chapter 14 14 and 15 probability.pdf · probability assignment rule: ... slide 14 - 15 for each of...

From Randomness

to Probability

Copyright © 2010, 2007, 2004 Pearson Education, Inc.

Chapter 14

Probability Rules!

Copyright © 2010, 2007, 2004 Pearson Education, Inc.

Chapter 15

Dealing with Random Phenomena

Slide 14 - 3

A random phenomenon is a situation in which we know what outcomes could happen, but we don’t know which particular outcome did or will happen.

In general, each occasion upon which we observe a random phenomenon is called a trial.

At each trial, we note the value of the random phenomenon, and call it an outcome.

When we combine outcomes, the resulting combination is an event.

The collection of all possible outcomes is called the sample space.

The Law of Large Numbers

Slide 14 - 4

First a definition . . .

When thinking about what happens with combinations of

outcomes, things are simplified if the individual trials are

independent.

Roughly speaking, this means that the outcome of one trial

doesn’t influence or change the outcome of another.

For example, coin flips are independent.

The Law of Large Numbers (cont.)

Slide 14 - 5

The Law of Large Numbers (LLN) says that the long-run relative frequency of repeated independent events gets closer and closer to a single value.

We call the single value the probability of the event.

Because this definition is based on repeatedly observing the event’s outcome, this definition of probability is often called empirical probability.

The Nonexistent Law of Averages

Slide 14 - 6

The LLN says nothing about short-run behavior.

Relative frequencies even out only in the long run, and this

long run is really long (infinitely long, in fact).

The so called Law of Averages (that an outcome of a random

event that hasn’t occurred in many trials is “due” to occur)

doesn’t exist at all.

Ex.

Baseball player is “due” for a hit

heads it “due” after 10 tails

Roulette: red is “due” after 5 blacks

Modeling Probability

Slide 14 - 7

When probability was first studied, a group of French mathematicians looked at games of chance in which all the possible outcomes were equally likely. They developed mathematical models of theoretical probability.

It’s equally likely to get any one of six outcomes from the roll of a fair die.

It’s equally likely to get heads or tails from the toss of a fair coin.

However, keep in mind that events are not always equally likely.

A skilled basketball player has a better than 50-50 chance of making a free throw.

Modeling Probability (cont.)

Slide 14 - 8

The probability of an event is the number of outcomes in the

event divided by the total number of possible outcomes.

outcomes possible of #

Ain outcomes of #)( AP

Personal Probability

Slide 14 - 9

In everyday speech, when we express a degree of uncertainty

without basing it on long-run relative frequencies or

mathematical models, we are stating subjective or personal

probabilities.

Personal probabilities don’t display the kind of consistency

that we will need probabilities to have, so we’ll stick with

formally defined probabilities.

The First Three Rules of Working with

Probability

Slide 14 - 10

We are dealing with probabilities now, not data, but the three

rules don’t change.

Make a picture.

Make a picture.

Make a picture.

The First Three Rules of Working with

Probability (cont.)

Slide 14 - 11

The most common kind of picture to make is called a Venn

diagram.

We will see Venn diagrams in practice shortly…

Formal Probability

Slide 14 - 12

1. Two requirements for a probability:

A probability is a number between 0 and 1.

For any event A, 0 ≤ P(A) ≤ 1.

Formal Probability (cont.)

Slide 14 - 13

2. Probability Assignment Rule:

The probability of the set of all possible outcomes

of a trial must be 1.

P(S) = 1 (S represents the set of all possible

outcomes.)


Slide 14 - 14

3. Complement Rule:

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

called the complement of A, denoted AC.

