Chapter Probability © 2010 Pearson Prentice Hall. All rights reserved 3 5

Chapter

Probability

© 2010 Pearson Prentice Hall. All rights reserved

35

Section 5.1 Probability Rules

5-2© 2010 Pearson Prentice Hall. All rights reserved

Probability is a measure of the likelihood of a random phenomenon or chance behavior. Probability describes the long-term proportion with which a certain outcome will occur in situations with short-term uncertainty.

Use the probability applet to simulate flipping a coin 100 times. Plot the proportion of heads against the number of flips. Repeat the simulation.


Probability deals with experiments that yield random short-term results or outcomes, yet reveal long-term predictability.

The long-term proportion with which a certain outcome is observed is the probability of that outcome.


The Law of Large Numbers

As the number of repetitions of a probability experiment increases, the proportion with which a certain outcome is observed gets closer to the probability of the outcome.

The Law of Large Numbers

As the number of repetitions of a probability experiment increases, the proportion with which a certain outcome is observed gets closer to the probability of the outcome.


In probability, an experiment is any process that can be repeated in which the results are uncertain.

The sample space, S, of a probability experiment is the collection of all possible outcomes.

An even is any collection of outcomes from a probability experiment. An event may consist of one outcome or more than one outcome. We will denote events with one outcome, sometimes called simple events, ei. In general, events are denoted using capital letters such as E.


Consider the probability experiment of having two children.

(a) Identify the outcomes of the probability experiment.(b) Determine the sample space.(c) Define the event E = “have one boy”.

EXAMPLE Identifying Events and the Sample Space of a Probability Experiment

EXAMPLE Identifying Events and the Sample Space of a Probability Experiment

(a) e1 = boy, boy, e2 = boy, girl, e3 = girl, boy, e4 = girl, girl(b) {(boy, boy), (boy, girl), (girl, boy), (girl, girl)}(c) {(boy, girl), (girl, boy)}



A probability model lists the possible outcomes of a probability experiment and each outcome’s probability. A probability model must satisfy rules 1 and 2 of the rules of probabilities.


EXAMPLE A Probability ModelEXAMPLE A Probability Model

In a bag of peanut M&M milk chocolate candies, the colors of the candies can be brown, yellow, red, blue, orange, or green. Suppose that a candy is randomly selected from a bag. The table shows each color and the probability of drawing that color. Verify this is a probability model.

Color Probability

Brown 0.12

Yellow 0.15

Red 0.12

Blue 0.23

Orange 0.23

Green 0.15

• All probabilities are between 0 and 1, inclusive.

• Because 0.12 + 0.15 + 0.12 + 0.23 + 0.23 + 0.15 = 1, rule 2 (the sum of all probabilities must equal 1) is satisfied.


If an event is a certainty, the probability of the event is 1.

If an event is impossible, the probability of the event is 0.

An unusual event is an event that has a low probability of occurring.


Copyright © 2010 Pearson Education, Inc.

True or false. The following represents a probability model.

A. True

B. False

Cell Phone Provider

Probability

AT&T 0.271

Sprint 0.236

T–Mobile 0.111

Verizon 0.263

Slide 5- 12


True or false. The following represents a probability model.

A. True

B. False

Cell Phone Provider

Probability

AT&T 0.271

Sprint 0.236

T–Mobile 0.111

Verizon 0.263

Slide 5- 13



Pass the PigsTM is a Milton-Bradley game in which pigs are used as dice. Points are earned based on the way the pig lands. There are six possible outcomes when one pig is tossed. A class of 52 students rolled pigs 3,939 times. The number of times each outcome occurred is recorded in the table at right. (Source: http://www.members.tripod.com/~passpigs/prob.html)

EXAMPLE Building a Probability Model EXAMPLE Building a Probability Model

Outcome Frequency

Side with no dot 1344

Side with dot 1294

Razorback 767

Trotter 365

Snouter 137

Leaning Jowler 32

(a) Use the results of the experiment to build a probability model for the way the pig lands.

