Casinos are big business. If you don’t think so, research the top ten casinos in the world and check out their buildings. They are huge. Some people all over the world like to gamble. They like the anticipation of winning, not to mention the financial benefit when they do. The question is, if casinos “give away” so much prize money, how can they stay in business? How can they afford to build and maintain those extravagant entertainment complexes? Does it come down to randomness and probability?

So now is your chance! Think about a casino game and look beyond the glimmer of the lights and free drinks! What is behind the gambling? Is it really a “numbers game?”

Conduct some research on a casino game or other game of chance. If you aren’t into that sort of thing, research a state or country lottery or games where you can’t control the results. If you need help selecting a game, reach out to your instructor. Remember to include a summary of your research

Summarize the articles that you read and then answer the following questions:

Do you think the organization makes a profit even if there is a “payoff” to some lucky customer?

Do you believe the game is a worthwhile investment?

Would you “put money down” if given a chance?

Chapter 5:

Randomness

and

Probability

1

5.1 Random Phenomena and Probability

With random phenomena, we canâ€™t predict the individual

outcomes, but we can hope to understand

characteristics of their long-run behavior.

For any random phenomenon, each attempt, or trial,

generates an outcome.

We use the more general term event to refer to

outcomes or combinations of outcomes.

5-2

5.1 Random Phenomena and Probability

The sample space is the collection of all possible outcomes.

We denote the sample space S or sometimes Î©.

The probability of an event is its long-run relative frequency.

Independence means that the outcome of one trial doesnâ€™t

influence or change the outcome of another.

5-3

5.1 Random Phenomena and Probability

The Law of Large Numbers (LLN) states that if the events

are independent, then as the number of trials increases, the

long-run relative frequency of an event gets closer and

closer to a single value.

Empirical probability is based on repeatedly observing the

eventâ€™s outcome.

5-4

5.1 Random Phenomena and Probability

Many people confuse the Law of Large numbers with the

so-called Law of Averages.

Many people believe that an outcome of a random event

that hasnâ€™t occurred in many trials is â€œdueâ€ to occur.

The Law of Averages doesnâ€™t exist.

5-5

5.2 The Nonexistent Law of Averages

Many people confuse the Law of Large numbers with the

so-called Law of Averages.

Many people believe that an outcome of a random event

that hasnâ€™t occurred in many trials is â€œdueâ€ to occur.

The Law of Averages doesnâ€™t exist.

5-6

5.3 Different Types of Probability

Model-Based (Theoretical) Probability

Example: Pew Research reports that of 10,190 randomly

generated working phone numbers, the initial results of

the calls were as follows:

Since the phone numbers were generated randomly, each

was equally likely, and the probability of a contact is

simply 7400 / 10,190 = 0.7262.

5-7

5.3 Different Types of Probability

Personal Probability

A subjective, or personal probability expresses your

uncertainty about the outcome.

Although personal probabilities may be based on

experience, they are not based either on long-run

relative frequencies or on equally likely events.

5-8

5.4 Probability Rules

Rule 1

If the probability of an event occurring is 0, the event canâ€™t

occur.

If the probability is 1, the event always occurs.

For any event A,

0 ï‚£ P( A) ï‚£.1

5-9

5.4 Probability Rules

Rule 2: The Probability Assignment Rule

The probability of the set of all possible outcomes must

be 1.

P(S) = 1

where S represents the set of all possible outcomes and

is called the sample space.

5-10

5.4 Probability Rules

Rule 3: The Complement Rule

The probability of an event occurring is 1 minus the

probability that it doesnâ€™t occur.

P( A) = 1 âˆ’ P( AC )

where the set of outcomes that are not in event A is

called the â€œcomplementâ€ of A, and is denoted AC.

5-11

5.4 Probability Rules

For Example: Leeâ€™s Lights sell lighting fixtures. Lee

records the behavior of 1000 customers entering the store

during one week. Of those, 300 make purchases. What is

the probability that a customer doesnâ€™t make a purchase?

