Transcript Basic Business Statistics: Concepts & Applications

Basic Probability
1
Sample Spaces
Collection of all possible outcomes
e.g.: All six faces of a die:
e.g.: All 52 cards
a deck of
bridge cards
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Events and Sample Spaces
An event is a set of basic outcomes from
the sample space, and it is said to occur
if the random experiment gives rise to
one of its constituent basic outcomes.
Simple Event
Outcome With 1 Characteristic
Joint Event
2 Events Occurring Simultaneously
Compound Event
One or Another Event Occurring
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Simple Event
A: Male
B: Over age 20
C: Has 3 credit cards
D: Red card from a deck of bridge cards
E: Ace card from a deck of bridge cards
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Joint Event
D and E, (DE):
Red, ace card from a bridge deck
A and B, (AB):
Male, over age 20
among a group of
survey respondents
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Intersection
• Let A and B be two events in the sample
space S. Their intersection, denoted AB,
is the set of all basic outcomes in S that
belong to both A and B.
• Hence, the intersection AB occurs if and
only if both A and B occur.
• If the events A and B have no common
basic outcomes, their intersection AB is
said to be the empty set.
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Compound Event
D or E, (DE):
Ace or Red card from bridge deck
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Union
• Let A and B be two events in the
sample space S. Their union, denoted
AB, is the set of all basic outcomes
in S that belong to at least one of these
two events.
• Hence, the union AB occurs if and
only if either A or B or both occurs
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Event Properties
Mutually Exclusive
Two outcomes that cannot
occur at the same time
E.g. flip a coin, resulting in
head and tail
Collectively Exhaustive
One outcome in sample space
must occur
E.g. Male or Female
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Special Events
Null Event
Club & Diamond on
1 Card Draw
Null Event

Complement of Event
For Event A, All
Events Not In A:
A' or A
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What is Probability?
1.Numerical measure of
1
likelihood that the event
will occur
Simple Event
Joint Event
Compound
Certain
.5
2.Lies between 0 & 1
3.Sum of events is 1
0
Impossible
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Concept of Probability
A Priori classical probability, the probability of
success is based on prior knowledge of the process
involved.
i.e. the chance of picking a black card from a
deck of bridge cards
Empirical classical probability, the outcomes are
based on observed data, not on prior knowledge of a
process.
i.e. the chance that individual selected at random
from the Kardan employee survey if satisfied
with his or her job. (.89)
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Concept of Probability
Subjective probability, the chance of
occurrence assigned to an event by a
particular individual, based on his/her
experience, personal opinion and
analysis of a particular situation.
i.e. The chance of a newly designed style of
mobile phone will be successful in market.
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Computing Probabilities
• The probability of an event E:
number of event outcomes
P( E ) 
total number of possible outcomes in the sample space
X