(b) Estimate the probability that a thrown pig lands on the “side with dot”. (c) Would it be unusual to throw a “Leaning Jowler”?


(a) Outcome Probability

Side with no dot

Side with dot 0.329

Razorback 0.195

Trotter 0.093

Snouter 0.035

Leaning Jowler 0.008

13440.341

3939

(b) The probability a throw results in a “side with dot” is 0.329. In 1000 throws of the pig, we would expect about 329 to land on a “side with dot”.

(c) A “Leaning Jowler” would be unusual. We would expect in 1000 throws of the pig to obtain “Leaning Jowler” about 8 times.



The classical method of computing probabilities requires equally likely outcomes.

An experiment is said to have equally likely outcomes when each simple event has the same probability of occurring.


EXAMPLE Computing Probabilities Using the Classical MethodEXAMPLE Computing Probabilities Using the Classical Method

Suppose a “fun size” bag of M&Ms contains 9 brown candies, 6 yellow candies, 7 red candies, 4 orange candies, 2 blue candies, and 2 green candies. Suppose that a candy is randomly selected.

(a) What is the probability that it is yellow?

(b) What is the probability that it is blue?

(c) Comment on the likelihood of the candy being yellow versus blue.

(a) There are a total of 9 + 6 + 7 + 4 + 2 + 2 = 30 candies, so N(S) = 30. (yellow)

(yellow)( )

6

300.2

NP

N S

(b) P(blue) = 2/30 = 0.067.

(c) Since P(yellow) = 6/30 and P(blue) = 2/30, selecting a yellow is three times as likely as selecting a blue.



A box contains 6 twenty-five watt light bulbs, 9 sixty-watt light bulbs, and 5 hundred-watt light bulbs. What is the probability a randomly selected light bulb is sixty-watts?

A. 0.45

B. 0.3

C. 0.05

D. 0.25

Slide 5- 21


A box contains 6 twenty-five watt light bulbs, 9 sixty-watt light bulbs, and 5 hundred-watt light bulbs. What is the probability a randomly selected light bulb is sixty-watts?

A. 0.45

B. 0.3

C. 0.05

D. 0.25

Slide 5- 22


Use the probability applet on your calculator (instructor will show you how) to simulate throwing a 6-sided die 100 times. Approximate the probability of rolling a 4. How does this compare to the classical probability? Repeat the exercise for 1000 throws of the die.

EXAMPLE Using SimulationEXAMPLE Using Simulation



The subjective probability of an outcome is a probability obtained on the basis of personal judgment.

For example, an economist predicting there is a 20% chance of recession next year would be a subjective probability.


In his fall 1998 article in Chance Magazine, (“A Statistician Reads the Sports Pages,” pp. 17-21,) Hal Stern investigated the probabilities that a particular horse will win a race. He reports that these probabilities are based on the amount of money bet on each horse. When a probability is given that a particular horse will win a race, is this empirical, classical, or subjective probability?

EXAMPLE Empirical, Classical, or Subjective ProbabilityEXAMPLE Empirical, Classical, or Subjective Probability

Subjective because it is based upon people’s feelings about which horse will win the race. The probability is not based on a probability experiment or counting equally likely outcomes.


Section 5.2 Probability Rules



Two events are disjoint if they have no outcomes in common. Another name for disjoint events is mutually exclusive events.


We often draw pictures of events using Venn diagrams. These pictures represent events as circles enclosed in a rectangle. The rectangle represents the sample space, and each circle represents an event. For example, suppose we randomly select a chip from a bag where each chip in the bag is labeled 0, 1, 2, 3, 4, 5, 6, 7, 8, 9. Let E represent the event “choose a number less than or equal to 2,” and let F represent the event “choose a number greater than or equal to 8.” These events are disjoint as shown in the figure.



The probability model to the right shows the distribution of the number of rooms in housing units in the United States.

