probability gen

8/12/2019 Probability Gen

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PROBABILITY



Managers often base their decisions on an

analysis of Uncertainties such as

following:

What are chances that sales will decrease if

we increase prices?

How likely is it that the project will finish ontime?

What is the chance that a new investment will

be profitable?



Definition:

Probability is a numerical measure of the

likelihood that an event will occur.



Probabilities values are always assigned on ascale from 0-1.

A probability near zero indicates an event is

unlikely to occur A probability near one indicates an event is

likely to occur

Other probabilities between 0 & 1 representsdegrees of likelihood that an event willoccur.



EXPERIMENT

An experiment is a process that generates

well-defined outcomes.

On any single repetition of an experiment,

one and only one of the possible experimental

outcomes will occur.



Example

EXPERIMENT EXPERIMENTAL

OUTCOME

Toss a coin Head, Tail

Select a part for inspection Defective, Non-Defective

Conduct a sales call Purchase, No Purchase

Roll a die 1, 2, 3, 4, 5, 6

Play a football game Win, Lose, Tie



Sample Space

The sample space for an experiment is the

set of all experimental outcomes.

Sample Points: Experimental outcomes are

also called sample points.



Sample Space for the first

experiment in the previous table

Experiment 1

S = {H, T}

Experiment 2

S = {Defective, Non Defective}

Experiment 4

S = {1,2, 3, 4, 5, 6 }



Example 1: Gender of Two

Children

A newly married couple plan to have two

children. Naturally, they are curious about

whether their children will be boys or girls.

Therefore, we consider the experiment of

having two children.

In order to find the sample space let

B denote the child is a boy

G denote the child is a girl



Therefore, the sample space is

S = { BB , BG , GB , GG } In order to consider the probabilities of these

outcomes, suppose that boys and girls are

equally likely each time a child is born.

Intuitively, this says that each of the samplespace outcomes is equally likely.

This says that there is a 25 % chance of these

outcomes will occur. These probabilities sum to

1.

1( ) ( ) ( ) ( )

4

P BB P BG P GB P GG



A Tree Diagram of the Genders of

Two Children



Events

An event is a set (or collection) of sample

space outcomes

The probability of an event is the sum of the

probabilities of the sample space outcomes

that correspond to the event



Example 1: Gender of Two Children

Experimental Outcomes:

BB, BG, GB, GG

All outcomes equally likely:

P(BB) = … = P(GG) = ¼

P(one boy and one girl) =

P(BG) + P(GB) = ¼ + ¼ = ½

P(at least one girl) =

P(BG) + P(GB) + P(GG) = ¼+¼+¼ = ¾



Probabilities: Equally Likely

Outcomes

If the sample space outcomes (orexperimental outcomes) are all equally

likely, then the probability that an event will

occur is equal to the ratio: The number of ways the event can occur

Over the total number of outcomes

outcomesspacesampleof numberTotal

eventthetocorrespondthatoutcomesspacesampleof Number



Example 2: Newspapers Subscribers

Suppose that 650,000 of the 1,000,000 households in an

eastern U.S. city subscribe to a newspaper called the

Atlantic Journal, and consider randomly selecting one of

the households in this city. That is, consider selecting

one household by giving each and every household in

the city the same chance of being selected.

Let A be the event that the randomly selected

households subscribes to the Atlantic Journal. Then

because the sample space of this experiment consists of

1,000,000 equally likely sample space outcomes

(households), it follows that



the number of households that subscribes to the Atlantic Journal

the total number of households in the city

650,0001,000,000

.65

This says that the probability that the randomly selected

Household subscribes to the Atlantic Journal is .65



Counting Rules

Being able to identify and count the

experimental outcomes is a necessary step in

assigning probabilities.



Multiple – Step Experiments

Consider the experiment of tossing two coins.

How many experimental outcomes are possible for

this experiment?

Step 1: Tossing of the first coin.

Step 2: Tossing of the second coin.

