basic probability. uncertainties managers often base their decisions on an analysis of uncertainties...

Basic Probability

Uncertainties

Managers often base their decisions on an analysis of uncertainties such as the following:

Application in Business

Probability in Manufacturing Manufacturing businesses can use probability to determine the

cost-benefit ratio or the transfer of a new manufacturing technology process by addressing the likelihood of improved profits.

In other instances, manufacturing firms use probability to determine the possibility of financial success of a new product when considering competition from other manufacturers, market demand, market value and manufacturing costs.

Other instances of probability in manufacturing include determining the likelihood of producing defective products, and regional need and capacity for certain fields of manufacturing.

Application in Business Scenario AnalysisProbability distributions can be used to create scenario analyses. For

example, a business might create three scenarios: worst-case, likely and best-case. The worst-case scenario would contain some value from the lower end of the probability distribution; the likely scenario would contain a value towards the middle of the distribution; and the best-case scenario would contain a value in the upper end of the scenario.

Risk EvaluationIn addition to predicting future sales levels, probability distribution can

be a useful tool for evaluating risk. Consider, for example, a company considering entering a new business line. If the company needs to generate $500,000 in revenue in order to break even and their probability distribution tells them that there is a 10 percent chance that revenues will be less than $500,000, the company knows roughly what level of risk it is facing if it decides to pursue that new business line.

Application in Business

Sales ForecastingOne practical use for probability distributions and scenario

analysis in business is to predict future levels of sales. It is essentially impossible to predict the precise value of a future sales level; however, businesses still need to be able to plan for future events. Using a scenario analysis based on a probability distribution can help a company frame its possible future values in terms of a likely sales level and a worst-case and best-case scenario. By doing so, the company can base its business plans on the likely scenario but still be aware of the alternative possibilities.

Important Terms

Probability – the chance that an uncertain event will occur (always between 0 and 1)

Event – Each possible outcome of a variable Simple Event – an event that can be described by a

single characteristic e.g. Tossing a coin: Getting head is one event

Getting tail is another eventHere, “Tossing a coin” is referred as ‘EXPERIMENT’

If we collect all the outcomes or events of an experiment that is called sample space. Sample Space – the collection of all possible eventse.g. Sample space can be written as : S = {Head tail}

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:

Events Simple event

An outcome from a sample space with one characteristic e.g., A red card from a deck of cards

Complement of an event A (denoted A’) All outcomes that are not part of event A e.g., All cards that are not diamonds

Joint event Involves two or more characteristics simultaneously e.g., An ace that is also red from a deck of cards

Favorable event Events in which we are interested e.g. Interest is have even number when rolling a die

Equally likely events Non preference of occurring of any event e.g. tossing a coin, event can be head or tail

Visualizing Events

Contingency Tables

Purchasing a big screen TV

No 100 650 750

Yes 200 50 250

Total 300 700 1000

Yes No Total

Sample Space

Actually purchased

Pla

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Bradley has invested in two stocks, Markley Oil and Collins Mining. Bradley has determined that thepossible outcomes of these investments three monthsfrom now are as follows.

Investment Gain or Loss in 3 Months (in $000)

Markley Oil Collins Mining

10 5 0-20

8-2

Example: Bradley Investments

An Experiment and Its Sample Space

Subjective MethodSubjective Method

An analyst made the following probability estimates.

Exper. Outcome Net Gain or Loss Probability

(10, 8)(10, -2)(5, 8)(5, -2)(0, 8)(0, -2)(-20, 8)(-20, -2)

$18,000 Gain $8,000 Gain $13,000 Gain $3,000 Gain $8,000 Gain $2,000 Loss $12,000 Loss $22,000 Loss

.20

.08

.16

.26

.10

.12

.02

.06


Probability as a Numerical Measureof the Likelihood of Occurrence

0 1.5

Increasing Likelihood of Occurrence

Probability:

The eventis veryunlikelyto occur.

The occurrenceof the event is just as likely asit is unlikely.

The eventis almostcertainto occur.

