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PROBABILITY

Business Statistics

Dr. Gunjan MalhotraAssistant Professor

Institute of Management Technology, Ghaziabadgmalhotra@imt.edu

mailforgunjan@gmail.com

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Probability is:

A quantitative measure of uncertainty A measure of the strength of belief in the

occurrence of an uncertain event A measure of the degree of chance or

likelihood of occurrence of an uncertain event Measured by a number between 0 and 1 (or

between 0% and 100%)

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Probability as a Numerical Measureof the Likelihood of Occurrence

0 1.5

Increasing Likelihood of Occurrence

Probability:

The eventis veryunlikelyto occur.

The eventis veryunlikelyto occur.

The occurrenceof the event is

just as likely asit is unlikely.

The occurrenceof the event is

just as likely asit is unlikely.

The eventis almostcertain

to occur.

The eventis almostcertain

to occur.

Uses of probability

❚ Classical probability models used instatistical theory

❙ Sensitivity, Specificity, Predictive value ❙ Relative risk ❙ Interpretation of other rates

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Structure of Probability

• Experiment• Event• Elementary Events• Sample Space• Unions and Intersections• Collectively Exhaustive Events• Complementary Events

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• Process that leads to one of several possible outcomes *, e.g.:

Coin toss• Heads, Tails

Throw die• 1, 2, 3, 4, 5, 6

Pick a card– AH, KH, QH, ...

Introduce a new product• Each trial of an experiment has a single observed

outcome.• The precise outcome of a random experiment is

unknown before a trial.

Experiment

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• Simple event– An outcome from a sample space with one

characteristic/Collection of outcomes having a common characteristic

– e.g., A red card from a deck of cards E.g.: Even number

– A = {2,4,6}

• Complement of an event A (denoted A’)– All outcomes that are not part of event A/ Event A occurs if an

outcome in the set A occurs– e.g., All cards that are not diamonds

Events

Event A A’

SampleSpace S

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• Joint event– Involves two or more characteristics simultaneously

• Example: An ace that is also red from a deck of cards

Sample Space or Event Set– Set of all possible outcomes (universal set) for a given

experiment Example: Roll a regular six-sided die

– S = {1,2,3,4,5,6}

• Elementary Event– cannot be decomposed or broken down into other events/

An outcome from a sample space with one characteristic• Example: A red card from a deck of cards

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Elementary Events

• A automobile consultant records fuel type and vehicle type for a sample of vehicles

2 Fuel types: Gasoline, Diesel3 Vehicle types: Truck, Car, SUV

6 possible elementary events: e1 Gasoline, Truck e2 Gasoline, Car e3 Gasoline, SUV e4 Diesel, Truck e5 Diesel, Car e6 Diesel, SUV

Gasoline

Diesel

CarTruck

Truck

Car

SUV

SUV

e1

e2

e3

e4

e5

e6

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Sample Space

The Sample Space is the collection of all possible events

e.g. All 6 faces of a die:

e.g. All 52 cards of a bridge deck:

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Intersection (And)– a set containing all elements in both A and B

Union (Or)– a set containing all elements in A or B or both

A B

A B

Unions and Intersection

BA

A B

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Collectively Exhaustive Events

• A set of events is collectively exhaustive if one of the events must occur

• Example: head or tails

E1 E2 E3

Sample Space with three collectively exhaustive events

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Complementary Events

• All elementary events not in the event ‘A’ are in its complementary event.

SampleSpace A

P Sample Space( ) 1

P A P A( ) ( ) 1A

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An example

• An experiment involving a single coin toss• There are two possible outcomes, Heads and Tails• Sample space S is {H,T}• If coin is fair, should assign equal probabilities to 2

outcomes• Since they have to sum to 1

– P({H}) = 0.5– P({T}) = 0.5– P({H,T}) = P({H})+P({T}) = 1.0

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Another example

• Experiment involving 3 coin tosses• Outcome is a 3-long string of H or T

– S ={HHH,HHT,HTH,HTT,THH,THT,TTH,TTT}• Assume each outcome is equally likely (equiprobable)

– “Uniform distribution”• What is probability of the event A that exactly 2 heads

occur?– A = {HHT,HTH,THH}– P(A) = P({HHT})+P({HTH})+P({THH})– = 1/8 + 1/8 + 1/8– =3/8

Probability Concepts• Mutually Exclusive Events/Disjoint Sets

– If E1 (black card) occurs, then E2 (red card) cannot occur

– E1 and E2 have no common elements• Example: A card cannot be Black and Red at the same time.