If P(Purchase) = 0.30

then P(no purchase) = 1 â€“ 0.30 = 0.70

5-12

5.4 Probability Rules

Rule 4: The Multiplication Rule

For two independent events A and B, the probability that

both A and B occur is the product of the probabilities of the

two events.

P( A and B) = P( A) ï‚´ P(B)

provided that A and B are independent.

5-13

5.4 Probability Rules

For Example: If we can assume that customers behave

independently, what is the probability that the next two

customers entering Leeâ€™s Lights both make purchases?

We can use the multiplication rule because the events are

independent:

P( A and B) = P( A) ï‚´ P(B)

P(first customer makes a purchase and second customer makes a purchase)

= P(purchase) x P(purchase)

= 0.30 x 0.30

= 0.09

5-14

5.4 Probability Rules

For Example: If we can assume that customers behave

independently, what is the probability that the next two

customers entering Leeâ€™s Lights both make purchases?

We can use the multiplication rule because the events are

independent:

P( A and B) = P( A) ï‚´ P(B)

P(first customer makes a purchase and second customer makes a purchase)

= P(purchase) x P(purchase)

= 0.30 x 0.30

= 0.09

5-15

5.4 Probability Rules

For Example: Some customers prefer to see the

merchandise but then make their purchase online. Lee

determines that thereâ€™s an 8% chance of a customer

making a purchase in this way. We know that about 30%

of customers make purchases when they enter the store.

What is the probability that a customer who enters the

store makes no purchase at all?

P (purchase in the store or online)

= P (purchase in store) + P (purchase online)

= 0.30 + 0.09 = 0.39

P(no purchase) = 1 â€“ 0.39 = 0.61

5-16

5.4 Probability Rules

Rule 6: The General Addition Rule

The General Addition Rule calculates the probability that

either of two events occurs. It does not require that the

events be disjoint.

P( A or B) = P( A) + P(B) âˆ’ P( A and B)

5-17

5.4 Probability Rules

For Example: Lee notices that when two customers enter

the store together, their purchases are not disjoint. In fact,

thereâ€™s a 20% theyâ€™ll both make a purchase. When two

customers enter the store together, what is the probability

that at least one of them will make a purchase?

P(at least one purchases) =

P(A purchases or B purchases) =

P(A purchases) + P(B purchases)

â€“ P(A and B both purchase) =

0.30 + 0.30 â€“ 0.20 = 0.40

5-18

5.4 Probability Rules

Example: Car Inspections

You and a friend get your cars inspected. The event of your

carâ€™s passing inspection is independent of your friendâ€™s car. If

75% of cars pass inspection what is the probability that

Your car passes inspection?

Your car doesnâ€™t pass inspection?

Both cars pass inspection?

At least one of two cars passes?

Neither car passes inspection?

5-19

5.4 Probability Rules

Example (continued): Car Inspections

You and a friend get your cars inspected. The event of your

carâ€™s passing inspection is independent of your friendâ€™s car.

If 75% of cars pass inspection what is the probability that

Your car passes inspection?

P(Pass) = 0.75

Your car doesnâ€™t pass inspection?

P(PassC) = 1-0.75 = 0.25

Both cars pass inspection?

P(Pass)P(Pass) = (0.75)(0.75) = 0.5625

5-20

5.4 Probability Rules

Example (continued): Car Inspections

You and a friend get your cars inspected. The event of

your carâ€™s passing inspection is independent of your

friendâ€™s car. If 75% of cars pass inspection what is the

probability that

At least one of two cars passes?

1 â€“ (0.25)2 = 0.9375 OR

0.75 + 0.75 â€“ 0.5625 = 0.9375

Neither car passes inspection?

1 â€“ 0.9375 = 0.0625

5-21

5.5 Joint Probability and

Contingency Tables

Events may be placed in a contingency table such as the

one in the example below.