Visualizing Events

Venn Diagrams Let A = Planned to purchase Let B = Actually purchased

A

B

A ∩ B = Planned and actually purchased

A U B = Planned or actually purchased

A

B

Planned

Actually Purchased

Fundamental Concepts

1. The probability, P, of any event or state of nature occurring is greater than or equal to 0 and less than or equal to 1. That is:

0 P (event) 1

2. The sum of the simple probabilities for all possible outcomes of an activity must equal 1

Diversey Paint Example

Demand for white latex paint at Diversey Paint and Supply has always been either 0, 1, 2, 3, or 4 gallons per day

Over the past 200 days, the owner has observed the following frequencies of demand

QUANTITY DEMANDED NUMBER OF DAYS PROBABILITY

0 40 0.20 (= 40/200)

1 80 0.40 (= 80/200)

2 50 0.25 (= 50/200)

3 20 0.10 (= 20/200)

4 10 0.05 (= 10/200)

Total 200

Total 1.00 (= 200/200)

Diversey Paint Example

Demand for white latex paint at Diversey Paint and Supply has always been either 0, 1, 2, 3, or 4 gallons per day

Over the past 200 days, the owner has observed the following frequencies of demand

QUANTITY DEMANDED NUMBER OF DAYS PROBABILITY

0 40 0.20 (= 40/200)

1 80 0.40 (= 80/200)

2 50 0.25 (= 50/200)

3 20 0.10 (= 20/200)

4 10 0.05 (= 10/200)

Total 200

Total 1.00 (= 200/200)

Notice the individual probabilities are all between 0 and 1

0 ≤ P (event) ≤ 1And the total of all event probabilities equals 1

∑ P (event) = 1.00

Determining objective probabilityobjective probability Relative frequency

Typically based on historical data

Types of Probability

P (event) =Number of occurrences of the event

Total number of trials or outcomes

Classical or logical method Logically determine probabilities without

trials

P (head) = 12

Number of ways of getting a head

Number of possible outcomes (head or tail)

Types of Probability

Subjective probabilitySubjective probability is based on the experience and judgment of the person making the estimate

Opinion polls Judgment of experts Delphi method Other methods

Example

Taking Stats

Not Taking Stats

Total

Male 84 145 229

Female 76 134 210

Total 160 279 439

Find the probability of selecting a male taking statistics from the population described in the following table:

191.0439

84

people ofnumber total

stats takingmales ofnumber Stats Taking Male ofy Probabilit

Mutually Exclusive Events

Events are said to be mutually exclusivemutually exclusive if only one of the events can occur on any one trial

Mutually exclusive events Events that cannot occur together

example: Tossing a coin will result in either a head or a

tail

Events A and B are mutually exclusive

A = defective; B = non-defective

Collectively Exhaustive Events Events are said to be collectively exhaustivecollectively exhaustive if the

list of outcomes includes every possible outcome

One of the events must occur The set of events covers the entire sample

space

example:Sample : 400 units are there in one lotExperiment : drawing an unit from the lotSimple event: A = Defective; B = Non defective Collectively exhaustive events : 350 are non defective and 50 are defective units.

Joint ProbabilityP (Planned and purchased) =

P (Planned and did not purchase) =

P (not Planned and purchased) =

Similarly others…

Marginal Probability

P (Planned to purchase) =

Yes No TotalYes 200 50 250No 100 650 750

Total 300 700 1000

Actually purchased

Pla

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Computing Joint and Marginal Probabilities

1000

200

households of no. total

purchased and planned households of No.

1000

250

households of no. total

planned households of No.

1000

50

1000

100

Computing Joint and Marginal Probabilities

The probability of a joint event, A and B:

Computing a marginal (or simple) probability:

Where B1, B2, …, Bk are k mutually exclusive and collectively exhaustive events

outcomeselementaryofnumbertotal

BandAsatisfyingoutcomesofnumber)BandA(P

)BdanP(A)BandP(A)BandP(AP(A) k21

Bradley has invested in two stocks, Markley Oil and Collins Mining. Bradley has determined that thepossible outcomes of these investments three monthsfrom now are as follows.

Investment Gain or Loss in 3 Months (in $000)

Markley Oil Collins Mining

10 5 0-20

8-2


An Experiment and Its Sample Space

Subjective MethodSubjective Method

An analyst made the following probability estimates.

Exper. Outcome Net Gain or Loss Probability

(10, 8)(10, -2)(5, 8)(5, -2)(0, 8)(0, -2)(-20, 8)(-20, -2)

$18,000 Gain $8,000 Gain $13,000 Gain $3,000 Gain $8,000 Gain $2,000 Loss $12,000 Loss $22,000 Loss

.20

.08

.16

.26

.10

.12

.02

.06


The union of events A and B is denoted by A B

The union of events A and B is the event containing all sample points that are in A or B or both.