Number of Rooms in Housing Unit

Probability

One 0.010

Two 0.032

Three 0.093

Four 0.176

Five 0.219

Six 0.189

Seven 0.122

Eight 0.079

Nine or more 0.080

Source: American Community Survey, U.S. Census Bureau

EXAMPLE The Addition Rule for Disjoint EventsEXAMPLE The Addition Rule for Disjoint Events

(a) Verify that this is a probability model.

All probabilities are between 0 and 1, inclusive.

0.010 + 0.032 + … + 0.080 = 1



Probability

One 0.010

Two 0.032

Three 0.093

Four 0.176

Five 0.219

Six 0.189

Seven 0.122

Eight 0.079

Nine or more 0.080

(b) What is the probability a randomly selected housing unit has two or three rooms?

P(two or three)

= P(two) + P(three)

= 0.032 + 0.093

= 0.125


(c) What is the probability a randomly selected housing unit has one or two or three rooms?


Probability

One 0.010

Two 0.032

Three 0.093

Four 0.176

Five 0.219

Six 0.189

Seven 0.122

Eight 0.079

Nine or more 0.080

P(one or two or three)

= P(one) + P(two) + P(three)

= 0.010 + 0.032 + 0.093

= 0.135



The data shows the distance employees of a company travel to work. One of these employees is randomly selected. Determine the probability the employee travels between 10 and 29 miles to work.

A. 0.401

B. 0.566

C. 0.334

D. 0.735

Slide 5- 36

Distance (miles) Number of employees

0 – 9 124

10 – 19 309

20 – 29 257

30 – 39 78

40 – 49 2


The data shows the distance employees of a company travel to work. One of these employees is randomly selected. Determine the probability the employee travels between 10 and 29 miles to work.

A. 0.401

B. 0.566

C. 0.334

D. 0.735

Slide 5- 37

Distance (miles) Number of employees

0 – 9 124

10 – 19 309

20 – 29 257

30 – 39 78

40 – 49 2


The table shows the favorite pizza topping for a sample of students. One of these students is selected at random. Find the probability the student is female or prefers sausage.

A. 0.458

B. 0.583

C. 0.125

D. 0.556

Slide 5- 38

Cheese Pepperoni Sausage TotalMale 8 5 2 15Female 2 4 3 9Total 10 9 5 24


The table shows the favorite pizza topping for a sample of students. One of these students is selected at random. Find the probability the student is female or prefers sausage.

A. 0.458

B. 0.583

C. 0.125

D. 0.556


Slide 5- 39




Suppose that a pair of dice are thrown. Let E = “the first die is a two” and let F = “the sum of the dice is less than or equal to 5”. Find P(E or F) using the General Addition Rule.

EXAMPLE Illustrating the General Addition RuleEXAMPLE Illustrating the General Addition Rule

( )( )

( )

6

361

6

N EP E

N S

( )( )

( )

10

365

18

N FP F

N S

( and )( and )

( )

3

361

12

N E FP E F

N S

( or ) ( ) ( ) ( and )

6 10 3

36 36 3613

36

P E F P E P F P E F



Complement of an Event

Let S denote the sample space of a probability experiment and let E denote an event. The complement of E, denoted EC, is all outcomes in the sample space S that are not outcomes in the event E.


Complement Rule

If E represents any event and EC represents the complement of E, then

P(EC) = 1 – P(E)


According to the American Veterinary Medical Association, 31.6% of American households own a dog. What is the probability that a randomly selected household does not own a dog?

P(do not own a dog) = 1 – P(own a dog)

= 1 – 0.316

= 0.684

EXAMPLE Illustrating the Complement RuleEXAMPLE Illustrating the Complement Rule


The data to the right represent the travel time to work for residents of Hartford County, CT.

(a) What is the probability a randomly selected resident has a travel time of 90 or more minutes?

EXAMPLE Computing Probabilities Using Complements EXAMPLE Computing Probabilities Using Complements

Source: United States Census Bureau

There are a total of

24,358 + 39,112 + … + 4,895 = 393,186

residents in Hartford County, CT.

