Probability

Goals After completing the concept of

probability, you should be able to: • Explain three approaches to assessing

probabilities• Apply common rules of probability• Use Bayes’ Theorem for conditional

probabilities• Compute the expected value and

standard deviation for a discrete probability distribution

Important Terms• Experiment – a process of

obtaining outcomes for uncertain events

• Elementary Event – the most basic outcome possible from a simple experiment

• Sample Space – the collection of all possible elementary outcomes

Examples for Experiment• Recording 1000 citizens opinions about

the government economic policy • Tossing a coin three times and

observing the upward face on each toss• Choosing a chairperson and a secretary

from a committee of 10 people• Observing the fraction of union

members in favor of a new contract.

Examples for Event• From the example of experiment

choosing a chairperson of 10 persons each person can be denoted by an event.

• In tossing of a coin experiment head is an event and tail is one event

• In a departmental store purchasing by a customer is an event and not purchasing by the customer is also another event

Events

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

cards

• Event – May involve two or more outcomes simultaneously

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

Sample SpaceThe Sample Space is the collection of all possible outcomese.g. All 6 faces of a die:e.g. All 52 cards of a bridge deck:

Visualizing Events

• Contingency Tables

• Tree Diagrams

Red 2 24 26

Black 2 24 26

Total 4 48 52

Ace Not Ace Total

Full Deck of 52 Cards

Red Card

Black Card

Not an Ace

Ace

Ace

Not an Ace

Sample Space

Sample Space2

24

2

24

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

Chap 4-10

• A credit card customer at Big Bazar can use Visa (V), MasterCard (M), or American Express (A). The merchandise may be books (B), Electronic Media (E), or other (O).

• a) Define the Experiment• b) Enumerate the elementary events in the

sample space related to the experiment• c) would each elementary event be equally likely

• A survey asked tax accounting firms their business from ( S= sole proprietorship, P =Partnership, C= corporation) and type of risk insurance they carry ( L = liability only, T = property loss only, B = both liability and property).

• 1. Enumerate the Elementary events in the sample space

• 2. Would these elementary events in the sample space be equally likely ? Explain

Probability Concepts

• Mutually Exclusive Events

– If E1 occurs, then E2 cannot occur

– E1 and E2 have no common elements

Black Cards

Red Cards

A card cannot be Black and Red at the same time.

E1

E2

• Independent and Dependent Events

– Independent: Occurrence of one does not influence the probability of occurrence of the other

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

Probability Concepts

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

Assigning Probability

• Classical Probability Assessment

Relative Frequency of Occurrence

Subjective Probability Assessment

P(Ei) = Number of ways Ei can occur

Total 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

For Each of The Following, Indicate Whether The type Of Probability Involved is an Example of a Priori,

Classical, Empirical or subjective Probability

• The result of the next toss of a fair coin is Head

• Italy will win soccer’s world cup the next time the competition is held

• The sum of the faces of two dice will be 7• The possibility of happening a fire accident in

a factory is 1%• TDP will win the next assembly election in AP

Rules of Probability

Rules for Possible Values and Sum

Individual Values Sum of All Values

0 ≤ P(ei) ≤ 1For any event ei

1)P(ek

1ii

where:k = Number of elementary events in the sample space

ei = ith elementary event

Addition Rule for Elementary Events

• The probability of an event Ei is equal to the sum of the probabilities of the elementary events forming Ei.

•

That is, if:

Ei = {e1, e2, e3}

then:

P(Ei) = P(e1) + P(e2) + P(e3)

Complement Rule

• The complement of an event E is the collection of all possible elementary events not contained in event E. The complement of event E is represented by E.

• Complement Rule:

P(E)1)EP( E

E

1)EP(P(E) Or,

Addition Rule for Two Events

P(E1 or E2) = P(E1) + P(E2) - P(E1 and E2)

E1 E2


Don’t count common elements twice!

■ Addition Rule:

E1 E2+ -

Addition Rule for Mutually Exclusive Events

• If E1 and E2 are mutually exclusive, then

P(E1 and E2) = 0 So


= P(E1) + P(E2)

= 0

E1 E2

if mutually

exclusive

•A card is draw from a well shuffled pack of 52 cards. What is the probability of getting an ace or a red card

Addition Rule 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

1. The employees of a certain company have elected 5 of their number to represent them on the employee-management productivity council. Profiles of the 5 are as follows:

Gender Age1. Male 302. Male 323. Female 454. Female 205. Male 40

This group decides to elect a spokesperson by drawing a name from a chit. What is the Pbt the spokesperson will be either female or over 35?

