1 chapter 4 – probability an introduction. 2 chapter outline – part 1 experiments, counting...

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Chapter 4 – Probability

An Introduction

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Chapter Outline – Part 1

Experiments, Counting Rules, and Assigning Probabilities Events and Their Probability Some Basic Relationships of Probability Conditional Probability

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Understand the Uncertainties

A decision-making process often involves analyses of uncertainties, such as:

What are the likelihood that the Federal Reserve will change its current monetary policy?

What are the chances that the oil price will go up by 5% in the summer?

What are the What are the oddsodds that a company’s investment in that a company’s investment in R&D will be profitable?R&D will be profitable?

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Probability

Likelihood, chances, and odds are all referring to probability. In statistics,

Probability is a numerical measure of the likelihood that an is a numerical measure of the likelihood that an event will occur.event will occur. The value range of probability is always between 0 and

1; A probability near zero indicates that an event is very A probability near zero indicates that an event is very

much unlikely to occur.much unlikely to occur. A probability near one indicates that an event is almostA probability near one indicates that an event is almost

certain to occur.certain to occur.

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Probability as A Numerical Measure of

the Likelihood of Occurrence

00 11..55

The eventThe eventis veryis veryunlikelyunlikelyto occur.to occur.

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 occurrenceThe occurrenceof the event isof the event is

just as likely asjust as likely asit is unlikely.it is unlikely.

The eventThe eventis almostis almostcertaincertain

to occur.to occur.

The eventThe eventis almostis almostcertaincertain

to occur.to occur.

ProbabilitProbability:y:

Increasing Likelihood of OccurrenceIncreasing Likelihood of Occurrence

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Statistical Experiments

Statistical experiments investigate the odds of a particular outcome of a study.

Experiments are conducted repeatedly and in the same manner, but the outcomes may vary.

Therefore, we call the outcomes of statistical experiments random.

Each outcome has its own probability to occur.

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An Experiment and Its Sample Space

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

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

An experimental outcome is also called a sample point.

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An Experiment and Its Sample Space

ExperimentExperiment

Toss a coinToss a coinInspect a partInspect a partConduct a sales callConduct a sales callRoll a dieRoll a diePlay a football gamePlay a football game

Experiment OutcomesExperiment Outcomes

Head, tailHead, tailDefective, non-defectiveDefective, non-defectivePurchase, no purchasePurchase, no purchase1, 2, 3, 4, 5, 61, 2, 3, 4, 5, 6Win, lose, tieWin, lose, tie

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An Experiment and Its Sample Space Example: Travel Routes Joe is to travel from city A, stop in city B, and then arrive at his

destination city C. There are two ways to travel from city A to city B, and three ways from city B to city C. Question: how many different routes can Joe take to travel from city A to city C?

City A

City B

City C

1

2

1

2

3

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An Experiment and Its Sample Space Example: Travel Routes We can use a Tree Diagram to list all the possible routes

(outcomes) of Joe’s travel.

A to B

Step 1

B to C

Step 2

Routes

AB1

AB2

AB3

BC1

BC2

BC3

So, the total number of routes is 6.

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A Counting Rule for Multiple-Step Experiments

If an experiment consists of a sequence of If an experiment consists of a sequence of kk steps steps in which there are in which there are nn11 possible results for the first possible results for the first

step, step, nn22 possible results for the second step, and so possible results for the second step, and so

on, then the total number of experimental on, then the total number of experimental outcomes is given by outcomes is given by nn1 1 nn22 . . . . . . nnkk..

Apply this rule to our example, the total # of routes Apply this rule to our example, the total # of routes Joe can take is calculated as follows:Joe can take is calculated as follows:

nn1 1 nn2 2 = 2 = 2 3 = 6 3 = 6

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Q4.1 – Student ID

• A student ID code consists of 5 digits. The first digit (read from the left) is a letter, and the other digits can be any number from 0 to 9. How many possible student ID codes are there? What if it is required that no repetitions are allowed among the four numbers?