48950.012

393,186

The probability a randomly selected resident will have a commute time of “90 or more minutes” is


(b) Compute the probability that a randomly selected resident of Hartford County, CT will have a commute time less than 90 minutes.

P(less than 90 minutes) = 1 – P(90 minutes or more)

= 1 – 0.012

= 0.988


Section 5.3 Independence and Multiplication Rule



Two events E and F are independent if the occurrence of event E in a probability experiment does not affect the probability of event F. Two events are dependent if the occurrence of event E in a probability experiment affects the probability of event F.


EXAMPLE Independent or Not?EXAMPLE Independent or Not?

(a) Suppose you draw a card from a standard 52-card deck of cards and then roll a die. The events “draw a heart” and “roll an even number” are independent because the results of choosing a card do not impact the results of the die toss.

(b) Suppose two 40-year old women who live in the United States are randomly selected. The events “woman 1 survives the year” and “woman 2 survives the year” are independent.

(c) Suppose two 40-year old women live in the same apartment complex. The events “woman 1 survives the

year” and “woman 2 survives the year” are dependent.




The probability that a randomly selected female aged 60 years old will survive the year is 99.186% according to the National Vital Statistics Report, Vol. 47, No. 28. What is the probability that two randomly selected 60 year old females will survive the year?

EXAMPLE Computing Probabilities of Independent EventsEXAMPLE Computing Probabilities of Independent Events

The survival of the first female is independent of the survival of the second female. We also have that P(survive) = 0.99186.

First survives and second survives First survives Second survives

(0.99186)(0.99186)

0.9838

P P P


A manufacturer of exercise equipment knows that 10% of their products are defective. They also know that only 30% of their customers will actually use the equipment in the first year after it is purchased. If there is a one-year warranty on the equipment, what proportion of the customers will actually make a valid warranty claim?

EXAMPLE Computing Probabilities of Independent EventsEXAMPLE Computing Probabilities of Independent Events

We assume that the defectiveness of the equipment is independent of the use of the equipment. So,

defective and used defective used

(0.10)(0.30)

0.03

P P P



The probability that a randomly selected female aged 60 years old will survive the year is 99.186% according to the National Vital Statistics Report, Vol. 47, No. 28. What is the probability that four randomly selected 60 year old females will survive the year?

P(all four survive)

= P (1st survives and 2nd survives and 3rd survives and 4th survives)

= P(1st survives) . P(2nd survives) . P(3rd survives) . P(4th survives)

= (0.99186) (0.99186) (0.99186) (0.99186)

= 0.9678

EXAMPLE Illustrating the Multiplication Principle for Independent EventsEXAMPLE Illustrating the Multiplication Principle for Independent Events



The probability that a randomly selected female aged 60 years old will survive the year is 99.186% according to the National Vital Statistics Report, Vol. 47, No. 28. What is the probability that at least one of 500 randomly selected 60 year old females will die during the course of the year?

P(at least one dies) = 1 – P(none die)

= 1 – P(all survive)

= 1 – 0.99186500

= 0.9832

EXAMPLE Computing “at least” ProbabilitiesEXAMPLE Computing “at least” Probabilities



Forty-four percent of college students have engaged in binge drinking. If five college students are randomly selected, what is the probability that at least one of the five has engaged in binge drinking?

A. 0.055

B. 0.216

C. 0.945

D. 0.016

Slide 5- 62


Forty-four percent of college students have engaged in binge drinking. If five college students are randomly selected, what is the probability that at least one of the five has engaged in binge drinking?

A. 0.055

B. 0.216

C. 0.945

D. 0.016

Slide 5- 63


Section 5.4 Conditional Probability and the General Multiplication Rule



Conditional Probability

The notation P(F | E) is read “the probability of event F given event E”. It is the probability of an event F given the occurrence of the event E.


EXAMPLE An Introduction to Conditional ProbabilityEXAMPLE An Introduction to Conditional Probability

Suppose that a single six-sided die is rolled. What is the probability that the die comes up 4? Now suppose that the die is rolled a second time, but we are told the outcome will be an even number. What is the probability that the die comes up 4?