• Independent events– Occurrence of one does not influence the probability

of occurrence of the other

• Dependent: – Occurrence of one affects the probability of the other

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Independent EventsE1 = heads on one flip of fair coin

E2 = heads on second flip of same coin

Result of second flip does not depend on the result of the first flip.

Dependent EventsE1 = rain forecasted on the news

E2 = take umbrella to work

Probability of the second event is affected by the occurrence of the first event

Independent vs. Dependent Events

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Assigning Probability

• There are three approaches to assess the probability of an uncertain event:

• Objective or Classical Probability Assessment

Relative Frequency of Occurrence

Subjective Probability Assessment

P(Ei) =Number of ways Ei can occurTotal number of elementary events

Relative Freq. of Ei =Number of times Ei occurs

N

An opinion or judgment by a decision maker about the likelihood of an event

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Objective or Classical Probability

based on equally-likely eventsbased on long-run relative frequency of eventsnot based on personal beliefs is the same for all observers (objective) If an experiment has n possible outcomes, this method would assign

a probability of 1/n to each outcome.examples:

toss a coin, throw a die, pick a card gambling models• applicable to games of chance

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Relative frequency theorybased on long-run relative frequency of eventsFifty percent chance of occurrence of a particular event Is objective as no personal judgment is involvedexample: toss a coin – head or tale

Number ofPolishers Rented

01234

Numberof Days

4 61810 240

Probability

.10 .15 .45 .25 .051.00

4/404/40

Each probability assignment is given by dividing the frequency (number of days) by the total frequency (total number of days).

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Subjective Probability

based on personal beliefs, experiences, prejudices, intuition - personal judgment

different for all observers (subjective) When economic conditions and a company’s circumstances change rapidly,

then we can use any data available as well as our experience and intuition, but ultimately a probability value should express our degree of belief that the experimental outcome will occur.

examples: Super Bowl, elections, new product introduction, snowfall, price movements of a particular stock, selecting a leader.

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Four Types of Probability

Marginal

The probability of X occurring

Union

The probability of X or Y occurring

Joint

The probability of X and Y occurring

Conditional

The probability of X occurring given that Y has occurred

YX YX

Y

X

P X( ) P X Y( ) P X Y( ) P X Y( | )

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General Law of Addition

The addition law provides a way to compute the probability of event A, or B, or both A and B occurring.

P(A U B) = P (A or B) = P(A) + P(B) - P(A ∩ B)

Mutually exclusive events:

If A and B are mutually exclusive, then

P(B)P(A)B)P(A so 0B)P(A

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Marginal Probability Example

BlackColor

Type Red Total

Ace 2 2 4

Non-Ace 24 24 48

Total 26 26 52

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4

52

2

52

2)()Re(P(Ace) BlackandAcePdandAceP

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Addition Rule for Events That are Mutually Exclusive - Example

• P (A or B) = P(A) + P(B)

• P(Red or Ace) = P(Red) +P(Ace)

= 26/52 + 4/52 = 30/52

BlackColor

Type Red Total

Ace 2 2 4

Non-Ace 24 24 48

Total 26 26 52

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Addition Rule for Events That are Not Mutually Exclusive - Example

P (A or B) = P(A) + P(B) - P(A and B)

P(Red or Ace) = P(Red) +P(Ace) - P(Red and Ace)

= 26/52 + 4/52 - 2/52 = 28/52Don’t count the two red aces twice!

BlackColor

Type Red Total

Ace 2 2 4

Non-Ace 24 24 48

Total 26 26 52

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Conditional Probability

• Conditional Probability - Probability of A given B

Independent events:

0)( ,)(

)()( BPwhereBPBAPBAP

P AB P A

P B A P B

( ) ( )

( ) ( )

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Rules of conditional probability:Rules of conditional probability:

If events A and D are statistically independent:

so

so

P AB P A BP B

( ) ( )( )

P A B P AB P B

P B A P A

( ) ( ) ( )

( ) ( )

P AD P A

P D A P D

( ) ( )

( ) ( )

P A D P A P D( ) ( ) ( )

Conditional Probability

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• What is the probability that a car has a CD player, given that it has AC ?

i.e., we want to find P(CD | AC)

Conditional Probability Example (Ques)

• Of the cars on a used car lot, 70% have air conditioning (AC) and 40% have a CD player (CD). 20% of the cars have both.

