( Badlands)
(0.12)(0.35)

 0.447
(0.08)(0.65)  (0.12)(0.35)
P(Pr airie | d ) 
Business Statistics, Can. Ed. © 2010 John Wiley & Sons Canada, Ltd.
Revision of Probabilities
with Bayes’ Rule: Ribbon Problem
Prior
Probability
Event
Prairie
P ( Ei )
0.65
Conditional
Probability
Joint
Probability
Revised
Probability
P(d| Ei )
P(Ei  d) P( Ei| d )
0.08
0.052
0.052
0.094
=0.553
Badlands
0.35
0.12
0.042
0.042
0.094
0.094
=0.447
Business Statistics, Can. Ed. © 2010 John Wiley & Sons Canada, Ltd.
Probability for a Sequence
of Independent Trials
• 25 percent of a bank’s customers are commercial (C) and 75 percent are retail (R). • Experiment: Record the category (C or R) for each of the next three customers arriving at the bank.
• Sequences with 1 commercial and 2 retail customers.
– C1
R2
R3
– R1
C2
R3
– R1
R2
C3
Business Statistics, Can. Ed. © 2010 John Wiley & Sons Canada, Ltd.
Probability for a Sequence
of Independent Trials
• Probability of specific sequences containing 1 commercial and 2 retail customers, assuming the events C and R are independent
 1   3  3 9
P(C1  R2  R3)  P(C) P( R) P( R)        
 4   4   4  64
 3  1   3 9
P( R1  C2  R3)  P( R) P(C) P( R)        
 4   4   4  64
 3  3  1  9
P( R1  R2  C3)  P( R) P( R) P(C)        
 4   4   4  64
Business Statistics, Can. Ed. © 2010 John Wiley & Sons Canada, Ltd.
Check whether
independent
X
and
Y
in
the
following
table
are
Y
|
|
0
1
p(x)
----|-----------------------1 |
1/8
1/8
1/4
|
|
X
0 |
2/8
2/8
2/4
|
|
1 |
1/8
1/8
1/4
|
P(y)
2/4
2/4
1
Anaswer: P(Xi and Yi) = P(Xi) P(Yi) for all i , hence they
are indepwndent.
textbook example page 141
E1=Prairie E2=Badlands D=side effects
P(E1)=0.65 P(E2)=0.35 P(D/E1)=0.08 P(D/E2)=0.12
P(E1/D)= P(D/E1)P(E1)/(P(D/E1)P(E1)+P(D/E2)P(E2))
= (0.08 * 0.65) /(0.08*0.65+0.12*0.35) = 0.553
P(E2/D)= P(D/E2)P(E2)/(P(D/E1)P(E1)+P(D/E2)P(E2))
= (0.12 * 0.35) /(0.08*0.65+0.12*0.35) = 0.447
Ch4 probability examples
Consider an ordinary deck of playing cards. What is the probability of
drawing, in a random drawing of a single card, a face card (jack, queen,
or king) or a diamond?
F=face card
D=diamond
P(F) = 12/52
P(D) = 13/52
P(F and D)=3/52
P(F or D)=P(F) + P(D) - P(F and D)
=12/52 + 13/52 - 3/52 = 11/26
Suppose two cards are dealt from a well-shuffled deck. They are placed side-by-side face
down on the table, the first card to the left of the second.
a) What is the probability that at least one of the cards is an ace?
P(A1 or A2) = P(A1 and A2)+P(notA1 and A2)+(A1 and notA2)
= (4/52)*(3/51)+(48/52)*(4/51)+(4/52)*(48/51)
= 33/221
b) If the first card is turned over and seen to be an ace, what is now
the probability that the second card is an ace?
P(A2/A1) = P(A1 and A2)/P(A1)
= (4/52)*(3/51)/(4/52)
= 3/51 = 1/17
Suppose Mr. Jones chooses at random one of the integers 1,2, or 3.
Then he throws as many dice as indicated by the chosen number. What
is the probability that he will score a total of 4 points?
P(4)=P(1d and 4)+P(2d and 4) +P(3d and 4)
1d => 4
2d => (2,2)(1,3) (3,1)
3d => (1,1,2) (1,2,1) (2,1,1)
P(4) = (1/3)*(1/6+3/36+3/216) = 19/216
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