Therefore,

S = {(H, H), (H, T), (T, H), (T, T)}



Counting Rule for multiple –

step experiments

If an experiment can be described as a

sequence of k steps with n1, possible

outcomes on the first step, n2 possible

outcomes on the second step, and so on,then the total number of experimental

outcomes is given by (n1).(n2)……(nk)



Counting Rule for:

Tossing two coins

(2) (2) = 4

Tossing three coins(2) (2) (2) = 8

Rolling two die

(6) (6) = 36



Case:

How the counting rule for multiple step experiments

can be used in the analysis of a capacity expansion

project for the Kentucky Power and Light Company

(KP&L). KP&L is starting a project designed toincrease the generating capacity of one of its plants

in Northern Kentucky. The project is divided into

sequential stages

Stage 1: DesignStage 2: Construction



An analysis of similar construction projects

reveled possible completion times for the

design stage of 2, 3, or 4 months andpossible completion times for the construction

stage of 6, 7, or 8 months.

Because of critical need for additional

electrical power , management set a goal of

10 months for the completion of the entire

project



Experimental Outcomes for KP&L Project

Stage 1

(Design)

Stage 2

(Construction)

Notation for

ExperimentalOutcome

Total Project

Completiontime

2 6 (2, 6) 8

2 7 (2, 7) 9

2 8 (2, 8) 10

3 6 (3, 6) 9

3 7 (3, 7) 10

3 8 (3, 8) 114 6 (4, 6) 10

4 7 (4, 7) 11

4 8 (4, 8) 12



Assigning Probabilities

Classical Method

Relative Frequency Method

Subjective Method



Basic Requirements for

Assigning Probabilities

The probability assigned to each experimental

outcome must be between 0 and 1

The sum of the probabilities for all the

experimental outcomes must equal 1.0. For n

experimental outcomes, this requirement can be

written as

iall for E P i 10

1............21 n

E P E P E P



Classical Method

The classical method of assigningprobabilities is appropriate when all theexperimental outcomes are equally likely.

If n experimental outcomes are possible, aprobability of 1/n is assigned to eachexperimental outcome.

Example: Tossing of a fair coin.

Rolling a die



The Relative Frequency

Method

Appropriate when data are available toestimate the proportion of the time theexperimental outcome will occur if the

experiment is repeated to large number oftimes.

Example: Consider a study of waiting times inthe X-ray department for a local hospital. A

clerk recorded the number of patients waitingfor service at 9:00 a.m. on 20 successivedays obtained the following results.



Number Waiting Number of Days

Outcome Occurred

0 2

1 5

2 6

3 4

4 3

Total 20



Subjective Method

Appropriate when one cannot realistically

assume that the experimental outcomes are

equally likely and when little relevant data are

available.

Any available information may be used such

as our experience or intuition.



Example:

Consider the case in which Tom and Jack make an

offer to purchase a house.

Two possible outcomes are:

= their offer is accepted. = their offer is rejected.

If Jack believes that the probability their offer will be

Accepted is 0.8.

Thus

1 E

2 E

2.0

8.0

2

1

E P

E P



Event and their Probabilities

An event is a collection of sample points.

Let C denote the event that the project is

completed in ten months or less.

C = {(2, 6), (2, 7), (2, 8), (3, 6), (3, 7), (4, 6)}

L = The event that the project is completed in

less than 10 months

{(2, 6), (2, 7), (3,6)}

Completion Time (months)



Stage 1

(Design)

Stage 2

(Construction)

Sample

Point

No. of Past Projects

having these

Completion times2 6 (2, 6) 6

2 7 (2, 7) 6

2 8 (2, 8) 2

3 6 (3, 6) 4

3 7 (3, 7) 8

3 8 (3, 8) 2

4 6 (4, 6) 2

4 7 (4, 7) 4

4 8 (4, 8) 6

Completion Time (months)