First roll: S = {1, 2, 3, 4, 5, 6} 1( )

6P S

Second roll: S = {2, 4, 6} 1( )

3P S



EXAMPLE Conditional Probabilities on Belief about God and Region of the CountryEXAMPLE Conditional Probabilities on Belief about God and Region of the Country

A survey was conducted by the Gallup Organization conducted May 8 – 11, 2008 in which 1,017 adult Americans were asked, “Which of the following statements comes closest to your belief about God – you believe in God, you don’t believe in God, but you do believe in a universal spirit or higher power, or you don’t believe in either?” The results of the survey, by region of the country, are given in the table below.

Believe in God

Believe in universal spirit

Don’t believe in either

East 204 36 15

Midwest 212 29 13

South 219 26 9

West 152 76 26

(a) What is the probability that a randomly selected adult American who lives in the East believes in God?

(b) What is the probability that a randomly selected adult American who believes in God lives in the East?


Believe in God

Believe in universal spirit

Don’t believe in either

East 204 36 15

Midwest 212 29 13

South 219 26 9

West 152 76 26

(a) What is the probability that a randomly selected adult American who lives in the East believes in God?

believe in God and live in the eastbelieves in God lives in the east

live in the east

NP

N

204

204 36 15

0.8(b) What is the probability that a randomly selected adult American who believes

in God lives in the East?

believe in God and live in the eastlives in the east believes in God

believes in God

NP

N

204

204 212 219 152

0.26


In 2005, 19.1% of all murder victims were between the ages of 20 and 24 years old. Also in 1998, 16.6% of all murder victims were 20 – 24 year old males. What is the probability that a randomly selected murder victim in 2005 was male given that the victim is 20 - 24 years old?

EXAMPLE Murder VictimsEXAMPLE Murder Victims

male and 20 24male 20 24

20 24

PP

P

0.166

0.191

= 0.869



The table shows the favorite pizza topping for a sample of students. What is the probability that a randomly selected student who was male preferred pepperoni?

A. 0.333

B. 0.375

C. 0.6

D. 0.556

Slide 5- 73



The table shows the favorite pizza topping for a sample of students. What is the probability that a randomly selected student who was male preferred pepperoni?

A. 0.333

B. 0.375

C. 0.6

D. 0.556

Slide 5- 74


Section 5.5 Counting Techniques


76

One, two, three, we’re…

Counting

77

Counting problems are of the following kind:“How many different 8-letter passwords are there?”“How many possible ways are there to pick 11 soccer players out of a 20-player team?”Most importantly, counting is the basis for computing probabilities of discrete events.“What is the probability of winning the lottery?”

BASIC COUNTING PRINCIPLES

78

The sum rule:

If a task can be done in n1 ways and a second task in n2 ways, and if these two tasks cannot be done at the same time, then there are n1 + n2 ways to do either task.


Example: The department will award a free computer to either a student or a teacher. How many different choices are there, if there are 530 students and 45 teachers?There are 530 + 45 = 575 choices.

79

Generalized sum rule: If we have tasks T1, T2, …, Tm that can be done in n1, n2, …, nm ways, respectively, and no two of these tasks can be done at the same time, then there are n1 + n2 + … + nm ways to do one of these tasks.


80

The product rule:Suppose that a procedure can be broken down into two successive tasks. If there are n1 ways to do the first task and n2 ways to do the second task after the first task has been done, then there are n1n2 ways to do the procedure.


81

Example: How many different license plates are there that contain exactly three English letters ?Solution: There are 26 possibilities to pick the first letter, then 26 possibilities for the second one, and 26 for the last one. So there are 26⋅26⋅26 = 17576 different license plates.


82

Generalized product rule:If we have a procedure consisting of sequential tasks T1, T2, …, Tm that can be done in n1, n2, …, nm ways, respectively, then there are n1 ⋅ n2 ⋅ … ⋅ nm ways to carry out the procedure.