As part of a Pick Your Prize Promotion, a store invited

customers to choose which of three prizes theyâ€™d like to win.

The responses could be placed in the following contingency

table:

5-22

5.5 Joint Probability and

Contingency Tables

Marginal probability depends only on totals found in

the margins of the table.

5-23

5.5 Joint Probability and

Contingency Tables

In the table below, the probability that a respondent

chosen at random is a woman is a marginal probability.

P(woman) = 251/478 = 0.525.

5-24

5.5 Joint Probability and

Contingency Tables

Joint probabilities give the probability of two events

occurring together.

P(woman and camera) = 91/478 = 0.190.

5-25

5.5 Joint Probability and

Contingency Tables

Each row or column shows a conditional distribution given

one event.

In the table above, the probability that a selected customer

wants a bike given that we have selected a woman is:

P(bike | woman) = 30/251 = 0.120.

5-26

5.6 Conditional Probability and the

General Multiplication Rule

In general, when we want the probability of an event from a

conditional distribution, we write P(B|A) and pronounce it

â€œthe probability of B given A.â€

A probability that takes into account a given condition is

called a conditional probability.

P(A and B)

P(B | A) =

P(A)

5-27

5.6 Conditional Probability and the

General Multiplication Rule

Rule 7: The General Multiplication Rule

The General Multiplication Rule calculates the

probability that both of two events occurs. It does not

require that the events be independent.

P( A and B) = P( A) ï‚´ P(B | A)

5-28

5.6 Conditional Probability and the

General Multiplication Rule

What is the probability that a randomly selected customer

wants a bike if the customer selected is a woman?

P ( A and B)

P (B | A ) =

= P(bike|woman)

P( A)

P (bike and woman) 30 478

=

=

= 0.12

P(woman)

251 478

5-29

5.6 Conditional Probability and the

General Multiplication Rule

Events A and B are independent whenever P(B|A) = P(B).

This means knowing event A has occurred has no impact on

the probability of event B occurring.

5-30

5.6 Conditional Probability and the

General Multiplication Rule

Are Prize preference and Sex independent?

If so, P(bike | woman) will be the same as P(bike).

P(bike | woman)= 30/251 = 0.12

P(bike) = 90/478 = 0.265

0.12 â‰ 0.265

Since the two probabilities are not equal, Prize preference and

Sex and not independent.

5-31

5.6 Conditional Probability and the

General Multiplication Rule

Independent vs. Disjoint

For all practical purposes, disjoint events cannot be

independent.

Donâ€™t make the mistake of treating disjoint events as if they

were independent and applying the Multiplication Rule for

independent events.

5-32

5.6 Conditional Probability and the

General Multiplication Rule

5-33

5.7 Constructing Contingency Tables

If youâ€™re given probabilities without a contingency table,

you can often construct a simple table to correspond to the

probabilities and use this table to find other probabilities.

5-34

5.7 Constructing Contingency Tables

A survey classified homes into two price categories (Low

and High). It also noted whether the houses had at least 2

bathrooms or not (True or False). 56% of the houses had

at least 2 bathrooms, 62% of the houses were Low

priced, and 22% of the houses were both.

Translating the percentages to probabilities, we have:

5-35

5.7 Constructing Contingency Tables

The 0.56 and 0.62 are marginal probabilities, so they go in

the margins.

The 22% of houses that were both Low priced and had at

least 2 bathrooms is a joint probability, so it belongs in the

interior of the table.

5-36

5.7 Constructing Contingency Tables

Because the cells of the table show disjoint events, the

probabilities always add to the marginal totals going

across rows or down columns.

5-37

5.8 Probability Trees

Some business decisions involve more subtle evaluation

of probabilities.

Given the probabilities of various states of nature, we can

use a picture called a probability tree or tree diagram to

help think through the decision-making process.

5-38

5.8 Probability Trees

Some business decisions involve more subtle evaluation

of probabilities.

