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CHAPTER 3CHAPTER 3

ELEMENTS OF PROBABILITYELEMENTS OF PROBABILITY

Probability as a Numerical Measureof the Likelihood of Occurrence

00 11..55

Increasing Likelihood of OccurrenceIncreasing Likelihood of Occurrence

ProbabilitProbability:y:

The eventThe eventis veryis veryunlikelyunlikelyto occur.to occur.

The occurrenceThe occurrenceof the event isof the event is just as likely asjust as likely asit is unlikely.it is unlikely.

The eventThe eventis almostis almostcertaincertainto occur.to occur.

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INTRODUCTION

• A probability is a numerical value expressing the degree of uncertainty regarding the occurrence of an event.

•Random experiment is a process or course of action, whose outcome is uncertain.

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Examples

– Experiment Outcomes

• Flip a coin Heads and Tails• Record a statistics test marks Numbers between 0 and 100• Measure the time to assemble Numbers from zero and above

a computer

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Assigning probabilities to Events

• Performing the same random experiment repeatedly, may result in different outcomes, therefore, the best we can do is consider the probability of occurrence of a certain outcome.

• To determine the probabilities we need to define and list the possible outcomes first

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• Determining the outcomes– Build an exhaustive list of all possible

outcomes.– Make sure the listed outcomes are mutually

exclusive.

• The set of all possible outcomes of an experiment is called a sample space and denoted by S.

Sample Space

An Experiment and Its Sample Space

An experiment is any process that generates well-defined outcomes. An experiment is any process that generates well-defined outcomes.

The sample space for an experiment is the set of all experimental outcomes. The sample space for an experiment is the set of all experimental outcomes.

An experimental outcome is also called a sample point. An experimental outcome is also called a sample point.

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EXAMPLES

• Tossing a coin experiment

S : {Head, Tail}

• Rolling a dice experiment

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

• Determination of the sex of a newborn child

S : {girl, boy}

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EXAMPLES

• Examine 3 fuses in sequence and note the results of each experiment, then an outcome for the entire experiment is any sequence of N’s (non-defectives) and D’s (defectives) of length 3. Hence, the sample space is

S : { NNN, NND, NDN, DNN, NDD, DND, DDN, DDD}

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Sample Space: S = {E1, E2,…,Ek}

Sample Spacea sample space of a random experimentis a list of all possible outcomes of the experiment. The outcomes must be mutually exclusive and exhaustive.

Simple EventsThe individual outcomes are called simple events.

EventAn event is any collectionof one or more simple events

Our objective is to determine P(A), the probability that event A will occur.

Our objective is to determine P(A), the probability that event A will occur.

E1 E2

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– Given a sample space S={E1,E2,…,Ek}, the following characteristics for the probability P(Ei) of the simple event Ei must hold:

– Probability of an event: The probability P(A) of event A is the sum of the probabilities assigned to the simple events contained in A.

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1. 0 1

2. 1

i

k

i

i

P E for each i

P E

1

1. 0 1

2. 1

i

k

i

i

P E for each i

P E

Assigning Probabilities

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Assigning Probabilities

• P(A) is the proportion of times the event A is observed.

total outcomes in A( )

total outcomes in SP A

•What is the probability of drawing an ace from a well-shuffled deck of 52 playing cards?

Assigning Probabilities There are three (3) approaches to solve for probabilities:

1- Classical Method

2- Relative Frequency Method

3- Subjective Method

Assigning probabilities based on the assumption of equally likely outcomes

Assigning probabilities based on experimentation or historical data

Assigning probabilities based on judgment

1- Classical Method

If an experiment has n possible outcomes, this method

would assign a probability of 1/n to each outcome.

Experiment: Rolling a die

Sample Space: S = {1, 2, 3, 4, 5, 6}

Probabilities: Each sample point has a 1/6 chance of occurring

Example

2- Relative Frequency Method

Number ofNumber ofPolishers RentedPolishers Rented

NumberNumberof Daysof Days

0011223344

44 6618181010 22

Lucas Tool Rental would like to assignprobabilities to the number of car polishersit rents each day. Office records show the followingfrequencies of daily rentals for the last 40 days.

Example: Lucas Tool Rental

Each probability assignment is given bydividing the frequency (number of days) by the

total frequency (total number of days).