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Q4.1 – Student ID

• The first digit can be any letter from A to Z, and any of the remaining 4 digits can be any number from 0 to 9. So, applying the multiple-step rule, the total number of student ID codes is 261010 1010=260,000.

• If no repetitions are allowed for the 4 numbers, the total number of student ID codes is 2610987=131,040.

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Counting Rule for Combinations

This rule calculates the number of combinations of picking n from N objects at a time when the order of selection does not matter.

CN

nN

n N nnN

!

!( )!C

N

nN

n N nnN

!

!( )!

where: where: NN! = ! = NN((NN 1)( 1)(NN 2) . . . (2)(1) 2) . . . (2)(1) nn! = ! = nn((nn 1)( 1)(nn 2) . . . (2)(1) 2) . . . (2)(1) 0! = 10! = 1

! is called factorial.! is called factorial.

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Counting Rule for Combinations

Example: Pick 3 out of 8 students to form a team for a group project. Question: how many different teams can be formed?

5678!5123

!5678

!5!3

!8

!38!3

!8

3

883

C

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Counting Rule for Permutations

This rule calculates the number of permutations of picking n from N objects at a time when the order of selection does matter.

where: where: NN! = ! = NN((NN 1)( 1)(NN 2) . . . (2)(1) 2) . . . (2)(1) nn! = ! = nn((nn 1)( 1)(nn 2) . . . (2)(1) 2) . . . (2)(1) 0! = 10! = 1

! is called factorial.! is called factorial.

!!

!nN

N

n

NnP N

n

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Counting Rule for Permutations

Example: Pick 3 out of 8 students to form a committee which is comprised of three different positions – President, Secretary, and Treasurer. Question: how many different committees can be formed?

336678!5

!5678

!5

!8

!38

!8

3

883

P

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Assigning Probabilities

Basic requirements for assigning probabilities 1. The probability assigned to each 1. The probability assigned to each

experimental outcome must be between 0 experimental outcome must be between 0 and 1, inclusively.and 1, inclusively.

0 0 << PP((EEii) ) << 1 for all 1 for all ii

where: where: EEii is the is the iith experimental outcometh experimental outcomeand and PP((EEii) is its probability) is its probability

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Assigning Probabilities

Basic requirements for assigning probabilities 2. The sum of the probabilities for all 2. The sum of the probabilities for all

experimentalexperimental outcomes must equal 1.outcomes must equal 1.

where: where: nn is the number of experimental outcomes is the number of experimental outcomes

PP((EE11) + ) + PP((EE22) + . . . + ) + . . . + PP((EEnn) = 1) = 1

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Assigning Probabilities

Classical MethodClassical MethodClassical MethodClassical Method

Relative Frequency MethodRelative Frequency MethodRelative Frequency MethodRelative Frequency Method

Subjective MethodSubjective MethodSubjective MethodSubjective Method

Assigning probabilities based on the assumptionAssigning probabilities based on the assumption of of equally likely outcomesequally likely outcomes

Assigning probabilities based on Assigning probabilities based on experimentationexperimentation or historical dataor historical data

Assigning probabilities based on Assigning probabilities based on judgmentjudgment

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Classical Method

Experiment: Rolling a dieExperiment: Rolling a die

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

Probabilities: Each sample point has aProbabilities: Each sample point has a 1/6 chance of occurring1/6 chance of occurring

If an experiment has If an experiment has nn possible possible outcomes, theoutcomes, the

classical method would assign a probability classical method would assign a probability of 1/of 1/nn

to each outcome.to each outcome.

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Relative Frequency Method

Example: Monthly Sales Volume of 50 Starbucks Stores in NYC The probability that a Starbuck store’s sales volume falls into a

certain range is given by dividing the frequency by the sample points. As learned from Chapter 2, the assigned probability is simply the relative frequency.

Sales Volume ($1000) Frequency Probability61-70 2 0.0471-80 6 0.1281-90 11 0.22

91-100 17 0.34101-110 11 0.22111-120 3 0.06

50

6/506/506/506/50

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Subjective Method

In the business world, as economic conditions and a firm’s circumstances change, the patterns revealed from historical data might not apply onto forecasting the future outcomes.