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Conditional Probability Example (Ans.)

No CDCD Total

AC 0.2 0.5 0.7

No AC 0.2 0.1 0.3

Total 0.4 0.6 1.0

• Of the cars on a used car lot, 70% have air conditioning (AC) and 40% have a CD player (CD).20% of the cars have both.

0.28570.7

0.2

P(AC)

AC)andP(CDAC)|P(CD

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Conditional Probability Example

No CDCD Total

AC 0.2 0.5 0.7

No AC 0.2 0.1 0.3

Total 0.4 0.6 1.0

• Given AC, we only consider the top row (70% of the cars). Of these, 20% have a CD player. 20% of 70% is about 28.57%.

0.28570.7

0.2

P(AC)

AC)andP(CDAC)|P(CD

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Independent Events

• Two events are independent if and only if:

• Events A and B are independent when the probability of one event is not affected by the other event

P(A)B)|P(A

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Independent Events- ExampleTest the matrix for the 200 executive responses to determine whether industry type is independent of geographic location.

Geographic Location

NortheastD

SoutheastE

MidwestF

WestG

Finance A .12 .05 .04 .07 .28

Manufacturing B .15 .03 .11 .06 .35

Communications C .14 .09 .06 .08 .37

.41 .17 .21 .21 1.00

28.0)(33.0)|(

28.0)( 33.021.0

07.0

)(

)()|(

APGAP

APGP

GAPGAP

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Multiplication Rules

• Multiplication rule for two events A and B:

P(B)B)|P(AB)andP(A

P(A)B)|P(A Note: If A and B are independent, thenand the multiplication rule simplifies to

P(B)P(A)B)andP(A

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Event M = Markley Oil ProfitableEvent C = Collins Mining Profitable

We know: P(M) = .70, P(C|M) = .5143

Multiplication Law-example

M C = Markley Oil Profitable and Collins Mining Profitable

Thus: P(M C) = P(M)P(M|C)= (.70)(.5143)= .36

Practice Problem

• Question 1.• Given the following contigency table:

What is the probability ofa. event A?b. event A’?c. event A and B?d. event A or B? 36

B B’

A 10 20

A' 20 40

Practice Problem

• Answer 1.• (a) 30/90 = 1/3 = 0.33 • (b) 60/90 = 2/3 = 0.67• (c) 10/90 = 1/9 = 0.11

• (d) 30/90 + 30/90 - 10/90 = 50/90 = 5/9 = 0.556

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• Question 2:• Are large companies less likely to offer board members stock

options than small- to mid sized companies? A survey conducted by the Segal Company of New York found that in a sample of 189 large companies, 40 offered stock options to their board members as part of their non-cash compensation packages. For small-to mid sized companies, 43 of the 180 surveyed indicated that they offer stock options as part of their non- cash compensation packages to their board members (Kemba J. Dunham, “The Jungle: Focus on Recruitment, Pay and Getting Ahead”, The Wall Street Journal, August 21, 2001, B6). Construct a contingency table to evaluate the probabilities. If a company is selected at random, what is the probability that the companya. offered stock options to their board members?b. is small-to mid-sized and did not offer stock options to their board members?c. is small- to mid-sized or offered stock options to their board members?d. explain the difference in the results in (b) and (c ).

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Answer 2:

• (a) P(offered stock options) = 83/369 = 0.2249• (b) P(small-to-midsized and did not offer stock options) = 137/369 =

0.3713• (c) P(small-to-midsized or offered stock options) = (180 + 83 – 43)/369

= 0.5962• (d) The probability of “small-to-midsized or offered stock options”

includes the probability of “small-to-midsized and offered stock options”, the probability of “small-to-midsized but did not offer stock options” and the probability of “large and offered stock options”.

Company Size

Large Small-to-mid sized

Stock Yes 40 43 83

Options No 149 137 286

189 180 369

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• Question 3:

A particular brand of women’s jeans is available in seven different sizes, three different colors, and three different styles. How many different jeans does the store manager need to order to have one pair of each type?

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• Answer 3:

(b) (7)(3)(3) =63

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Question 4.