Example: Relative Frequency Method

4/404/404/404/40

ProbabilityProbabilityNumber ofNumber ofPolishers RentedPolishers Rented

NumberNumberof Daysof Days

0011223344

44 6618181010 224040

.10.10 .15.15 .45.45 .25.25 .05.051.001.00

3- Subjective Method When economic conditions and a company’s circumstances change rapidly it might be inappropriate to assign probabilities based solely on historical data. 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. The best probability estimates often are obtained by combining the estimates from the classical or relative frequency approach with the subjective estimate. Example: There is 75% chance that England will adopt

the Euro currency by 2012.

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Intersection

• The intersection of event A and B is the event that occurs when both A and B occur.

• The intersection of events A and B is denoted by (A and B) OR AB.

• The joint probability of A and B is the probability of the intersection of A and B, which is denoted by P(A and B) OR P(AB).

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Union

• The union event of A and B is the event that occurs when either A or B or both occur.

• At least one of the events occur.

• It is denoted “A or B” OR AB

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Complement Rule

• The complement of event A (denoted by AC) is the event that occurs when event A does not occur.

• The probability of the complement event is calculated by

P(AC) = 1 - P(A)P(AC) = 1 - P(A)A and AC consist of all the simple events in the sample space. Therefore,P(A) + P(AC) = 1

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MUTUALLY EXCLUSIVE EVENTS

• Two events A and B are said to be mutually exclusive or disjoint, if A and B have no common outcomes. That is,

A and B = (empty set)

•A small city has 3 automobile dealerships: A GM dealer sellingChevrolets, Pontiacs and Buicks; a Ford dealer selling FordsAnd Mercurys; and a Chrysler dealer selling Plymouths and Chryslers. If an experiment consists of observing the brand of the next car sold, then events A={Chevrolet, Pontiac, Buick} and B={Ford, Mercury} are mutually exclusivemutually exclusive because the next car sold cannot be both a GM product and a Ford product.

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EXAMPLE

• The number of spots turning up when a six-sided die is tossed is observed. Consider the following events.

A: The number of observed is at most 2.

B: The number observed is an even number.

C: The number 4 turns up.

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VENN DIAGRAM

• A graphical representation of the sample space.

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S

1

3 5

4 6

A

BC

AB

1

4 6

A

B2

AB

1

4 6

A

B22

AC = A and C are mutually exclusive

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AXIOMS OF PROBABILTY

1) For any event A, 0 P(A) 1.

2) P(S) = 1.

3) For any sequence of mutually exclusive events E1, E2,…

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( ), 1,2,...,n n

i i

ii

P E P E n

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EXAMPLE

• A student has a box containing 25 computer disks of which 15 are blank and the other 10 are not. If she randomly selects disks one by one, what is the probability that at least 2 must be selected in order to find one that is blank?

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EQUALLY LIKELY OUTCOMES

• If there are N possible outcomes, then the

probability assigned to each is 1/N. The

same probability is assigned to each

simple event.

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For any two events A and B

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

Addition Rule

•Suppose that a metal fabrication process yields output with faulty boundings (FB) 10% of the timeand excessive oxidation (EO) in 25% of all segments,5% of the output has both faults. What is the probability that either of the events will occur?

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ODD EVENTS

• The odd event of A is defined by

( ) ( )( ) 1 ( )C

P A P AP A P A

•It tells us how much more likely to see the occurrence of event A.

•P(A)=3/4P(AC)=1/4 P(A)/P(AC) = 3. That is, the odds is 3. It is 3 times likely that A occurs as it is that it does not.

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COUNTING TECHNIQUES• Methods to determine how many subsets

can be obtained from a set of objects are called counting techniques.

BASIC PRINCIPLE OF COUNTINGSuppose that two experiments are to be

performed. If experiment 1 can result in any one of m possible and if, for each outcome of experiment 1, there are n possible outcomes of experiment 2, then together there are mn possible outcomes of the two experiments.

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EXAMPLE

• A travel agency offers weekend trips to 12 different cities by air, rail or bus. In how many different ways such a trip be arranged?

Tree Diagram:Cities

12...

12ARB

ARB

ARB

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GENERALIZED BASIC RULE FOR COUNTING

• If r experiments that are to be performed are such that the first one may results in any of n1 possible outcomes, and if for each of these n1 possible outcomes there are n2 possible outcomes of the second experiment, and if for each of the possible outcomes of the first two experiment there are n3 possible outcomes of the third experiment, and if …,then there are a total of nn11nn22…n…nrr possible outcomes of the r experiments.

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EXAMPLE

• If a test consists of 12 true-false questions, in how many different ways can a student mark the test paper with one answer to each question?

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THE FACTORIAL• The number of different sequences that

might be possible when observations are taken one at a time.

n! = n(n-1)(n-2)…2.10! = 1, 1! = 1

Example: Among 4 people how many different ways are there to choose a president, a vice president, a treasurer and a secretary?