Experience, intuition, and judgment should be utilized to evaluate the likelihood that a particular outcome will occur.

For instance, the performance of stock market is difficult, if not impossible, to predict based solely on historical data. The subjective evaluations from analysts are often combined with the classical or relative frequency methods to form better probability estimates.

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Events and Their Probabilities

An An eventevent is a collection of sample points.is a collection of sample points.

The The probability of any eventprobability of any event is equal to the sum of is equal to the sum of the probabilities of the sample points in the event.the probabilities of the sample points in the event.

If we can identify all the sample points of an If we can identify all the sample points of an experiment and assign a probability to each, we experiment and assign a probability to each, we can compute the probability of an event.can compute the probability of an event.

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Events and Their Probabilities

Example: Monthly Sales Volume of 50 Starbucks Stores

Event Event MM = Monthly sales no more than $100,000 = Monthly sales no more than $100,000

M = M = {(61,70),(71,80),(81,90),(91,100)}{(61,70),(71,80),(81,90),(91,100)}

P(M) = PP(M) = P(61,70)+(61,70)+PP(71,80)+(71,80)+PP(81,90)+(81,90)+PP(91,100)(91,100) = 0.04+0.12+0.22+0.34= 0.04+0.12+0.22+0.34 = 0.72= 0.72

Event Event MM = Monthly sales no more than $100,000 = Monthly sales no more than $100,000

M = M = {(61,70),(71,80),(81,90),(91,100)}{(61,70),(71,80),(81,90),(91,100)}

P(M) = PP(M) = P(61,70)+(61,70)+PP(71,80)+(71,80)+PP(81,90)+(81,90)+PP(91,100)(91,100) = 0.04+0.12+0.22+0.34= 0.04+0.12+0.22+0.34 = 0.72= 0.72

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Events and Their Probabilities

Example: Roll a die

Event Event KK = The number of dots on the side facing = The number of dots on the side facing

facing up is more than 3.facing up is more than 3.

K = K = {4,5,6}{4,5,6}

P(K) = PP(K) = P(4)+(4)+PP(5)+(5)+PP(6)(6) = 1/6+1/6+1/6= 1/6+1/6+1/6 = 3/6= 3/6 = 0.5= 0.5

Event Event KK = The number of dots on the side facing = The number of dots on the side facing

facing up is more than 3.facing up is more than 3.

K = K = {4,5,6}{4,5,6}

P(K) = PP(K) = P(4)+(4)+PP(5)+(5)+PP(6)(6) = 1/6+1/6+1/6= 1/6+1/6+1/6 = 3/6= 3/6 = 0.5= 0.5

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Q4.2 – Black and White Balls

• In a bag, there are one white ball and two black balls. Randomly pick two balls at a time out of the bag, what is the probability that the two balls are one white and one black?

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Q4.2 – Black and White Balls

• First, let’s list all the sample points of two out of three balls (W, B1, B2):– W & B1

– W & B2

– B1 & B2

• So, the probability of one white and one black is 2/3.

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Basic Relationships of Probability

There are some There are some basic probability relationshipsbasic probability relationships that can be used to compute the probability of an that can be used to compute the probability of an event without knowledge of all the sample point event without knowledge of all the sample point probabilities.probabilities.

Complement of An Event Union of Two Events Intersection of Two Events (joint events) Mutually Exclusive Events

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Venn Diagram A Venn diagram is to graphically illustrate the probability

relationships. The rectangular represents the whole sample space () and a circle inside

the rectangular represents an event.

Event Event AA

Sample Space

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Complement of An Event

The The complementcomplement of event of event A A is defined to be the event consisting of all is defined to be the event consisting of all sample points that are not in sample points that are not in A.A.