Total = 40



Sample Point Project

Completion

Time

Probability of

Sample Point

(2, 6) 8 months P(2, 6) = 6/40 =.15

(2, 7) 9 months P(2, 7) = 6/40 =.15

(2, 8) 10 months P(2, 8) = 2/40 =.05

(3, 6) 9 months P(3, 6) = 4/40 =.10

(3, 7) 10 months P(3, 7) = 6/40 =.20

(3, 8) 11 months P(3, 8) = 6/40 =.05

(4, 6) 10 months P(4, 6) = 6/40 =.05(4, 7) 11 months P(4, 7) = 6/40 =.10

(4, 8) 12 months P(4, 8) = 6/40 =.15

Total = 1.00



Probability of event C (i.e. project is completed in 10

months or less) is given by:

Similarly, probability that the project is completed in

less than 10 months is given by:

( ) (2,6) (2,7) (2,8) (3,6) (3,7) (4,6)

= 0.15 0.15 0.05 0.10 0.20 0.05 0.70

P C P P P P P P

( ) (2,6) (2,7) (3,6)

= 0.15 0.15 0.10 0.40

P L P P P



Complement of an Event

Given an event A, the complement of A is

defined to be the event consisting of all sample

points that are not in A. The complement of A is

denoted byP(A) + P( ) = 1

c

A c A

Event A c A

Sample Space S

Complement of Event A



Computing Probability Using the

Complement

P(A) = 1 – P( )

Example:

Consider the case of a sales manager who, after

reviewing sales reports, states that 80% of newcustomer contacts result in no sale.

By allowing A = the event of a sale and

= the event of no sale

Using the above eqn

P(A) = .20

c A

c A



Example 2:

The probability that a randomly selected household

In an eastern U.S. city subscribers to the Atlantic Journal is

.65.It follows that the probability of the complement of this event

Is 1 - .65 = .35



Addition Law

The addition law is helpful when we are

interested in knowing the probability that at

least one of the two occurs.

That is, with events A and B we are

interested in knowing the probability that

event A or event B or both occur.



Union of two Events

The union of A and B is the event containing

all sample points belonging to A or B or both.

The union is denoted by A U B.

Sample Space S

Event AEvent B



Intersection of Two Events

Given two events A and B, the intersection of A and

B is the event containing the sample points

belonging to both A and B. The intersection is

denoted by A B.

ADDITION LAW

P(A U B) = P(A) + P(B) – P(A B )



Example:

There is a small assembly plant with 50 employees .Each worker is expected to complete workassignments on time and in such a way that theassembled product will pass a final inspection. On

occasion, some of the workers failed to meet theperformance standards by completing work lateor assembling a defective product. At the end of aperformance evaluation period, the production

manager found that 5 of 50 workers completed worklate, 6 of 50 workers assembled a defective product,and 2 of the 50 workers both completed work lateand assembled a defective product .

Let



L = the event that the work is completed late

D = the event that the assembled product is

defective

The relative frequency information leads to the following

probabilities

After reviewing the performance data, the production manager

decided to assign a poor performance rating to any employee whose

work was either late or defective; thus the event of interest is L U D.

50

04.2D)P(L

50

12.6P(D) 50

10.5)(

L P



What is the probability that the production

manager assigned an employee a poor

performance rating?

P(L U D) = P(L) + P(D) – P(L D)

P(L U D) = .10 + .12 - .04 = .18

This calculation tells us that there.18 is a probabilitythat a randomly selected employee received a poor

performance rating.



Example 2: Newspaper Subscribers

Define events:

A = event that a randomly selected household

subscribes to the Atlantic Journal

B = event that a randomly selected householdsubscribes to the Beacon News

Given:

total number in city, N = 1,000,000

number subscribing to A, N(A) = 650,000

number subscribing to B, N(B) = 500,000

number subscribing to both, N(A∩B) = 250,000



Example: Newspaper Subscribers

Use the relative frequency method to assign

probabilities

25.0000,000,1

000,250

50.0000,000,1

000,500

65.0000,000,1

000,650

B A P

B P

A P



Define events:

A = event that a randomly selected household

subscribes to the Atlantic Journal.

= event that a randomly selected household doesnot subscribes to the Atlantic Journal.