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EXAMPLE

A class in statistics consists of 6 men and 4 women. An examination is given and the students are ranked according to their performances. Assume that no students obtain the same score.

a) How many different rankings are possible?

b) If the men are ranked among themselves and the women are ranked among themselves, how many different rankings are possible?

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EXAMPLE

• How many ways can the letter in the word COMPUTER be arranged in a row?

• In how many arrangements O is just after C?

• How many start with CO?

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PERMUTATIONS

• Any ordered sequence of r objects taken from a set of n distinct objects is called a permutation of size r of the objects.

,

!( 1)...( 1)

( )!r n

nP n n n r

n r

•Both composition and order are important.

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EXAMPLE

• Suppose that a random sample of 5 students is taken one at a time without replacement from the 38 members of the Tau Beta Pi. What is the number of sample outcomes?

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EXAMPLE

• Consider the case where 5 of 7 students are to be seated in a row. In how many ways these students can be seated?

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COMBINATION

• Given a set of n distinct objects, any unordered subset of size r of the objects is called a combination.

,

!!( )!r n

n nC

r n rr

Properties

1, 1, 0

n n n n

n r n r

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EXAMPLE

• In how many different ways can 3 of 20 laboratory assistants be chosen to assist with an experiment?

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EXAMPLE

A carton of 12 rechargeable batteries contains one that is defective. In how many ways can an inspector choose 3 of the batteries and

a) Get the one that is defective?

b) Not get the one that is defective?

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EXAMPLE

We have 24 light bulbs, 2 defective. If 2 of the bulbs are chosen at random what are the probabilities that

a) Neither bulb will be defective?

b) One of the bulb will be defective?

c) Both bulbs will be defective?

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CONTINGENCY (CROSS-TABULATION) TABLES

• Presents counts of two or more variables

A1 A2 Total

B1 a b a+b

B2 c d c+d

Total a+c b+d n = a+b+c+d

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Joint, Marginal, and Conditional Probability

• Joint probability is the probability that two events will occur simultaneously.

• Marginal probability is the probability of the occurrence of the single event.

A1 A2 Total

B1 a/n b/n (a+b)/n

B2 c/n d/n (c+d)/n

Total (a+c)/n (b+d)/n 1The marginal probability of A1.

The joint prob. of A2 and B1

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– A potential investor examined the relationship between the performance of mutual funds and the school. The fund manager earned his/her MBA.

– The following table describes the joint probabilities.

EXAMPLE

Mutual fund outperform the market

Mutual fund doesn’t outperform the

market

Top 20 MBA program .11 .29

Not top 20 MBA program .06 .54

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• The joint probability of [mutual fund outperform…] and […from a top 20 …] = .11– The joint probability of

[mutual fund outperform…] and […not from a top 20 …] = .06

EXAMPLE

Mutual fund outperforms the market

(B1)

Mutual fund doesn’t outperform the market (B2)

Top 20 MBA program (A1)

.11 .29

Not top 20 MBA program (A2)

.06 .54

P(A1 and B1)

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Marginal Probability

• These probabilities are computed by adding across rows and down columns

Mutual fund outperforms the market

(B1)

Mutual fund doesn’t

outperform the market (B2)

Marginal Prob.

P(Ai)

Top 20 MBA program (A1) .11 .29 .40

Not top 20 MBA program (A2)

.06 .54 .60

Marginal Probability P(Bj)

+ =

+ =

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Marginal Probability

• These probabilities are computed by adding across rows and down columns

Mutual fund outperforms the market

(B1)

Mutual funddoesn’t

outperform the market (B2)

Marginal Prob.

P(Ai)

Top 20 MBA program (A1) .11 .29 .40

Not top 20 MBA program (A2)

.06 .54 .60

Marginal Probability P(Bj) .17 .83

+ +

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• Find the conditional probability that a randomly selected fund is managed by a “Top 20 MBA Program graduate”, given that it did not outperform the market.

• Solution:P(A1|B2) = P(A1 and B2) = .29 = .3949

P(B2) .83

Conditional Probability

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CONDITIONAL PROBABILITY

• The probability that event A will occur given that or on the condition that, event B has already occurred. It is denoted by P(A|B).

( )( | ) , ( ) 0

( )P A B

P A B P BP B

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EXAMPLES1. What is the probability that the card drawn

from a deck of 52 cards is a Jack, given that it is a Face card?

2. The director of an insurance company’s computing center estimates that the company’s computer has a 20% chance of catching a computer virus. However, she feels that there is only a 6% chance of the computer’s catching a virus that will completely disable its operating system. If the company’s computer should catch a virus, what is the probability that the operating system will be completely disabled?