The complement of The complement of AA is denoted by is denoted by AAcc.. P(AP(Acc) = P() = P( ) – P(A) = 1 – P(A)

AAAAcc

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Complement of An Event

The seemingly very simple Complement rule proves to be very useful.The seemingly very simple Complement rule proves to be very useful. When it’s more difficult to figure out P(A) than P(AWhen it’s more difficult to figure out P(A) than P(ACC), we can use the ), we can use the

Complement rule to calculate P(A).Complement rule to calculate P(A).P(A) = 1 – P(AC)

AAAAcc

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Complement of An Event

Example: Example: Monthly Sales Volume of 50 Starbucks Stores When it’s more difficult to figure out P(A) than When it’s more difficult to figure out P(A) than

P(AP(ACC), we can use the Complement rule to calculate ), we can use the Complement rule to calculate P(A).P(A).

P(A) = 1 – P(AC)

Event Event MM = Monthly sales more than $70,000 = Monthly sales more than $70,000

MMcc = Monthly sales less than or equal to $70,000 = Monthly sales less than or equal to $70,000

= {(61,70)}= {(61,70)}

P(M) = 1 – PP(M) = 1 – P((MMcc)) = 1 – 0.04= 1 – 0.04 = 0.96= 0.96

Event Event MM = Monthly sales more than $70,000 = Monthly sales more than $70,000

MMcc = Monthly sales less than or equal to $70,000 = Monthly sales less than or equal to $70,000

= {(61,70)}= {(61,70)}

P(M) = 1 – PP(M) = 1 – P((MMcc)) = 1 – 0.04= 1 – 0.04 = 0.96= 0.96

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Union of Two Events

The The unionunion of events of events AA and and BB is the event containing all sample points that are in is the event containing all sample points that are in A A oror B B or or both.both.

The union of events The union of events AA and and BB is denoted by is denoted by AA B B ((‘‘’ is read as ‘or’)’ is read as ‘or’)

AAAAAA BB

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Intersection of Two Events

The The intersectionintersection of events of events AA and and BB is the set of all sample points that are in both is the set of all sample points that are in both A A and and BB. . The intersection of events The intersection of events AA and and BB is denoted by is denoted by AA ((‘‘’ is read as ‘and’)’ is read as ‘and’)

AAAAAA BB

Intersection of Intersection of AA and and BBIntersection of Intersection of AA and and BB

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Law of Addition

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

PP((AA BB) = ) = PP((AA) + ) + PP((BB) ) PP((AA BBPP((AA BB) = ) = PP((AA) + ) + PP((BB) ) PP((AA BB

NoteNote: Because both : Because both PP((AA) and ) and PP((BB) ) include the common part [ include the common part [ PP((AA BB, subtracting , subtracting PP((AA BB from the sum of from the sum of PP((AA) ) and and PP((BB) ) is to avoid double-counting. is to avoid double-counting.

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Law of Addition

Example: From a standard deck of 52 playing cards, what is the probability that a card drawn from the deck is a King or a Heart?

Event A = The card is a King; Event B = The card is a Heart Event (A he card is a King or a Heart Event (A he card is both a King and a Heart, i.e. the King of Heart

P(A P(A) + P(B) – P(A

–

Note: The King of Heart is included in both event A and event B.

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Mutually Exclusive Events

Two events are said to be Two events are said to be mutually exclusivemutually exclusive if the events if the events have no sample points in common, i.e. when one event have no sample points in common, i.e. when one event occurs, the other cannot occur.occurs, the other cannot occur.

AA BB

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Mutually Exclusive Events

If events If events AA and and BB are mutually exclusive are mutually exclusive, , PP((AA BB = = 0.0.

The addition law for mutually exclusive events is:The addition law for mutually exclusive events is:

PP((AA BB) = ) = PP((AA) + ) + PP((BB))PP((AA BB) = ) = PP((AA) + ) + PP((BB))

There is no need toinclude “ P(A

B”

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Conditional Probability

The probability of an event given that another eventThe probability of an event given that another event has occurred is called a has occurred is called a conditional probabilityconditional probability.. The probability of an event given that another eventThe probability of an event given that another event has occurred is called a has occurred is called a conditional probabilityconditional probability..