Given the probabilities of various states of nature, we can

use a picture called a probability tree or tree diagram to

help think through the decision-making process.

5-39

5.8 Probability Trees

Example:

Personal electronic devices, such as smart phones and

tablets, are getting more capable all the time. Microscopic

and even submicroscopic flaws that can cause intermittent

performance failures can develop during their manufacture.

Defects will always occur, so the quality engineer in charge

of the production process must monitor the number of

defects and take action if the process seems out of control.

5-40

5.8 Probability Trees

Example:

The probability tree on the next slide shows that a

manufacturing problem can occur with the motherboard, the

memory, or the case alignment for an electronic device.

The next set of branches denote the probability that a minor

adjustment will fix each type of problem.

5-41

5.8 Probability Trees

5-42

5.8 Probability Trees

Example:

At the end of each branch, joint probabilities can be

calculated by applying the multiplication rule for

independent events.

5-43

5.8 Probability Trees

5-44

5.9 Reversing the Conditioning: Bayesâ€™

Rule

When you need to find reverse conditional probabilities,

we recommend drawing a tree and finding the

appropriate probabilities by using the definition of

conditional probability (see text).

5-45

5.9 Reversing the Conditioning: Bayesâ€™

Rule

In general, we can write for n events Ai that are mutually

exclusive (each pair is disjoint) and exhaustive (their union is

the whole sample space):

5-46

WHAT CAN GO WRONG?

â€¢

Beware of probabilities that donâ€™t add up to 1.

â€¢

Donâ€™t add probabilities of events if theyâ€™re not

disjoint.

â€¢ Donâ€™t multiply probabilities of events if theyâ€™re not

independent.

â€¢

Donâ€™t confuse disjoint and independent.

5-47

FROM LEARNING TO EARNING

Apply the facts about probability to determine whether

an assignment of probabilities is legitimate.

â€¢ Probability is long-run relative frequency.

â€¢ Individual probabilities must be between 0 and 1.

â€¢ The sum of probabilities assigned to all outcomes

must be 1.

Understand the Law of Large Numbers and that the

common understanding of the â€œLaw of Averagesâ€ is

false.

5-48

FROM LEARNING TO EARNING

Know the rules of probability and how to apply them.

â€¢

â€¢

â€¢

â€¢

â€¢

The Complement Rule says that P(A) = 1 â€“ P(Ac).

The Multiplication Rule for independent events say

that

P(A and B) = P(A)*P(B), provided A and B are

independent.

The General Multiplication Rule says that

P(A and B) = P(A)*P(B|A).

The Addition Rule for disjoint events says that

P(A or B) = P(A) + P(B), provided events A and B

are disjoint.

The General Addition Rule says that

P(A or B) = P(A) + P(B) â€“ P(A and B).

5-49

FROM LEARNING TO EARNING

Know how to construct and read a contingency table.

Know how to define and use independence.

â€¢ Events A and B are independent if P(B|

A) = P(B).

Know how to construct tree diagrams and use them to

calculate probabilities.

Know how to use Bayesâ€™ Rule to compute conditional

probabilities.

5-50

Chapter 6:

Random

Variables

and

Probability

Models

1

6.1 Expected Value of a Random

Variable

A variable whose value is based on the outcome of a

random event is called a random variable.

If we can list all possible outcomes, the random variable is

called a discrete random variable.

If a random variable can take on any value between two

values, it is called a continuous random variable.

6-2

6.1 Expected Value of a Random

Variable

For both discrete and continuous random variables, the set

of all the possible values and their associated probabilities is

called the probability model.

When the probability model is known, then the expected

value can be calculated:

E ( X ) = ïƒ¥ x ïƒ— P ( x ) (discrete random variable)

E ( X ) can be written as ï but never as x or y

6-3

6.1 Expected Value of a Random

Variable

The probability model for a particular life insurance policy

is shown. Find the expected annual payout on a policy.