B = event that a randomly selected household

subscribes to the Beacon News.

= event that a randomly selected household doesnot subscribes to the Beacon News.

A

B

Table 2 1: A Summary of the Number of



Table 2.1: A Summary of the Number of

Households Corresponding to the Events

A, , B, and A∩B

Events Subscribes to

Beccon News, B

Does Not

subscribe to

Beacon News,

Total

Subscribes to

Atlantic Journal, A

2,50,000 6,50,000

Does Not

Subscribes to

Atlantic Journal,

3,50,000

Total 5,00,000 5,00,000 10,00,000

B

A

A B



Table 2.2: The number of Households

corresponding to ( and ) A B


Beccon News, B

Does Not subscribe to

Beacon News,

Total

Subscribes to

Atlantic Journal, A

2,50,000 6,50,000 – 2,50,000 =

4,00,000

6,50,000

Does Not

Subscribes to

Atlantic Journal,

3,50,000

Total 5,00,000 5,00,000 10,00,000

B

A




corresponding to ( and B) A


Beccon News, B

Does Not subscribe

to Beacon News,

Total

Subscribes to

Atlantic Journal, A

2,50,000 6,50,000 – 2,50,000 =

4,00,000

6,50,000

Does Not

Subscribes to

Atlantic Journal,

5,00,000 – 2,50,000 =

2,50,000

3,50,000

Total 5,00,000 5,00,000 10,00,000

A

B




corresponding to ( and B) A


Beccon News, B

Does Not subscribe

to Beacon News,

Total

Subscribes to

Atlantic Journal, A

2,50,000 6,50,000 – 2,50,000 =

4,00,000

6,50,000

Does Not

Subscribes to

Atlantic Journal,

5,00,000 – 2,50,000 =

2,50,000

3,50,000 – 2,50,000 =

1,00,000

3,50,000

Total 5,00,000 5,00,000 10,00,000

A

B



Table 2.5: A Contingency Table

summarizing Subscription Data for the

Atlantic Journal and the Beacon News


Beccon News, B

Does Not subscribe

to Beacon News,

Total

Subscribes to

Atlantic Journal, A

2,50,000 4,00,000 6,50,000

Does Not

Subscribes to

Atlantic Journal,

2,50,000 1,00,000 3,50,000

Total 5,00,000 5,00,000 10,00,000

A

B




Refer to the contingency table in Table 2.5 for

all data

For example, the chance that a household

does not subscribe to either newspaper

Want , so from middle row and middle

column of Table 2.5

1000000001

000100BAP .

, ,

,

BAP




The chance that a household subscribes toeither newspaper:

Note that if the joint probability was notsubtracted it would have gotten 1.15, which is

greater than 1, which is absurd The subtraction avoids double counting the joint

probability

90.0

25.050.065.0

B) P(A B) B)=P(A)+P( P(A



Conditional Probability

Often, the probability of an event is

influenced by whether related event already

occurred.

P(A / B)

Indicate that we are considering the

probability of event A given the condition that

event B has occurred.



Example

Consider the situation of the promotion status of maleand female officers of a major metropolitan police forcein eastern united states. The police force consists of1200 officers, 960 men and 240 women. Over the past

years , 324 officers on the police force receivedpromotions. After reviewing the promotion record, acommittee of female officers raised a discrimination caseon the basis that 288 male officers had receivedpromotions but only 36 female officers had been

promoted. The police administration argued that therelatively low number of promotions for female officerswas due not to discrimination, but to the fact thatrelatively few females are members of police force.



Let

M = event an officer is a man.

W = event an officer is a woman.

A = event an officer is promoted.

A’ = event an officer is not promoted.



P(M ∩ A) = 288/1200 = .24 = prob. that a randomly

selected officer is a man and is promoted.

P(M ∩ A') = 672/1200 = .56 = prob. that a randomlyselected officer is a man and is not promoted.

P(W ∩ A) = 36/1200 = prob. that a randomly

selected officer is a woman and is promoted.