The probability of an event occurring is 1 minus

the probability that it doesn’t occur:

P(A) = 1 – P(AC)

Example (p. 338 #2)

Slide 14 - 15

For each of the following, list the sample space and tell

whether you think the events are equally likely.

a) Roll two dice; record the sum of the numbers

b) A family has 3 children; record each child’s sex in order of

birth

c) Toss four coins; record the number of tails

Example (p. 338 #8)

Slide 14 - 16

Commercial airplanes have an excellent safety record. Nevertheless, there are crashes occasionally, with the loss of many lives. In the weeks following a crash, airlines often report a drop in the number of passengers, probably because people are afraid to risk flying.

a) A travel agent suggests that since the law of averages makes it highly unlikely to have two plane crashes within a few weeks of each other, flying soon after a crash is the safest time. What do you think?

b) If the airline industry proudly announces that it has set a new record for the longest period of safe flights, would you be reluctant to fly? Are the airlines due to have a crash?

Example (p. 338 #11) The plastic arrow on a spinner for a child’s game stops rotating

to point at a color that will determine what happens next.

Which of the following probability assignments are possible?

Explain.

Probabilities of…

Red Yellow Green Blue

a) 0.25 0.25 0.25 0.25

b) 0.10 0.20 0.30 0.40

c) 0.20 0.30 0.40 0.50

d) 0 0 1.00 0

e) 0.10 0.20 1.20 -1.50Slide 14 - 17

Chapter 14 & 15 Homework

Slide 14 - 18

p. 338 #1,7,12


Slide 14 - 19

4. Addition Rule:

Events that have no outcomes in common (and,

thus, cannot occur together) are called disjoint (or

mutually exclusive).

Ex: flipping a coin and getting heads

and flipping a coin and getting tails


Slide 14 - 20

4. Addition Rule (cont.):

For two disjoint events A and B, the

probability that A or B occurs is the sum of the

probabilities of the two events.

P(A B) = P(A) + P(B), provided that A and

B are disjoint.

Or means add

Example (p. 339 #20)

Slide 14 - 21

In a large Introductory Statistics lecture hall, the professor reports that 55% of the students enrolled have never taken a Calculus course, 32% have taken one semester of Calculus, and the rest have taken two or more semesters of Calculus. The professor randomly assigns students to groups of three to work on a project for the course. What is the probability that the first groupmate you meet has studied

a) two or more semesters of Calculus?

b) some Calculus?

c) no more than one semester of Calculus?


Slide 14 - 22

5. 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 B) = P(A) P(B), provided that A

and B are independent.

And (or both) means multiply


Slide 14 - 23

5. Multiplication Rule (cont.):

Two independent events A and B are not disjoint,

provided the two events have probabilities greater

than zero as the intersection:


Slide 14 - 24

You are assigned to be part of a group of three students from

the Intro Stats class described in Exercise 20. What is the

probability that of your other two groupmates,

a) neither has studied Calculus?

b) both have studied at least one semester of Calculus?

c) at least one has had more than one semester of Calculus?

Formal Probability - Notation

Slide 14 - 25

Notation alert:

In this text we use the notation P(A B) and P(A B).

In other situations, you might see the following:

P(A or B) instead of P(A B)

P(A and B) instead of P(A B)

The General Addition Rule

Slide 15 - 26

When two events A and B are disjoint, we can use the

addition rule for disjoint events from Chapter 14:

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

However, when our events are not disjoint (can occur at the

same time), this earlier addition rule will double count the

probability of both A and B occurring. Thus, we need the

General Addition Rule.

Let’s look at a picture…

The General Addition Rule (cont.)

Slide 15 - 27

5. General Addition Rule:

For any two events A and B,

P(A B) = P(A) + P(B) – P(A B)

The following Venn diagram shows a situation

in which we would use the general addition rule:

The General Addition Rule (cont.)

Slide 15 - 28

5. General Addition Rule:


P(A B) = P(A) + P(B) – P(A B)

Ex: A survey of college students found that 56% live on

campus, 62% participate in a meal program, and 42%

do both. What is the probability that a randomly

selected student either lives or eats on campus? Draw

a picture!

Let L = {student lives on campus} and

M = {student has a campus meal plan}

It Depends…

Slide 15 - 29

Back in Chapter 3, we looked at contingency

tables and talked about conditional distributions.