A conditional probability is computed as follows :A conditional probability is computed as follows : A conditional probability is computed as follows :A conditional probability is computed as follows :

The conditional probability of The conditional probability of AA given given BB is denoted is denoted by by PP((AA||BB).). The conditional probability of The conditional probability of AA given given BB is denoted is denoted by by PP((AA||BB).).

( )( | )

( )P A B

P A BP B

( )

( | )( )

P A BP A B

P B

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Conditional Probability

Example – Investment in Two Stock Funds

The following table presents the joint and marginal probabilities of investing in two stock funds – Technology and Utility.

Utility Gain (T) Loss (Tc) TotalGain (U) 0.24 0.36 0.60

Loss (Uc) 0.28 0.12 0.40Total 0.52 0.48 1.00

Technology

Joint ProbabilitiesJoint Probabilities(appear in the (appear in the

bodybodyof the table)of the table)

Marginal Marginal ProbabilitiesProbabilities

(appear in the (appear in the marginsmargins

of the table)of the table)

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Conditional Probability

Example – Investment in Two Stock Funds

Event U = Gain in the Utility Fund Event T = Gain in the Technology Fund P(U|T) = Gain in the Utility Fund given the gain in the Technology Fund. Since P(U and P(T) = 0.52,

P (UP (

0.24

0.52P (U|T) = = = P (U|T) = = =

0.460.46

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Law of Multiplication The The multiplication lawmultiplication law provides a way to provides a way to

compute the probability of the intersection compute the probability of the intersection of two events.of two events.

The law is written as:The law is written as:

PP((AA BB) = ) = PP((BB))PP((AA||BB))

The law is written as:The law is written as:

PP((AA BB) = ) = PP((BB))PP((AA||BB))

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Law of Multiplication

Event U = Gain in the Utility Fund Event T = Gain in the Technology Fund P(U|T) = Gain in the Utility Fund given the gain in the Technology Fund. Since P(U and P(T) = 0.52,

P (U|P (P (U T) = = 0.52 0.46 = 0.24

Example: Investment in Two Stock Funds

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Independent Events

If the probability that event A happens is not affected by whether or not event B happens, we say that events A and B are independent.

Two events Two events AA and and BB are independent if: are independent if: Two events Two events AA and and BB are independent if: are independent if:

PP((AA||BB) = ) = PP((AA))PP((AA||BB) = ) = PP((AA)) PP((BB||AA) = ) = PP((BB))PP((BB||AA) = ) = PP((BB))oror

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Multiplication Law forIndependent Events

The following equation can be used to test if two events are independent.

PP((AA BB) = ) = PP((AA))PP((BB))PP((AA BB) = ) = PP((AA))PP((BB))

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Multiplication Law forIndependent Events

Example: Investment in Two Stock Funds

Event U = Gain in the Utility Fund Event T = Gain in the Technology Fund

Are events U and T independent? Let’s check if P(U P(U) P(T) ? Because P(U P(U) =0.6, P(T) = 0.52,

P(U)P(T) = 0.60.52 = 0.312 0.24

Therefore: U and T are not independent.

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Mutually Exclusive Events vs. Independent Mutually Exclusive Events vs. Independent EventsEvents

Two events with Two events with nonzerononzero probabilities cannot be both probabilities cannot be both mutually exclusive and independent.mutually exclusive and independent.

For two mutually exclusive events with nonzero For two mutually exclusive events with nonzero probabilities, when one occurs, the other cannot happen. probabilities, when one occurs, the other cannot happen. Thus, the two events are Thus, the two events are dependentdependent on each other. on each other.

Two events that are not mutually exclusive are also Two events that are not mutually exclusive are also independent independent only ifonly if at least one of the two events has a at least one of the two events has a zero probability to occur.zero probability to occur.

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Q4.3

If A and B are independent events with P(A) = 0.4 and P(B) = 0.6, then P(AB) is:

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

P(A) + P(B) - P(A)P(for A & B are independent