Basic Business Statistics: Concepts

1
Basic Business Statistics Concepts Applications

• Basic ProbabilityChapter 4

2
Learning Objectives

• Concepts Events, Sample Space, Intersection
  Union
• Tools Venn Diagram, Tree Diagram, Contingency
  Table
• Probability, Conditional Probability, Addition
  Rule Multiplication Rule
• Use Bayes Theorem
• Summarize the Rules of Permutation and Combination

3
Basic Concepts

• Random Experiment is a process leading to at
  least two possible outcomes with uncertainty as
  to which will occur.
• A coin is thrown
• A consumer is asked which of two products he or
  she prefers
• The daily change in an index of stock market
  prices is observed

4
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

5
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

6
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
7
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
8
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.

9
Compound Event
D or E, (D?E) Ace or Red card from bridge deck
10
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

11
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

12
Special Events
Null Event

• Null Event
• Club Diamond on 1 Card Draw
• 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
• likelihood that the event
• will occur
• Simple Event
• Joint Event
• Compound
• 2. Lies between 0 1
• 3. Sum of events is 1

1
Certain
.5
0
Impossible
14
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 Kalosha employee survey if
  satisfied with his or her job. (.89)

15
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
• Each of the outcomes in the sample space is
  equally likely to occur

e.g. P( ) 2/36
(There are 2 ways to get one 6 and the other 4)
17
Concept of Probability

• Experi- Number Number Relative
• menter of Trials of heads Frequency
• Demogen 2048 1061 0.5181
• Buffon 4040 2048 0.5069
• Pearson 12000 6019 0.5016
• Pearson 24000 12012 0.5005
• What conclusion can be drawn from the
  observations?

18
Presenting Probability Sample Space

• 1. Listing
• S Head, Tail
• 2. Venn Diagram
• 3. Tree Diagram
• 4. Contingency
• Table

19
Venn Diagram
Example Kalosha Employee Survey Event A
Satisfied, A Dissatisfied
A
A
S
P(A) 356/400 .89, P(A) 44/400 .11
20
Tree Diagram
Example Kalosha Employee Survey
Joint Probability
Advanced
.485
Satisfied
Not Advanced
.405
P(A).89
Kalosha Employee
Advanced
.035
Not Satisfied
Not Advanced
.075
P(A).11
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Joint Probability Using Contingency Table
Event
Event
B
B
Total
1
2
P(A
)
A
P(A
?
B
)
P(A
?
B
)
1
1
1

1
1

2
P(A
)
A
P(A
?
B
)
P(A
?
B
)
2
2
2

1
2

2
P(B
)
P(B
)
1
Total
1
2
Marginal (Simple) Probability
Joint Probability
22
Joint Probability Using Contingency Table
Kalosha Employee Survey
Joint Probability
Total
Satisfied
Not Satisfied
Advanced
.485
.035
.52
Not Advanced
.405
.075
.48
Total
.89
.11
1.00
Simple Probability
23
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 A, Complement of event A
• Fig. 3.4 and 3.5
• The events A?B and A?B are mutually exclusive,
  and their union is B.
• (A ? B) ? (A ? B) B

24
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)

25
Compound ProbabilityAddition 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(A) 1. So, P(A) 1 ? P(A)

26
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

27
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)gt0
  P(B)

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Example

• Recall the previous hamburger chain example,
  what is the probability that a ketchup user uses
  mustard?
• P(AB) P(A?B)/P(B)
• .65/.80 .8125
• Please pay attention to the difference from the
  joint event in wording of the question.

29
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
30
Conditional Probability Using Contingency Table
Conditional Event Draw 1 Card. Note Black Ace
Color
Revised Sample Space
Type
Red
Black
Total
2
2
4
Ace
24
24
48
Non-Ace
26
26
52
Total
P(Ace and Black) P(Black)
2/52 26/52
P(AceBlack)
2/26

31
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,
  still-unopened door.
• Do you think that switch to the other door would
  increase your chance of winning the car?

32
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)

33
Statistical Independence
Kalosha Employee Survey
Note (.52)(.89) .4628 ? .485
Total
Satisfied
Not Satisfied
Advanced
.485
.035
.52
Not Advanced
.405
.075
.48
Total
.89
.11
1.00
P(B1)
P(A1and B1)
P(A1)
34
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(BA)
  P(B)P(AB)
• 3. For Independent EventsP(A and B) P(A?B)
  P(A)P(B)

35
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.

36
Practice of Randomized Response

• Question
• Is the second last digit of your office phone
  number odd?
• 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 national emblem (??),
  answer question a) otherwise answer question
  b). Please circle Yes or No below as your
  answer.
•   Yes No

37
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
38
Bayes Theorem Formula
P(A
B
P(B
)
)
?
i
i
P(B

A)


i
P(A
P(B
...
B
P(B
)


P(A
B
)
)
)
?
?
1
k
k
1
P(B

A)
?
All Bis are the same event (e.g. B2)!
i
?

P(A)
Same Event
39
Bayess Theorem Example
Fifty percent of borrowers repaid their loans.
Out of those who repaid, 40 had a college
degree. Ten percent of those who defaulted had a
college degree. What is the probability that a
randomly selected borrower who has a college
degree will repay the loan?
B1 repay, B2 default, Acollege degree P(B1)
.5, P(AB1) .4, P(AB2) .1, P(B1A) ?
40
Bayes Theorem Example Table Solution
Event
Prior
Cond.
Joint
Post.
Prob
Prob
Prob
Prob
B
P(B
)
P(AB
)
P(B
A)
P(B

?
A)
i
i
i
i
i
B
.5
.4
.20
.20/.25 .8
X

1
B
.5
.1
.05
.05/.25 .2
2
1.0
P(A) 0.25
1.0
Default
Repay
P(College)
41
Random Testing for AIDS?

• In 1987, the US Secretary of Health proposed to
  test blood to estimate how many Americans, and
  which ones, were infected with AIDS. Continuing
  calls have been made for mandatory testing to
  identify persons with the disease. These calls
  met with considerable resistance from
  statisticians, who speak out against the
  mindless proposals for mandatory AIDS
  testing....
• Refer to Random Testing for AIDS? Chance, vol.
  1, no. 1, 1988

42
Permutation and Combination

• Counting Rule 1
• ExampleIn TV Series Kangxi Empire, Shilang
  tossed 50 coins, the number of outcomes is 22
  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 tonn n nr

43
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

44
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

45
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
• n(n -1)(n -2)?(2)(1) n!
• Where n! is called factorial and 0! is defined
  as 1.

46
Permutation and Combination

• Counting Rule 3 Permutation
• ExampleWhat 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

47
Permutation and Combination

• Counting Rule 4 Combination
• ExampleThe 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

48
Examples

• Applications NBA Draft (2002)
• Randomly choose 4 numbers from 14 numbers
• Example
• 1. World Cup 2002 how many games should a soccer
  team play in a group of four teams?
• 2. There are 10 cities, how many different
  non-stop tickets? How many different prices of
  the tickets?