When we want the probability of an event from a

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

pronounce it “the probability of B given A.”

A probability that takes into account a given

condition is called a conditional probability.

It Depends… (cont.)

Slide 15 - 30

To find the probability of the event B given the event A, we restrict our attention to the outcomes in A. We then find the fraction of thoseoutcomes B that also occurred.

Note: P(A) cannot equal 0, since we know that Ahas occurred.

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


Slide 15 - 31

Contingency Table on pg. 346

P(girl) =

P(girl and popular) =

P(sports) =

P(sports Ɩ girl) =

P (sports Ɩ boy) =

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

Grades Popular Sports Total

Boy 117 50 60 227

Girl 130 91 30 251

Total 247 141 90 478


Slide 15 - 32

Ex: A survey of college students found that 56% live on campus, 62%

participate in a meal program, and 42% do both.

While dining in a campus facility open only to students with meal plans, you

make a friend. What is the probability that your new friend lives on

campus?

Let L = {student lives on campus} and M = {student has a campus meal plan}

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

The General Multiplication Rule

Slide 15 - 33

When two events A and B are independent, we can use the multiplication rule for independent events from Chapter 14:

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

However, when our events are not independent, this earlier multiplication rule does not work. Thus, we need the General Multiplication Rule.

The General Multiplication Rule (cont.)

Slide 15 - 34

We encountered the general multiplication rule in the form of conditional probability.

Rearranging the equation in the definition for conditional probability, we get the 6. General Multiplication Rule:


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

or

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

Independence

Slide 15 - 35

Independence of two events means that the outcome of one

event does not influence the probability of the other.

With our new notation for conditional probabilities, we can

now formalize this definition:

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

(Equivalently, events A and B are independent whenever

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

IndependenceEx: A survey of college students found that 56% live on campus, 62%

participate in a meal program, and 42% do both.

Are living on campus and having a meal plan independent? Are they disjoint?

Let L = {student lives on campus} and M = {student has a campus meal plan}

Independent ≠ Disjoint

Slide 15 - 37

Disjoint events cannot be independent! Well, why not?

Since we know that disjoint events have no outcomes in

common, knowing that one occurred means the other didn’t.

Thus, the probability of the second occurring changed based

on our knowledge that the first occurred.

It follows, then, that the two events are not independent.

A common error is to treat disjoint events as if they were

independent, and apply the Multiplication Rule for independent

events—don’t make that mistake.

Depending on Independence

Slide 15 - 38

It’s much easier to think about independent events than to deal with conditional probabilities.

It seems that most people’s natural intuition for probabilities breaks down when it comes to conditional probabilities.

Don’t fall into this trap: whenever you see probabilities multiplied together, stop and ask whether you think they are really independent.

Drawing Without Replacement

Slide 15 - 39

Sampling without replacement means that once one individual is

drawn it doesn’t go back into the pool.

We often sample without replacement, which doesn’t matter

too much when we are dealing with a large population.

However, when drawing from a small population, we need to

take note and adjust probabilities accordingly.

Drawing without replacement is just another instance of working

with conditional probabilities.

Tree Diagrams

Slide 15 - 40

A tree diagram helps us think through conditional

probabilities by showing sequences of events as paths that

look like branches of a tree.

Making a tree diagram for situations with conditional

probabilities is consistent with our “make a picture” mantra.

Tree Diagrams (cont.)

Slide 15 - 41

Figure 15.5 is a nice example

of a tree diagram and shows

how we multiply the

probabilities of the branches

together.

All the final outcomes are

disjoint and must add up to

one.

We can add the final

probabilities to find

probabilities of compound

events.

Tree Diagrams (cont.)

Create a Tree diagram to represent tossing a fair coin and

then rolling one die.

What is the probability that you flip heads and then roll a 2?

What is the probability that your outcome consists of

rolling a 2?