2. An inspector of the Alaska pipeline has the task of comparing the reliability of two pumping stations. Each station is susceptible to two kinds of failure: pump failure and leakage. When either (or both) occur, the station must be shut down. The data at hand indicate that the following probabilities prevail:Station P (Pump failure) P (Leakage) P (Both)

1 0.07 0.10 02 0.09 0.12 0.06

Which station has the higher pbt of being shut down?

• The probability that a new marketing approach will be successful is 0.6. The probability that the expenditure for developing the approach can be kept with in the original budget is 0.5. The probability that both of these objectives will be achieved is 0.3. What is the probability that at least one of these objectives will be achieved?

• A problem is given to 3 Mangers A, B, and C. Their chances of solving it are 1/2, 1/3, 1/4 respectively. What is the probability that the problem will be solved?

Conditional Probability

• Conditional probability for any two events E1 , E2:

)P(E

)EandP(E)E|P(E

2

2121

0)P(Ewhere 2

• 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. 20% of the cars have both.


No CDCD Total

AC .2 .5 .7

No AC .2 .1 .3

Total .4 .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.

.2857.7

.2

P(AC)

AC)andP(CDAC)|P(CD

(continued)


No CDCD Total

AC .2 .5 .7

No AC .2 .1 .3

Total .4 .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%.

.2857.7

.2

P(AC)

AC)andP(CDAC)|P(CD

(continued)

For Independent Events:

• Conditional probability for independent events E1 , E2:

)P(E)E|P(E 121 0)P(Ewhere 2

)P(E)E|P(E 212 0)P(Ewhere 1

Multiplication Rules(Joint Probabilities)

• Multiplication rule for two events E1 and E2:

)E|P(E)P(E)EandP(E 12121

)P(E)E|P(E 212 Note: If E1 and E2 are independent, then

and the multiplication rule simplifies to

)P(E)P(E)EandP(E 2121

Tree Diagram Example

Diesel P(E2) = 0.2

Gasoline P(E1) = 0.8

Truck: P(E3|E1) = 0.2

Car: P(E4|E1) = 0.5

SUV: P(E5|E1) = 0.3

P(E1 and E3) = 0.8 x 0.2 = 0.16

P(E1 and E4) = 0.8 x 0.5 = 0.40

P(E1 and E5) = 0.8 x 0.3 = 0.24

P(E2 and E3) = 0.2 x 0.6 = 0.12

P(E2 and E4) = 0.2 x 0.1 = 0.02

P(E3 and E4) = 0.2 x 0.3 = 0.06

Truck: P(E3|E2) = 0.6

Car: P(E4|E2) = 0.1

SUV: P(E5|E2) = 0.3

A bag contains 32 marbles:4 are red, 9 are black, 12 are blue, 6 are yellow, and 1 is purple. Marbles are drawn one at a time with replacement. What is the probability that i. The second marble is yellow given the first was yellow?ii. The second marble is yellow given the first was black?iii. The third marble is purple given both the first and second were purple?

Ex:- Union shop steward Peter has drafted a set of wage and benefit demands to be presented to management. To get an idea of worker support for the package, he randomly polls the two largest groups of workers at his plant, The machinists (M) and the inspectors (I). He polls 30 of each group with the following results:Opinion of package M IStrongly support 9 10Mildly support 11 3Undecided 2 2Mildly oppose 4 8Strongly oppose 4 7

i. What is the probability that a machinist randomly selected from the polled group mildly supports the package?

ii. What is the probability that an inspector randomly selected from the polled group is undecided about the package?

iii. What is the probability that a worker (machinist or inspector) randomly selected from the polled group strongly or mildly supports the package?

iv. What types of probability are these?

4. The southeast regional manager of General Express, a private parcel – delivery firm, is worried about the likelihood of strikes by some of his employees. He has learned that the chance of a strike by his pilots is 0.75 and the chance of a strike by his drivers is 0.65. Further, he knows that if the drivers strike , there is a 90% chance that the pilots will strike in sympathy.

a) What is the pbt of both group’s striking?

b) If the pilots strike, what is the pbt that the drivers will strike in sympathy?

Posterior probabilities or Bayes’ theorem:-

• Often we begin probability analysis with initial or prior probabilities.

• Then, from a sample, special report, or a product test we obtain some additional information.

• Given this information, we calculate revised or posterior probabilities.

• Bayes’ theorem provides the means for revising the prior probabilities.

NewNewInformationInformation

NewNewInformationInformation

Applicationof Bayes’Theorem

Applicationof Bayes’Theorem

PosteriorPosteriorProbabilitiesProbabilities

PosteriorPosteriorProbabilitiesProbabilities

PriorPriorProbabilitiesProbabilities

PriorPriorProbabilitiesProbabilities

Bayes’ Theorem 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 and U = unsuccessful well

• P(S) = .4 , P(U) = .6 (prior probabilities)• Define the detailed test event as D• Conditional probabilities:

P(D|S) = .6 P(D|U) = .2

• Revised probabilities

Bayes’ Theorem Example

EventPriorProb.