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EXAMPLES

3. Of a company’s employees, 30% are women and 6% are married women. Suppose an employee is selected at random. If the employee selected is a woman, what is the probability that she is married?

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MULTIPLICATION RULE

( ) ( | ). ( )

( ) ( | ). ( )

P A B P A B P B

or

P A B P B A P A

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EXAMPLE• A computer manufacturer inspected memory chips 100%

before they enter assembly operations. Let

D: Defective chip

D*: Non-defective chip

A: A chip approved for assembly by inspector

A*: A chip not approved for assembly by inspector

From past experience, it is known that P(D)=0.10. Also, it is known that the probability of an inspector passing a chip given that it is defective is 0.005, while the corresponding probability, given that the chip is non-defective is 0.999.

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EXAMPLE (contd.)

a) Find the joint probability that a chip is defective and is approved for assembly.

b) Find the probability that a chip is acceptable and is approved for assembly.

c) Find the probability that a chip is approved by assembly.

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EXAMPLE• The accompanying contingency table gives

frequencies for a classification of the equipment used in a manufacturing plant.

Equipment use

Working status Low Moderate High Total

In working order 10 18 12 40

Under repair 2 6 8 16

Total 12 24 20 56

a) Find the probability that a randomly selected piece of equipment is a high-use item given that it is in working order.

b) Find the probability that a randomly selected piece of equipment is under repair given that it is a moderate use item.

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Bayes’ Formula

• Conditional probability is used to find the probability of an event given that one of its possible causes has occurred.

• We use Bayes’ formula to find the probability of the possible cause given that an event has occurred.

1

( | ) ( )( | ) , 1,2,...,

( | ) ( )

j jj k

i i

i

P B A P AP A B j k

P B A P A

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– Medical tests can produce false-positive or false-negative results.

– A particular test is found to perform as follows:• Correctly diagnose “Positive” 94% of the time.• Correctly diagnose “Negative” 98% of the time.

– It is known that 4% of men in the general population suffer from the illness.

– What is the probability that a man is suffering from the illness, if the test result were positive?

EXAMPLE

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– Define the following events• D = Has a disease • DC = Does not have the disease• PT = Positive test results• NT = Negative test results

– Build a probability tree

SOLUTION

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– The probabilities provided are:• P(D) = .04 P(DC) = .96• P(PT|D) = .94 P(NT|D)= .06• P(PT|DC) = .02 P(NT|DC) = .98

– The probability to be determined is )PT|D(P

Solution – Continued

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D D

P(PT|DC ) = .02

P( NT|D C) = .98

P(PT|D) = .94

P( NT|D) = .06

PT|DPT|D

PT|DPTPT|DPTPTPTPTPT

DDDD|

PT

P(D C) = .96

P(D) = .04

PT|D PT|DPT|D

P(D and PT)=.0376

P(DC and PT)=.0192

)PT|D(P

P(PT) =.0568

+ )PT(P)PTandD(P

6620.0568.0376.

Bayes’ Formula

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P(PT|DC ) = .02

P( NT|D C) = .98

P(PT|D) = .94

P( NT|D) = .06

P(D C) = .96

P(D) = .04

)PT|D(P 6620.0568.0376.

Bayes’ Law

Prior probabilities

Likelihoodprobabilities

Posterior probabilities

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EXAMPLE

• In a state of 25% of all cars emit excessive amounts of pollutants. When tested 99% of all cars that emit excessive amounts of pollutant will fail, but 17% of the cars that don not emit excessive amounts will also fail. What is the probability that a car fails the test actually emits excessive amounts of pollutants?

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Independence

• Independent events– Two events A and B are said to be

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

– That is, the probability of one event is not affected by the occurrence of the other event.

•Or A and B are independent iff P(AB)=P(A)P(B)

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EXAMPLE

• An inspector at a video game assembly plant is checking out a special shipment of 10 printed circuit chips. Suppose that only 7 are good (G), the rest are bad (B). Consider the events G1, R1 and G2, B2 for the characteristics of the first and second chip inspected (Testing destroys each inspected chip).

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EXAMPLE (contd.)

a) P(G1)=?

b) Suppose the first chip is good. Find the probability that the second chip is also good.

c) Suppose that the first chip is bad. Find P(G2|B1).

d) Are events G1 and G2 are independent?

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EXAMPLE

• Consider a system with components in parallel.

p1=0.1

p2=0.3

p3=0.4

P ( at least one of the switches works) =?