6-4

6.1 Expected Value of a Random

Variable

The probability model for a particular life insurance policy

is shown. Find the expected annual payout on a policy.

ïƒ¦ 1 ïƒ¶

ïƒ¦ 2 ïƒ¶

ïƒ¦ 997 ïƒ¶

ï = E ( X ) = $100, 000 ïƒ§

ïƒ· + $50, 000 ïƒ§

ïƒ· + $0 ïƒ§

ïƒ·

1000

1000

1000

ïƒ¨

ïƒ¸

ïƒ¨

ïƒ¸

ïƒ¨

ïƒ¸

= $200

We expect that the insurance company will pay out $200

per policy per year.

6-5

6.1 Expected Value of a Random

Variable

Standard Deviation of a discrete random variable:

ï³ = Var ( X ) = ïƒ¥ ( x âˆ’ ï ) ïƒ— P ( x )

2

2

ï³ = SD ( X ) = Var ( X )

6-6

6.2 Standard Deviation of a Random

Variable

The probability model for a particular life insurance policy is

shown. Find the standard deviation of the annual payout.

6-7

6.2 Standard Deviation of a Random

Variable

Example: Book Store Purchases

Suppose the probabilities of a customer purchasing 0, 1, or

2 books at a book store are 0.2, 0.4 and 0.4 respectively.

What is the expected number of books a customer will

purchase?

What is the standard deviation of the book purchases?

6-8

6.2 Standard Deviation of a Random

Variable

Example: Book Store Purchases

ï = 0 ( 0.2 ) + 1( 0.4 ) + 2 ( 0.4 )

= 1.2

ï³ 2 = ïƒ¥( x âˆ’ ï ) ïƒ— P ( x)

2

= (0 – 1.2) 2 (0.2) + (1 – 1.2) 2 (0.4) + (2 – 1.2) 2 (0.4)

= 0.288 + 0.016 + 0.256 = 0.56

ï³ = 0.56 = 0.748

6-9

6.3 Properties of Expected Values and

Variances

Adding a constant c to X:

Multiplying X by a constant a:

6-10

6.3 Properties of Expected Values and

Variances

Addition Rule for Expected Values of Random Variables

Addition Rule for Variances of (independent) Random

Variables

6-11

6.3 Properties of Expected Values and

Variances

The expected annual payout per insurance policy is

$200 and the variance is $14,960,000.

If the payout amounts are doubled, what are the new

expected value and variance?

E ( 2 X ) = 2 E ( X ) = 2 ïƒ— 200 = $400

Var ( 2 X ) = 22 Var ( X ) = 4 ïƒ—14,960,000 = 59,840,000

6-12

6.3 Properties of Expected Values and

Variances

Compare this to the expected value and variance on two

independent policies at the original payout amount.

E ( X + Y ) = E ( X ) + E (Y ) = 2 ïƒ— 200 = $400

Var ( X + Y ) = Var ( X ) + Var (Y ) = 2 ïƒ—14,960,000 = 29,920,00

Note: The expected values are the same but the variances are different.

6-13

6.3 Properties of Expected Values and

Variances

The association of two random variables can be measured

using the covariance of X and Y:

Cov ( X , Y ) = E ( ( X âˆ’ ï )(Y âˆ’ï® ) )

Then, the covariance gives us the extra information needed

to find the variance of the sum or difference of two random

variables when they are not independent:

Var ( X ï‚± Y ) = Var ( X ) + Var (Y ) ï‚± 2Cov ( X , Y )

6-14

6.3 Properties of Expected Values and

Variances

Covariance doesnâ€™t have to be between â€“1 and +1, which

makes it harder to interpret. To fix this â€œproblemâ€, we can

divide the covariance by each of the standard deviations to

get the correlation:

Corr ( X , Y ) =

Cov ( X , Y )

ï³ Xï³Y

6-15

6.4 Bernoulli Trials

Bernoulli trials have the following characteristics:

â€¢ There are only two outcomes per trial, called success

and failure.