T
e.g. P(
) = 2/36
(There are 2 ways to get one 6 and the other 4)
• Each of the outcomes in the sample space is
equally likely to occur
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Concept of Probability
Experimenter
Number
of Trials
Number
of heads
Relative
Frequency
AITBAR
2048
1061
0.5181
ABBASI
4040
2048
0.5069
WAHIDI 12000
6019
0.5016
12012
0.5005
TARIQ
24000
What conclusion can be drawn from the
observations?
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Presenting Probability &
Sample Space
1. Listing
S = {Head, Tail}
2. Venn Diagram
3. Tree Diagram
4. Contingency
Table
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Venn Diagram
Example: Kardan Employee Survey
Event: A = Satisfied, Ā = Dissatisfied
A
Ā
S
P(A) = 356/400 = .89, P(Ā) = 44/400 = .11
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Tree Diagram
Example: Kardan Employee Survey
Joint
Probability
Advanced .485
Satisfied
Kardan
Employee
P(A)=.89
Not
Satisfied
P(Ā)=.11
Not
.405
Advanced
Advanced .035
Not
.075
Advanced
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Joint Probability
Using Contingency Table
Event
Event
B1
B2
Total
A1
P(A1  B1) P(A1  B2) P(A1)
A2
P(A2  B1) P(A2  B2) P(A2)
Total
P(B1)
Joint Probability
P(B2)
1
Marginal (Simple)
Probability
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Joint Probability
Using Contingency Table
Kardan Employee Survey
Joint Probability
Satisfied
Advanced
Not
Advanced
Total
Not
Satisfied
Total
.485
.035
.52
.405
.075
.48
.89
.11
1.00
Simple Probability
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Use of Venn Diagram
Fig. 3.1: AB, Intersection of events A & B,
mutually exclusive
Fig. 3.2: AB, Union of events A & B
Fig. 3.3: Ā, Complement of event A
Fig. 3.4 and 3.5:
The events AB and ĀB are mutually
exclusive, and their union is B.
(A  B)  (Ā  B) = B
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Use of Venn Diagram
Let E1, E2,…, Ek be K mutually exclusive and
collective exhaustive events, and let A be
some other event. Then the K events E1  A,
E2  A, …, Ek  A are mutually exclusive,
and their union is A.
(E1  A)  (E2  A)  …  (Ek  A) = A
(See supplement Fig. 3.7)
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Compound Probability
Addition 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)
4. Probability of Complement
P(A) + P(Ā) = 1. So, P(Ā) = 1  P(A)
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Addition Rule: Example
A hamburger chain found that 75% of all customers
use mustard, 80% use ketchup, 65% use both.
What is the probability that a particular customer
will use at least one of these?
A = Customers use mustard
B = Customers use ketchup
AB = a particular customer will use at least one
of these
Given P(A) = .75, P(B) = .80, and P(AB) = .65,
P(AB) = P(A) + P(B)  P(AB)
= .75 + .80  .65= .90
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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(B)>0
P(B)
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Example
Recall the previous hamburger chain
example, what is the probability that a
ketchup user uses mustard?
P(A|B) = P(AB)/P(B)
= .65/.80 = .8125
Please pay attention to the difference
from the joint event in wording of the
question.
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Conditional Probability
Draw a card, what is the probability of black ace?
What is the probability of black ace when black happens?
Black
Event (Ace and Black)
Black
(S)
Ace
S
Black ‘Happens’: Eliminates All Other Outcomes and
Thus Increase the Conditional Probability
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Conditional Probability Using
Contingency Table
Conditional Event: Draw 1 Card. Note Black Ace
Color
Red
Black
Ace
2
2
Total
4
Non-Ace
24
24
48
Total
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26
52
Type
Revised
Sample
Space
P(Ace and Black)
2/52
P(Ace|Black) =
= 2/26
=
P(Black)
26/52
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Game Show
• Imagine you have been selected for a game
show that offers the chance to win an
expensive new car. The car sits behind one of
three doors, while monkeys reside behind the
other two.
• You choose a door, and the host opens one of
the remaining two, revealing a monkey.
• The host then offers you a choice: stick with
your initial choice or switch to the other, stillunopened door.
• Do you think that switch to the other door
would increase your chance of winning the car?
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Statistical Independence
1. Event Occurrence Does Not Affect
Probability of Another Event
e.g. Toss 1 Coin Twice, Throw 3 Dice
2. Causality Not Implied
3. Tests For Independence
P(A | B) = P(A), or P(B | A) = P(B),
or P(A and B) = P(A)P(B)
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Statistical Independence
Kardan Employee Survey
Note: (.52)(.89) = .4628  .485
Satisfied
Advanced
Not
Advanced
Total
P(A1and B1)
Not
Satisfied
Total
.485
.035
.52
.405
.075
.48
.89
.11
1.00
P(A1)
P(B1)
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Multiplication Rule
1. Used to Get Joint Probabilities for
Intersection of Events (Joint Events)
2. P(A and B) = P(A  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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Randomized Response
An approach for solicitation of honest answers
to sensitive questions in surveys.
• A survey was carried out in spring 1997 to
about 150 undergraduate students taking
Statistics for Business and Economics at
Peking University.
• Each student was faced with two questions.
• Students were asked first to flip a coin and
then to answer question a) if the result was
the national emblem and b) otherwise.
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Practice of Randomized Response
Question
a) Is the second last digit of your office phone
number odd?
b) Recall your undergraduate course work, have you
ever cheated in midterm or final exams?
Respondents, please do as follows:
1. Flip a coin.
2. If the result is the head, answer question a);
otherwise answer question b). Please circle “Yes”
or “No” below as your answer.
Yes
No
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Bayes’ Theorem
1. Permits Revising Old
Probabilities Based on
New Information
2. Application of
Conditional Probability
3. Mutually Exclusive
Events
Prior
Probability
New
Information
Apply Bayes'
Theorem
Revised
Probability
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Permutation and Combination
Counting Rule 1
Example：In TV Series: Kangxi Empire,
Shilang tossed 50 coins, the number of
outcomes is 2·2 ·… ·2 ＝ 250. What is the
probability of all coins with heads up?
If any one of n different mutually exclusive
and collectively exhaustive events can occur
on each of r trials, the number of possible
outcomes is equal to：n·n ·… ·n ＝ nr
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Permutation and Combination
Application: Lottery Post Card (Post of China)
Six digits in each group (2000 version)
Winning the first prize: No.035718
Winning the 5th prize: ending with number 3
If there are k1 events on the first trial, k2 events
on the second trial, and kr events on the rth
trial, then the number of possible outcomes
is:
k1· k2 ·… · kr
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Permutation and Combination
Applicatin: If a license plate consists of 3 letters
followed by 3 digits, the total number of
outcomes would be? (most states in the US)
Application: China License Plates
How many licenses can be issued?
Style 1992: one letter or digit plus 4 digits.
Style 2002: 1) three letters + three digits
2) three digits + three letters
3) three digits + three digits
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Permutation and Combination
Counting Rule 2
Example: The number of ways that 5 books could
be arranged on a shelf is: (5)(4)(3)(2)(1) = 120
The number of ways that all n objects can be
arranged in order is：
Pnn = n(n -1)(n -2)(2)(1) = n!
Where n! is called factorial and 0! is
defined as 1.
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Permutation and Combination
Counting Rule 3: Permutation
Example：What is the number of ways of
arranging 3 books selected from 5 books in
order? (5)(4)(3) = 60
The number of ways of arranging r objects
selected from n objects in order is：
n!
P 
(n  r )!
n
r
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Permutation and Combination
Counting Rule 4: Combination
Example：The number of combinations of 3 books
selected from 5 books is
(5)(4)(3)/[(3)(2)(1)] = 10
Note: 3! possible arrangements in order are irrelevant
The number of ways that arranging r objects selected
from n objects, irrespective of the order, is equal to
 n
n!
C    
 r  r!(n  r )!
n
r
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