Reversing the Conditioning

Slide 15 - 43

Reversing the conditioning of two events is rarely intuitive.

Suppose we want to know P(A|B), and we know only P(A), P(B),

and P(B|A).

We also know P(A B), since

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

From this information, we can find P(A|B):

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

Bayes’s Rule

Slide 15 - 44

7. When we reverse the probability from the conditional

probability that you’re originally given, you are actually

using Bayes’s Rule.

P B | A P A | B P B

P A | B P B P A | BC P BC

Class Practice

You roll a fair die three times. What is the probability that

a) you roll all 6’s?

b) you roll all odd numbers?

c) you roll at least one 5?

d) the numbers you roll are not all 5’s?

Slide 14 - 45


Slide 14 - 46

p. 338 # 17, 19, 21

Example (p. 361 #2)

Suppose the probability that a U.S. resident has traveled to

Canada is 0.18, to Mexico is 0.09, and to both countries is

0.04. What is the probability that an American chosen at

random has

a) traveled to Canada but not Mexico?

b) traveled to either Canada or Mexico?

c) not traveled to either country?

Slide 15 - 47


The American Red Cross says that about 45% of the U.S.

population has Type O blood, 40% Type A, 11% Type B and

the rest Type AB.

a) Someone volunteers to give blood. What is the probability

that this donor

i) has Type AB blood?

ii) has Type A or Type B?

iii) is not Type O?

b) Among four potential donors, what is the probability that

i) all are Type O?

ii) no one is Type AB?

iii) they are not all Type A?

iv) at least one person is Type B?Slide 14 - 48

Example (p. 362 #15) You are dealt a hand of three cards, one at a time (the cards are

not replaced). Find the probability of each of the following:

a) The first heart you get is the third card dealt.

b) Your cards are all red.

c) You get no spades.

d) You have at least one ace.

Slide 15 - 49

Example (p. 361 #5) The marketing research organization GfK Custom Research North America conducts

a yearly survey on consumer attitudes worldwide. They collect demographic

information on the roughly 1500 respondents from each country that they survey.

Here is a table showing the number of people with various levels of education in five

countries. What is the probability that the person we choose at random is from

the United States?

Slide 15 - 50

Educational Level by Country

Post

Graduate

College Some high

school

Primary or

less

No answer Total

China 7 315 671 506 3 1502

France 69 388 766 309 7 1539

India 161 514 622 227 11 1535

U.K. 58 207 1240 32 20 1557

USA 84 486 896 87 4 1557

Total 379 1910 4195 1161 45 7690

Example (p. 361 #5 cont) The marketing research organization GfK Custom Research North America conducts




countries. What is the probability that the person we choose at random

completed his or her education before college?

Slide 15 - 51


Post

Graduate

College Some high

school

Primary or

less

No answer Total

China 7 315 671 506 3 1502

France 69 388 766 309 7 1539

India 161 514 622 227 11 1535

U.K. 58 207 1240 32 20 1557

USA 84 486 896 87 4 1557

Total 379 1910 4195 1161 45 7690






France or did some post-graduate study?

Slide 15 - 52


Post

Graduate

College Some high

school

Primary or

less

No answer Total

China 7 315 671 506 3 1502

France 69 388 766 309 7 1539

India 161 514 622 227 11 1535

U.K. 58 207 1240 32 20 1557

USA 84 486 896 87 4 1557

Total 379 1910 4195 1161 45 7690






France and finished only primary school or less?

Slide 15 - 53


Post

Graduate

College Some high

school

Primary or

less

No answer Total

China 7 315 671 506 3 1502

France 69 388 766 309 7 1539

India 161 514 622 227 11 1535

U.K. 58 207 1240 32 20 1557

USA 84 486 896 87 4 1557

Total 379 1910 4195 1161 45 7690


A certain bowler can bowl a strike 70% of the time. What’s

the probability that she

a) goes her first three frames without a strike?

b) makes her first strike in the third frame?

c) has at least one strike in the first three frames?

d) bowls a perfect game (12 consecutive strikes)?