0

1

0

1010

1

0

1010

10183

Tree DiagramsHow many bit strings of length four do NOT have two consecutive 1s?

There are 8 strings.

0

1

1

0

0

0

1

1

1

0

0

1

0

1

84

The pigeonhole principle: If (k + 1) or more objects are placed into k boxes, then there is at least one box containing two or more of the objects.Example 1: If there are 11 players in a soccer team that wins 12-0, there must be at least one player in the team who scored at least twice.Example 2: If you have 6 classes from Monday to Friday, there must be at least one day on which you have at least two classes.

THE PIGEONHOLE PRINCIPLE

85

The generalized pigeonhole principle: If N objects are placed into k boxes, then there is at least one box containing at least N/k of the objects.Example 1: In a 60-student class, at least 12 students will get the same letter grade (A, B, C, D, or F).


86

Example 2: Assume you have a drawer containing a random distribution of a dozen brown socks and a dozen black socks. It is dark, so how many socks do you have to pick to be sure that among them there is a matching pair?

Solution: There are two types of socks, so if you pick at least 3 socks, there must be either at least two brown socks or at least two black socks.

Generalized pigeonhole principle: 3/2 = 2.



How many 4-letter television call signs are possible, if each sign must start with either a K or a W?

A. 35,152

B. 456,976

C. 16

D. 104

Slide 5- 87


How many 4-letter television call signs are possible, if each sign must start with either a K or a W?

A. 35,152

B. 456,976

C. 16

D. 104

Slide 5- 88


There are 15 dogs entered in a show. How many ways can first, second, and third place be awarded?

A. 45

B. 455

C. 2,730

D. 3,375

Slide 5- 89


There are 15 dogs entered in a show. How many ways can first, second, and third place be awarded?

A. 45

B. 455

C. 2,730

D. 3,375

Slide 5- 90


92

Permutations and CombinationsHow many ways are there to pick a set of 3 people from a group of 6?The answer to this depends on whether we want the order in which they are picked to matter or not. For example, picking person C, then person A, and then person E leads to the same group as first picking E, then C, and then A.There are 6 choices for the first person, 5 for the second one, and 4 for the third one, so there are 6⋅5⋅4 = 120 ways to do this.Since in the original statement, it does not seem that order is important. This is not the correct result!However, these cases are counted separately in the above equation.

93

So how can we compute how many different subsets of people can be picked (that is, we want to disregard the order of picking) ?To find out about this, we need to first look at permutations.A permutation of a set of distinct objects is an ordered arrangement of these objects.An ordered arrangement of r elements of a set is called an r-permutation.

Permutations and Combinations

94

Example: Let S = {1, 2, 3}.The arrangement 3, 1, 2 is a permutation of S.The arrangement 3, 2 is a 2-permutation of S.The number of r-permutations of a set with n distinct elements is denoted by P(n, r) or nPr .We can calculate P(n, n) with the product rule:P(n, n) = n⋅(n – 1)⋅(n – 2) ⋅…⋅3⋅2⋅1.(n choices for the first element, (n – 1) for the second one, (n – 2) for the third one…)


95

Example: 8P3 = (8⋅7⋅6⋅5⋅4⋅3⋅2⋅1)/(5⋅4⋅3⋅2⋅1) = 8⋅7⋅6 = 336General formula: P(n, r) = n!/(n – r)! = nPr

Knowing this, we can return to our initial question:How many ways are there to pick a set of 3 people from a group of 6 (disregarding the order of picking)?


A combination is an arrangement, without regard to order, of n distinct objects without repetitions. The symbol

nCr represents the number of combinations of n distinct objects taken r at a time, where r < n.


97

An r-combination of elements of a set is an unordered selection of r elements from the set.Thus, an r-combination is simply a subset of the set with r elements.Example: Let S = {1, 2, 3, 4}.Then {1, 3, 4} is a 3-combination from S.The number of r-combinations of a set with n distinct elements is denoted by C(n, r) or nCr.Example: C(4, 2) = 6, since, for example, the 2-combinations of a set {1, 2, 3, 4} are {1, 2}, {1, 3}, {1, 4}, {2, 3}, {2, 4}, {3, 4}.