Conditional Prob.

JointProb.

RevisedProb.

S (successful) .4 .6 .4*.6 = .24 .24/.36 = .67

U (unsuccessful)

.6 .2 .6*.2 = .12 .12/.36 = .33

Sum = .36

(continued)

Business Statistics: A Decision-Making Approach, 6e © 2005 Prentice-Hall, Inc.

Chap 4-42

• Given the detailed test, the revised probability of a successful well has risen to .67 from the original estimate of .4

Bayes’ Theorem Example

EventPriorProb.

Conditional Prob.

JointProb.

RevisedProb.

S (successful) .4 .6 .4*.6 = .24 .24/.36 = .67

U (unsuccessful)

.6 .2 .6*.2 = .12 .12/.36 = .33

Sum = .36

(continued)

1.In a bolt factory machines A, B and C manufactures respectively 25%,35% and 40% of the total output. Of their output 5%, 4%,2% are defective bolts. A bolt is drawn from the output and is found to be defective. What is the chance that it was produced by machine B?

LetE1 be the event of drawing a bolt at random manufactured by the machine AE2 be the event of drawing a bolt at random manufactured by the machine BE3 be the event of drawing a bolt at random manufactured by the machine CLet X be the event of its beingdefective

Prior pbts:P(E1)=25%P(E2)=35%P(E3)=40%

Likelihood pbts:P(X/E1)=5%P(X/E2)=4%P(X/E3)=2%

Additional information:A defective bolt was selected from the output.

To find the chance that it was produced by machine B we apply the Bayes’ theorem and is given by

)|(P)(P...)|(P)(P)|(P)(P

)|(P)(P)|(

2211 nn

iii EXEEXEEXE

EXEXEP

Events

(Ei

Prior Probabilities

P(Ei)Conditional Probabilities P(X|Ei)

Joint Probabilities P(Ei ∩ X)

Posterior Probabilities

P(Ei |X)

E1 0.25 0.05 0.0125 0.3623

E2 0.35 0.04 0.014 0.4058

E3 0.40 0.02 0.008 0.2319

Ex: A proposed shopping center will provide strong competition for downtown businesses like John Clothiers. If the shopping center is built, the owner of John Clothiers feels it would be best to relocate. The shopping center cannot be built unless a zoning change is approved by the town council. The planning board must first make a recommendation, for or against the zoning change, to the council.

Let: A1 = town council approves the zoning

changeA2 = town council disapproves the Zoning

change

• Prior Probabilities Using subjective judgment: P(A1) = 0.7, P(A2) = 0.3

• New Information The planning board has recommended against the zoning change. Let B denote the event of a negative recommendation by the planning board. Given that B has occurred, should John Clothiers revise the probabilities that the town council will approve or disapprove the zoning change?

• Conditional Probabilities Past history with the planning board and the town council indicates the following:

P(B|A1) = 0.2 P(B|A2) = 0.9

• To find the posterior probability that event Ai will occur given that event B has occurred we apply Bayes’ theorem.

• Bayes’ theorem is applicable when the events for which we want to compute posterior probabilities are mutually exclusive and their union is the entire sample space.

P A BA B A

A B A A B A A B Aii i

n n

( | )( ) ( | )

( ) ( | ) ( ) ( | ) ... ( ) ( | )

P P

P P P P P P1 1 2 2

P A BA B A

A B A A B A A B Aii i

n n

( | )( ) ( | )

( ) ( | ) ( ) ( | ) ... ( ) ( | )

P P

P P P P P P1 1 2 2

• Posterior ProbabilitiesGiven the planning board’s recommendation not to approve the zoning change, we revise the prior probabilities as follows.

= 0.34• Conclusion

The planning board’s recommendation is good news for John Clothiers. The posterior probability of the town council approving the zoning change is 0.34 versus a prior probability of 0.70

P A BA B A

A B A A B A( | )

( ) ( | )( ) ( | ) ( ) ( | )1

1 1

1 1 2 2

P PP P P P

P A BA B A

A B A A B A( | )

( ) ( | )( ) ( | ) ( ) ( | )1

1 1

1 1 2 2

P PP P P P

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

7 27 2 3 9

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

7 27 2 3 9

Tabular Approach:

Events

Ai

Prior Probabilities

P(Ai)

Conditional Probabilities P(B|

Ai)Joint Probabilities

P(Ai ∩ B)

Posterior Probabilities

P(Ai |B)

A1 0.7 0.2 0.14 0.3415

A2 0.3 0.9 0.27 0.6585

Probability

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