Union of Two Events

SampleSpace SSampleSpace SEvent A Event B

Union of Two Events

Event M = Markley Oil Profitable

Event C = Collins Mining Profitable

M C = Markley Oil Profitable or Collins Mining Profitable (or both)

M C = {(10, 8), (10, 2), (5, 8), (5, 2), (0, 8), (20, 8)}

P(M C) = P(10, 8) + P(10, 2) + P(5, 8) + P(5, 2) + P(0, 8) + P(20, 8)

= .20 + .08 + .16 + .26 + .10 + .02

= .82


The intersection of events A and B is denoted by A

The intersection of events A and B is the set of all sample points that are in both A and B.

SampleSpace SSampleSpace SEvent A Event B

Intersection of Two Events

Intersection of A and BIntersection of A and B

Intersection of Two Events



M C = Markley Oil Profitable and Collins Mining Profitable

M C = {(10, 8), (5, 8)}

P(M C) = P(10, 8) + P(5, 8)

= .20 + .16

= .36


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

Addition Law

The law is written as:

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

Event M = Markley Oil ProfitableEvent C = Collins Mining Profitable

M C = Markley Oil Profitable or Collins Mining Profitable

We know: P(M) = .70, P(C) = .48, P(M C) = .36

Thus: P(M C) = P(M) + P(C) P(M C)

= .70 + .48 .36= .82

Addition Law

(This result is the same as that obtained earlierusing the definition of the probability of an event.)



Two events are said to be mutually exclusive if the events have no sample points in common.

Two events are mutually exclusive if, when one event occurs, the other cannot occur.

SampleSpace SEvent A Event B


If events A and B are mutually exclusive, P(A B = 0.

The addition law for mutually exclusive events is:

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

There is no need toinclude “ P(A B”

Adding Not Mutually Exclusive Events

P (event A or event B) = P (event A) + P (event B)– P (event A and event B both occurring)

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

P(five or diamond) = P(five) + P(diamond) – P(five and diamond)

= 4/52 + 13/52 – 1/52

= 16/52 = 4/13

The equation must be modified to account for double counting

The probability is reduced by subtracting the chance of both events occurring together

Adding Mutually Exclusive Events

We often want to know whether one or a second event will occur

When two events are mutually exclusive, the law of addition is –

P (event A or event B) = P (event A) + P (event B)

P (spade or club) = P (spade) + P (club)

= 13/52 + 13/52

= 26/52 = 1/2 = 0.50 = 50%

Venn Diagrams

P (A) P (B)

Events that are mutually exclusive

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

Events that are not mutually exclusive

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

P (A) P (B)

P (A and B)

Example

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

Computing Conditional Probabilities

A conditional probability is the probability of one event, given that another event has occurred:

P(B)

B)andP(AB)|P(A

P(A)

B)andP(AA)|P(B

Where P(A and B) = joint probability of A and B

P(A) = marginal probability of A

P(B) = marginal probability of B

The conditional probability of A given that B has occurred

The conditional probability of B given that A has occurred



We know: P(M C) = .36, P(M) = .70

Thus:

Conditional Probability

( ) .36( | ) .5143

( ) .70P C M

P C MP M

( ) .36( | ) .5143

( ) .70P C M

P C MP M

( | )P C M( | )P C M


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

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.


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

(continued)


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

(continued)

Multiplication Law

The multiplication law provides a way to compute the probability of the intersection of two events.


P(A B) = P(B)P(A|B)


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

Multiplication Law

M C = Markley Oil Profitable and Collins Mining Profitable

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

(This result is the same as that obtained earlierusing the definition of the probability of an event.)


Independent Events

If the probability of event A is not changed by the existence of event B, we would say that events A and B are independent.

Two events A and B are independent if:

P(A|B) = P(A) P(B|A) = P(B)or

The multiplication law also can be used as a test to see if two events are independent.


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

Multiplication Law for Independent Events


We know: P(M C) = .36, P(M) = .70, P(C) = .48 But: P(M)P(C) = (.70)(.48) = .34, not .36

Are events M and C independent?DoesP(M C) = P(M)P(C) ?

Hence: M and C are not independent.