• Suppose that 47 % of all Americans have flown in an airplane at least once and that 28 % of all Americans have ridden on a train at least once. What is the probability that a randomly selected Americans has either ridden on a train or flown in an airplane? Can this problem be solved? Under what conditions can it be solved? If the problem cannot be solved, what information is needed to make it solvable?

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Answer 4.

• A = event - flown in an airplane at least once• T = event - ridden in a train at least once• • P(A) = .47 P(T) = .28 • • P (ridden either a train or an airplane) =• P(A U T) = P(A) + P(T) - P(A ∩ T) = .47 + .28 - P(A ∩ T)• • Cannot solve this problem without knowing the probability of the

intersection. • We need to know the probability of the intersection of A and T, the

proportion who have ridden both. •

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Question 5.

• A yield improvement study at a semiconductor manufacturing facility provided defect data for a sample of 40 wafers. The following table presents a summary of the responses to two questions: “Were particles found on the die that produced the wafer?” and “Is the wafer good or bad?”

• Condition of Die• Quality of Wafer No Particles Particles Totals• Good 320 14 334• Bad 80 36 116• Total 400 50 450

• a. Suppose you know that a wafer is bad. What then is the probability that it was produced from a die that had particles?

• b. Suppose you know that a wafer is good. What then is the probability that it was produced from a die that had particles?

• c. Are the two events, a good wafer and a die with no particle, statistically independent? Explain.

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Answer 5:

(a)P (had particles | bad) = 36/116 = 0.3103

(b) P (had particles | good) = 14/334 = 0.0419

(c) P (no particles | good ) = 320/334 = 0.9581

P (no particles) = 400/450 = 0.8889Since P (no particles | good ) ≠ P (no particles), “a good wafer” and “a die with noparticle” are not statistically independent.

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Question 6:

• A box of nine golf gloves contains two left-handed gloves and seven right-handed gloves.a. If two gloves are randomly selected from the box without replacement, what is the probability that both gloves selected will be right-handed?b. If two gloves are randomly selected from the box without replacement, what is the probability there will be one right-handed glove and one left-handed glove selected?c. If three gloves are selected with replacement, what is the probability that all three will be left-handed?d. If you are sampling with replacement, what would be the answer to (a) and (b)?

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Answer 6:

• (a) P(2 right-handed gloves) = (7/9) (6/8) = 42/72 = 7/12 = 0.5833

• (b) P(1 right-handed and 1 left-handed glove) = (7/9) (2/8) + (2/9) (7/8) = 28/72 = 7/18 = 0.3889

• (c) P(3 left-handed gloves) = (2/9)3

• (d) (a) P(2 right-handed gloves) = (7/9) (7/9) = 49/81= 0.6049

(b) P(1 right-handed and 1 left-handed glove) = (7/9) (2/9) + (2/9) (7/9) = 28/81 = 0.3456

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Question 7:

• The probability that a consumer entering a retail outlet for microcomputers and software packages will buy a computer of a certain type is 0.15. The probability that the consumer will buy a particular software package is 0.10. There is 0.05 probability that the consumer will buy both the computer and software package. What is the probability that the consumer will buy the computer or the software package or both?

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Answer 7:

• (C U S)= 0.15 + 0.10 – 0.05 = 0.20

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Question 8:

• The probability that a bank customer will default on a loan is 0.04 if the economy is high and 0.13 if the economy is not high. Suppose the probability that the economy will be high is 0.65. What is the probability that the person will default on the loan?

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Answer 8:

)H|P(D

Let D, H be the events: customer defaults, economy is high.P(D | H) = 0.04

= 0.13 P(H) = 0.65P(D) = P(D | H) P(H) +

= (.04)(.65) + (.13)(.35) = 0.0715

)H)P(H|P(D

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Question 9:

• In a takeover bid for a certain company, management of the raiding firm believes that the takeover has a 0.65 probability of success if a member of the board of the raided firm resigns, and a 0.30 chance of success if she does not resign. Management of the raiding firm further believes that the chances for a resignation of the member in question are 0.70. What is the probability of a successful takeover?

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Answer 9:

)R|P(T

Let T,R be the events: successful takeover, resignation of a board member. P(T | R) = 0.65

= 0.30 P(R) = 0.70P(T) = P(T | R)P(R) +

= (.65)(.7) + (.30)(.30) = 0.545

)R)P(R|P(T

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Questions???