P(W ∩ A') = 204/1200 = prob. that a randomlyselected officer is a woman and is not promoted.

Because each of these values gives the probability of the

intersection of two events, the probabilities are called jointprobabilities.



Promotion Status of Police officers Over the

Past Two Years

Men Women Totals

Promoted 288 36 324

Not Promoted 672 204 876

Totals 960 240 1200



Joint Probability Table for Promotions

Men (M) Women (W) Totals

Promoted (A) .24 .03 .27

Not Promoted (A') .56 .17 .73

Totals .80 .20 1.00

Joint probabilities

appear in the body

of the table.

Marginal probabilities

appear in the

margins of the table.



Conditional Probability

Therefore, conditional probability P(A/M) can be computed asthe ratio of the joint probability P(A∩M) to the marginal

probability P(M).

P(A / M) = P(A∩M) = .24 = .30

P(M) .80

P(A / W) = P(A∩W) = .03 = .15

P(W) .20

Thus, the conditional probability values support the argument presented by the female officers.



Independent Events

P(A) = .27, P(A / M) = .30 and P(A / W) = .15

Probability of a promotion (event A) is

affected or influenced by whether the officer

is a man or women.

Particularly because P(A / M) ≠ P(A), we

would say that prob. A and M are dependent

events.

Two events A & B are



Two events A & B are

independent if,

P(A / B) = P(A)

P(B / A) = P(B)

Otherwise the events are dependent.



Multiplication Law

It computes the probability of two events.

It is based on the definition of conditional

probability

P(A∩B) = P(B) . P(A/B)

Or P(A∩B) = P(A) . P(B/A)



Example:

Consider a newspaper circulation department where it is

known that 84% of the households in a particular

neighborhood subscribe to the daily edition of the paper.

If we let D denote the event that a household subscribes

to the daily edition P(D) = .84. In addition, it is known that

the probability that a household that already holds a daily

subscription also subscribes to the Sunday edition (event

S) is .75 i.e. P(S / D) = .75. What is the probability that a

household subscribes to both the Sunday and dailyeditions of the newspaper?



Solution:

P(S∩D) = P(D) . P(S / D)

= .84 × .75 = .63

Therefore 63% of the households subscribes

to both Sunday and daily edition.

Multiplication Law for



Multiplication Law for

Independent Events

P(A∩B) = P(A) . P(B)

Example: A boy wants to marry a girl having

qualities: white complexion, the probability of gettingsuch a girl is one in twenty; handsome dowry – the

probability of getting such a girl is one in fifty;

Westernized manners and etiquettes, the probability

here is one in hundred. Find out the probability ofhis getting married to such a girl when the

possession of these three attributes is independent.



Solution:

Prob. Of a girl with white complexionP(W) = 1/20 = 0.05

Prob. Of a girl with handsome dowry

P(D) = 1/50 = 0.02

Prob. Of a girl with westernized manners & etiquettes

P(E) = 1/100 = 0.01

Since the events are independent, the probability of

simultaneous occurrence of all these qualities.

P(M) = P(W) . P(D) . P(E)

= 0.05 . 0.02 . 0.01 = 0.00001

= 1/ 100000

Bayes’ Theorem



Bayes Theorem

In the discussion of conditional probability, we indicated

that revising probability when new information isobtained is an important phase of probability analysis.

Often analysis is begin with initial or prior probability

estimates for specific events of interest.

Then, from sources such as a sample, a special report,or a product test, we obtain additional information about

the events.

Given this new information, prior probability values gets

calculated and revised, referred to as posteriorprobabilities.

Bayes’ theorem provides a means for making these

probability calculations.



Prior

Probabilities

New

Information

Application

Of Baye’s Theorem

Posterior

Probabilities



Formula

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Example

Three machines producing 40%, 35%, and

25% of the total output are known to produce

with defective proportion of items as: 0.04,

0.06, and 0.03 respectively. On a particularday, a unit of output is selected at random,

and is found to be defective. What is the

probability that it was produced by thesecond machine?



Solution

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probability gen

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