Multiplication Lawfor Independent Events

Independent Events

1. A black ball drawn on first drawP (B) = 0.30 (a marginal probability)

2. Two green balls drawnP (GG) = P (G) x P (G) = 0.7 x 0.7 = 0.49

(a joint probability for two independent events)

A bucket contains 3 black balls and 7 green balls We draw a ball from the bucket, replace it, and

draw a second ball

Independent Events

3. A black ball drawn on second draw if the first draw is green

P (B | G) = P (B) = 0.30 (a conditional probability but equal to the marginal

because the two draws are independent events)

4. A green ball is drawn on the second if the first draw was green

P (G | G) = P (G) = 0.70(a conditional probability as in event 3)

A bucket contains 3 black balls and 7 green balls We draw a ball from the bucket, replace it, and

draw a second ball

When Events Are Dependent

Assume that we have an urn containing 10 balls of the following descriptions 4 are white (W) and lettered (L) 2 are white (W) and numbered (N) 3 are yellow (Y) and lettered (L) 1 is yellow (Y) and numbered (N)

P (WL) = 4/10 = 0.4 P (YL) = 3/10 = 0.3

P (WN) = 2/10 = 0.2 P (YN) = 1/10 = 0.1

P (W) = 6/10 = 0.6 P (L) = 7/10 = 0.7

P (Y) = 4/10 = 0.4 P (N) = 3/10 = 0.3


4 ballsWhite (W)

and Lettered (L)

2 ballsWhite (W)

and Numbered (N)

3 ballsYellow (Y)

and Lettered (L)

1 ball Yellow (Y)and Numbered (N)

Probability (WL) =4

10

Probability (YN) =1

10

Probability (YL) =3

10

Probability (WN) =2

10

Urn contains 10 balls


The conditional probability that the ball drawn is lettered, given that it is yellow, is

P (L | Y) = = = 0.75P (YL)

P (Y)

0.3

0.4

Verify P (YL) using the joint probability formula

P (YL) = P (L | Y) x P (Y) = (0.75)(0.4) = 0.3

Joint Probabilities for Dependent Events

P (MT) = P (T | M) x P (M) = (0.70)(0.40) = 0.28

If the stock market reaches 12,500 point by January, there is a 70% probability that Tubeless Electronics will go up

There is a 40% chance the stock market will reach 12,500

Let M represent the event of the stock market reaching 12,500 and let T be the event that Tubeless goes up in value

Marginal Probability

Marginal probability for event A:

Where B1, B2, …, Bk are k mutually exclusive and collectively exhaustive events

)P(B)B|P(A)P(B)B|P(A)P(B)B|P(A P(A) kk2211

Visualizing Events

Contingency Tables

Purchasing a big screen TV

No 100 650 750

Yes 200 50 250

Total 300 700 1000

Yes No Total

Sample Space

Actually purchased

Pla

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ed t

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pu

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ase

Posterior Probabilities

Bayes’ Process

Revising Probabilities with Bayes’ Theorem

Bayes’ theorem is used to incorporate additional information and help create posterior probabilitiesposterior probabilities

Prior Probabilities

New Information

Example

Let's say we want to know how a change in interest rates would affect the Value of a stock market index. All major stock market indexes have a plethoraof historical data available so you should have no problem finding theoutcomes for these events with a little bit of research. For this example we willuse the data below to find out how a stock market index will react to a rise ininterest rates.

P(SI) = the probability of the stock index increasingP(SD) = the probability of the stock index decreasingP(ID) = the probability of interest rates decreasingP(II) = the probability of interest rates increasing

Interest Rates

Stock Price

Decline Increase Unit Frequency

Decline 200 950 1150

Increase 800 50 850

Unit Frequency 1000 1000 2000

P(SD|II) =

( ) ( | )

( )

P SD P II SD

P II


A cup contains two dice identical in appearance but one is fair (unbiased), the other is loaded (biased)

The probability of rolling a 3 on the fair die is 1/6 or 0.166 The probability of tossing the same number on the loaded

die is 0.60 We select one by chance,

toss it, and get a result of a 3 What is the probability that

the die rolled was fair? What is the probability that

the loaded die was rolled?


We know the probability of the die being fair or loaded is

P (fair) = 0.50 P (loaded) = 0.50And that

P (3 | fair) = 0.166 P (3 | loaded) = 0.60

We compute the probabilities of P (3 and fair) and P (3 and loaded)

P (3 and fair) = P (3 | fair) x P (fair)= (0.166)(0.50) = 0.083

P (3 and loaded) = P (3 | loaded) x P (loaded)= (0.60)(0.50) = 0.300


We know the probability of the die being fair or loaded is

P (fair) = 0.50 P (loaded) = 0.50And that

P (3 | fair) = 0.166 P (3 | loaded) = 0.60

We compute the probabilities of P (3 and fair) and P (3 and loaded)

P (3 and fair) = P (3 | fair) x P (fair)= (0.166)(0.50) = 0.083

P (3 and loaded) = P (3 | loaded) x P (loaded)= (0.60)(0.50) = 0.300

The sum of these probabilities gives us the unconditional probability of tossing a 3