Slide 14 - 54

Example (p. 362 # 8)

In its monthly report, the local animal shelter states that it

currently has 24 dogs and 18 cats available for adoption.

Eight of the dogs and 6 of the cats are male. Find each of the

following conditional probabilities if an animal is selected at

random. (Hint: make a table)

a) The pet is male, given that it is a cat.

b) The pet is a cat, given that it is female.

c) The pet is female, given that it is a dog.

Slide 15 - 55

Death Penalty

Favor Oppose

Party

Republican 0.26 0.04

Democrat 0.12 0.24

Other 0.24 0.1

The table shows the political affiliations of American voters and their

positions on the death penalty.

a) What’s the probability that

i) a randomly chosen voter favors the death penalty?

ii) a Republican favors the death penalty?

iii) a voter who favors the death penalty is a Democrat?

b) A candidate thinks she has a good chance of getting the votes of anyone

who is a Republican or in favor of the death penalty. What portion of the

voters is that?


Slide 15 - 56


Given the table of probabilities from Exercise 10, are party

affiliation and position on the death penalty independent?

Explain.

Slide 15 - 57

Death Penalty

Favor Oppose

Party

Republican 0.26 0.04

Democrat 0.12 0.24

Other 0.24 0.1

Example (p. 363 # 22)

According to Exercise 2, the probability that a U.S. resident

has traveled to Canada is 0.18, to Mexico is 0.09, and to both

countries is 0.04.

a) What’s the probability that someone who has traveled to

Mexico has visited Canada too?

b) Are traveling to Mexico and to Canada disjoint events?

Explain.

c) Are traveling to Mexico and to Canada independent events?

Explain.

Slide 15 - 58


A private college report contains these statistics:

70% of incoming freshmen attended public schools

75% of public school students who enroll as freshmen

eventually graduate

90% of other freshmen eventually graduate

a) Is there any evidence that a freshman’s chances to graduate

may depend upon what kind of high school the student

attended? Explain.

b) What percent of freshman eventually graduate?

c) (p. 364 #36) What percent of students who graduate from

the college attended a public high school?Slide 15 - 59

Class Practice

You roll a fair die three times. What is the probability that

a) you roll all 6’s?

b) you roll all odd numbers?

c) none of your rolls gets a number divisible by 3?

d) you roll at least one 5?

e) the numbers you roll are not all 5’s?

f) the numbers you roll are all not 5’s?

Slide 14 - 60

Class Exercise

A company’s records indicate that on any given day about 1%

of their day-shift employees and 2% of the night-shift

employees will miss work. Sixty percent of the employees

work the day shift.

a) Is absenteeism independent of shift worked? Explain.

b) What percent of employees are absent on any given day?

c) What percent of the absent employees are on the night shift?

Slide 15 - 61

Class Exercise In April 2003, Science magazine reported on a new computer-based

test for ovarian cancer, “clinical proteomics,” that examines a blood

sample for the presence of certain patterns of proteins. Ovarian

cancer, though dangerous, is very rare, afflicting only 1 of every 5000

women. The test is highly sensitive, able to correctly detect the

presence of ovarian cancer in 99.97% of women who have the disease.

However, it is unlikely to be used as a screening test in the general

population because the test gave false positives 5% of the time. Why

are false positives such a big problem? Draw a tree diagram and

determine the probability that a woman who tests positive using this

method actually has ovarian cancer.

Slide 15 - 62


Slide 14 - 63

p. 338 # 31, 33, 35, 38

p. 361 # 1, 6 , 7, 9, 27, 33, 35

YOU MUST SHOW YOUR WORK!!

Teaching Probability

https://www.teachingchannel.org/videos/teaching-

probability-odds#video-sidebar_tab_video-guide-tab

https://www.teachingchannel.org/videos/teaching-probability-odds#video-sidebar_tab_video-guide-tab

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