â€¢ The probability of success, called p, is the same on

every trial (the probability of failure, 1 â€“ p, is often called

q).

â€¢ The trials are independent.

6-16

6.4 Bernoulli Trials

Examples of Bernoulli trials include tossing a coin, collecting

yes / no responses from a survey, or shooting free throws in

basketball.

When using Bernoulli trials to develop probability models, we

require that trial are independent.

We can check the 10% Condition: As long as the number of

trials or sample size is less than 10% of the population size,

we can proceed with confidence that the trials are

independent.

6-17

6.5 Discrete Probability Models

The Uniform Model

If X is a random variable with possible outcomes 1, 2, â€¦, n

and P ( X = i ) = 1 / n for each i, then we say X has a discrete

Uniform distribution U[1, â€¦, n].

When tossing a fair die, each number is equally likely to

occur. So tossing a fair die is described by the Uniform

model

U[1, 2, 3, 4, 5, 6], with P ( X = i ) = 1/ 6.

6-18

6.5 Discrete Probability Models

The Geometric Model

Predicting the number of Bernoulli trials required to achieve

the first success.

6-19

6.5 Discrete Probability Models

The Binomial Model

Predicting the number of successes in a fixed number of

Bernoulli trials.

6-20

6.5 Discrete Probability Models

The Binomial Model

Predicting the number of successes in a fixed number of

Bernoulli trials.

6-21

6.5 Discrete Probability Models

Example: Probability in customer acquisition

A venture capital firm has a list of potential investors who

have previously invested in new technologies. On average,

these investors invest about 5% of the time. A new client

of the firm is interested in finding investors for a mobile

phone application that enables financial transactions, an

application that is finding increasing acceptance in much of

the developing world. An analyst at the firm is about to start

calling potential investors.

6-22

6.5 Discrete Probability Models

Example: Closing Sales

A salesman normally closes a sale on 80% of his

presentations. Assuming the presentations are

independent:

What model should be used to determine the probability

that he closes his first presentation on the fourth attempt?

What is the probability he closes his first presentation on

the fourth attempt?

6-23

6.5 Discrete Probability Models

Example (continued): Closing Sales

What model should be used to determine the probability that

he closes his first presentation on the fourth attempt?

Geometric

What is the probability he closes his first presentation on the

fourth attempt?

P ( X = 4) = (0.20) (0.80) = 0.0064

3

6-24

6.5 Discrete Probability Models

Example: Professional Tennis

A tennis player makes a successful first serve 67% of the

time. Of the first 6 serves of the next match:

What model should be used to determine the probability that

all 6 first serves will be in bounds?

What is the probability that all 6 first serves will be in bounds?

How many first serves can be expected to be in bounds?

6-25

6.5 Discrete Probability Models

Example (continued): Professional Tennis

What model should be used to determine the probability

that all 6 first serves will be in bounds? Binomial

What is the probability that all 6 first serves will be

inbounds?

ïƒ¦ 6ïƒ¶

P( X = 6) = ïƒ§ ïƒ· (0.67)6 (0.33)0 = 0.0905

ïƒ¨ 6ïƒ¸

How many first serves can be expected to be inbounds?

E(X) = np = 6(0.67) = 4.02

6-26

6.5 Discrete Probability Models

Example (continued): Professional Tennis

What model should be used to determine the probability that

all 6 first serves will be in bounds? Binomial

What is the probability that all 6 first serves will be inbounds?

ïƒ¦ 6ïƒ¶

P( X = 6) = ïƒ§ ïƒ· (0.67)6 (0.33)0 = 0.0905

ïƒ¨ 6ïƒ¸

How many first serves can be expected to be inbounds?

E(X) = np = 6(0.67) = 4.02

6-27

6.5 Discrete Probability Models

Example (continued): Satisfaction Survey

What probability model would be used to model the selection

of a single number? Uniform, all numbers are equally likely.