P (3) = 0.083 + 0.300 = 0.383


P (loaded | 3) = = = 0.78P (loaded and 3)

P (3)0.300

0.383

The probability that the die was loaded is

P (fair | 3) = = = 0.22P (fair and 3)

P (3)0.083

0.383

If a 3 does occur, the probability that the die rolled was the fair one is

These are the revisedrevised or posteriorposterior probabilitiesprobabilities for the next roll of the die

We use these to revise our prior probabilityprior probability estimates

Bayes Calculations

Given event B has occurred

STATE OF NATURE

P (B | STATE OF NATURE)

PRIOR PROBABILITY

JOINT PROBABILITY

POSTERIOR PROBABILITY

A P(B | A) x P(A) = P(B and A) P(B and A)/P(B) = P(A|B)

A’ P(B | A’) x P(A’) = P(B and A’) P(B and A’)/P(B) = P(A’|B)

P(B)

Given a 3 was rolled

STATE OF NATURE

P (B | STATE OF NATURE)

PRIOR PROBABILITY

JOINT PROBABILITY

POSTERIOR PROBABILITY

Fair die 0.166 x 0.5 = 0.083 0.083 / 0.383 = 0.22

Loaded die 0.600 x 0.5 = 0.300 0.300 / 0.383 = 0.78

P(3) = 0.383

General Form of Bayes’ Theorem

)()|()()|()()|(

)|(APABPAPABP

APABPBAP

We can compute revised probabilities more directly by using

where

the complement of the event ; for example, if is the event “fair die”, then is “loaded die”

AA

A

A

Bayes’ theorem is the extension of conditional probability

General Form of Bayes’ Theorem

This is basically what we did in the previous example

If we replace with “fair die” Replace with “loaded die Replace with “3 rolled”We get

AAB

)|( rolled 3die fairP

)()|()()|()()|(

loadedloaded3fairfair3fairfair3

PPPPPP

22038300830

50060050016605001660

...

).)(.().)(.().)(.(

Example

A manufacturer claims that its drug test will detect steroid use (that is, show positive for an athlete who uses steroids) 95% of the time. Further, 15% of all steroid-free individuals also test positive. 10% of the rugby team members use steroids. Your friend on the rugby team has just tested positive. The probability that he uses steroids is?

Consider: E = the event that a rugby team member tests positiveF = the event that a rugby team member uses steroidsP(E|F) = 0.95; P(E|F') = 0.15; P(F) = 0.1; P(F') = 0.9

( ) ( | ) ( )( | )

( ) ( | ) ( ) ( | ') ( ')

P E F P E F P FP F E

P E P E F P F P E F P F

Example

A drilling company has estimated a 40% chance of striking oil for their new well.

A detailed test has been scheduled for more information. Historically, 60% of successful wells have had detailed tests, and 20% of unsuccessful wells have had detailed tests.

Given that this well has been scheduled for a detailed test, what is the probability

that the well will be successful?

Let S = successful well

U = unsuccessful well P(S) = 0.4 , P(U) = 0.6 (prior

probabilities)

Define the detailed test event as D

Conditional probabilities:

P(D|S) = 0.6 P(D|U) = 0.2

Goal is to find P(S|D)

Example(continued)

0.6670.120.24

0.24

(0.2)(0.6)(0.6)(0.4)

(0.6)(0.4)

U)P(U)|P(DS)P(S)|P(D

S)P(S)|P(DD)|P(S

Example(continued)

Apply Bayes’ Theorem:

So the revised probability of success, given that this well has been scheduled for a detailed test, is 0.667

Given the detailed test, the revised probability of a successful well has risen to 0.667 from the original estimate of 0.4

Example

EventPriorProb.

Conditional Prob.

JointProb.

RevisedProb.

S (successful) 0.4 0.6 (0.4)(0.6) = 0.24 0.24/0.36 = 0.667

U (unsuccessful)

0.6 0.2 (0.6)(0.2) = 0.12 0.12/0.36 = 0.333

Sum = 0.36

(continued)

Application Questions

An automobile dealer has kept records on the customers who visited his showroom. Forty percent of the people who visited his dealership were women. Furthermore, his records show that 37% of the women who visited his dealership purchased an automobile, while 21% of the men who visited his dealership purchased an automobile.

a. What is the probability that a customer entering the showroom will buy an automobile?

b. Suppose a customer visited the showroom and purchased a car. What is the probability that the customer was a woman?

c. Suppose a customer visited the showroom but did not purchase a car. What is the probability that the customer was a man?

basic probability. uncertainties managers often base their decisions on an analysis of uncertainties...

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