What is the probability the number selected will be

an even number? 0.5

What is the probability the number selected will

end in 000?

0.001

6-28

6.5 Discrete Probability Models

Example: Probability in customer acquisition

A venture capital firm has a list of potential investors who

have previously invested in new technologies. On average,

these investors invest about 5% of the time. A new client

of the firm is interested in finding investors for a mobile phone

application that enables financial transactions, an application

that is finding increasing acceptance in much of the

developing world. An analyst at the firm is about to start

calling potential investors.

6-29

6.5 Discrete Probability Models

Example: (Continued) Probability in customer acquisition

What is the probability that the first person she calls will want

to invest?

How many investors will she have to call, on average, to find

someone interested?

6-30

6.5 Discrete Probability Models

Example: (Continued) Probability in customer acquisition

What is the probability that the first person she calls will want

to invest?

Each investor has a 5% or 1/20 chance of wanting to invest,

so the chance that the first person she calls is interested is

1/20.

How many investors will she have to call, on average, to find

someone interested?

This uses a Geometric model. Let X = number of people she

calls until the first interested person.

E( X ) = 1 p – 1/ (1/ 20) = 20 people.

6-31

6.5 Discrete Probability Models

Example: (Continued) Probability in customer acquisition

If she calls 10 investors, what is the probability that exactly 2

of them will be interested?

Using the Binomial model, let Y = number of people

interested in 10 calls, then

10 2

10 Â´ 9

8

P(Y = 2) = ( ) p (1- p) =

(1/ 20)2 (19 / 20)8 = 0.0746

2

2

What assumptions are you making to answer these

questions?

We are assuming that the trials are independent and that the

probability of being interested in investing is the same for all

potential investors.

6-32

WHAT CAN GO WRONG?

â€¢ Probability distributions are still just models.

â€¢ If the model is wrong, so is everything else.

â€¢ Watch out for variables that arenâ€™t independent.

â€¢ Donâ€™t write independent instances of a random

variable with notation that looks like they are the

same variables.

6-33

WHAT CAN GO WRONG?

â€¢ Donâ€™t forget: Variances of independent random

variables add. Standard deviations donâ€™t.

â€¢ Donâ€™t forget: Variances of independent random

variables add, even when youâ€™re looking at the

difference between them.

â€¢ Be sure you have Bernoulli trials.

6-34

FROM LEARNING TO EARNING

Apply the facts about probability to determine whether

an assignment of probabilities is legitimate.

â€¢ Probability is long-run relative frequency.

â€¢ Individual probabilities must be between 0 and 1.

â€¢ The sum of probabilities assigned to all outcomes

must be 1.

Understand the Law of Large Numbers and that the

common understanding of the â€œLaw of Averagesâ€ is

false.

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FROM LEARNING TO EARNING

Understand how probability models relate values to

probabilities.

â€¢ For discrete random variables, probability models

assign a probability to each possible outcome.

Know how to find the mean, or expected value, of a

discrete probability model and the standard

deviation:

E ( X ) = ïƒ¥ x ïƒ— P ( x)

ï³ =

ïƒ¥ (x âˆ’ ï ) ïƒ— P (x )

2

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FROM LEARNING TO EARNING

Foresee the consequences of shifting and scaling

random variables:

E(X Â± c) = E(X) Â± c

E(aX) = aE(X)

Var(X Â± c) = Var(X)

Var(aX) = a2Var(X)

SD(X Â± c) = SD(X)

SD(aX) = |a|SD(X)

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FROM LEARNING TO EARNING

Understand that when adding or subtracting random

variables the expected values add or subtract well:

E(X Â± Y) = E(X) Â± E(Y).

However, when adding or subtracting independent

random variables, the variances add:

Var(X Â±Y) = Var(X) + Var(Y)

Be able to explain the properties and parameters of the

Uniform, the Binomial, the Geometric, and the Poisson

distributions.

6-38

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