PERMUTATIONS AND COMBINATIONS

98

How can we calculate C(n, r)?Consider that we can obtain the r-permutation of a set in the following way:First, we form all the r-combinations of the set(there are C(n, r) such r-combinations).Then, we generate all possible orderings in each of these r-combinations (there are P(r, r) such orderings in each case).Therefore, we have:P(n, r) = C(n, r)⋅P(r, r)


99

C(n, r) = nCr = P(n, r)/P(r, r) = n!/(n – r)!/(r!/(r – r)!) = n!/(r!(n – r)!)Now we can answer our initial question:How many ways are there to choose a set of 3 people from a group of 6 (disregarding the order of picking)?C(6, 3) = 6!/(3!⋅3!) = 720/(6⋅6) = 720/36 = 20There are 20 different ways, that is, 20 different groups to be picked.


100

Corollary: Let n and r be nonnegative integers with r ≤ n.Then C(n, r) = C(n, n – r).Note that “choosing a group of r people from a group of n people” is the same as “splitting a group of n people into a group of r people and another group of (n – r) people”.


101

Example:A soccer club has 8 female and 7 male members. For today’s match, the coach wants to have 6 female and 5 male players on the grass. How many possible configurations are there?

8C6 ⋅ 7C5

= 28⋅21 = 588



Determine the value of 9C3.


The United States Senate consists of 100 members. In how many ways can 4 members be randomly selected to attend a luncheon at the White House?



There are 13 students in a club. How many ways can four students be selected to attend a conference?

A. 17,160

B. 52

C. 28,561

D. 715

Slide 5- 105


There are 13 students in a club. How many ways can four students be selected to attend a conference?

A. 17,160

B. 52

C. 28,561

D. 715

Slide 5- 106

Section 5.6 Bayes’s Rule

The material in this section is available on the CD that accompanies the text.




EXAMPLE Introduction to the Rule of Total ProbabilityEXAMPLE Introduction to the Rule of Total Probability

At a university55% of the students are female and 45% are male15% of the female students are business majors20% of the male students are business majors

What percent of students, overall, are business majors?

● The percent of the business majors in the university contributed by females 55% of the students are female 15% of those students are business majors Thus 15% of 55%, or 0.55 • 0.15 = 0.0825 or 8.25% of the total

student body are female business majors

● Contributed by males In the same way, 20% of 45%, or 0.45 • 0.20 = .09 or 9% are

male business majors


● Altogether 8.25% of the total student body are female business majors 9% of the total student body are male business majors

● So … 17.25% of the total student body are business majors



• Another way to analyze this problem is to use a tree diagram

Female

Male

0.55

0.45




Female

Male

0.55

0.45

Business

Not Business

Business

Not Business

0.55•0.15

0.55•0.85

0.45•0.20

0.45•0.80



Multiply out, and add the two business branches


Business

Not Business

Business

Not Business

Female

Male

0.55

0.45

0.55•0.15

0.55•0.85

0.45•0.20

0.45•0.80

0.0825

0.0900

0.4675

0.3600

0.0825

0.0900

Total = 0.1725

0.4675

0.3600

• This is an example of the Rule of Total Probability

P(Bus) = 55% • 15% + 45% • 20%= P(Female) • P(Bus |

Female) + P(Male) • P(Bus | Male)

• This rule is useful when the sample space can be divided into two (or more) disjoint parts


● A partition of the sample space S are twonon-empty sets A1 and A2 that divide up S

● In other words A1 ≠ Ø

A2 ≠ Ø

A1 ∩ A2 = Ø (there is no overlap)

A1 U A2 = S (they cover all of S)


● Let E be any event in the sample space S

● Because A1 and A2 are disjoint, E ∩ A1 andE ∩ A2 are also disjoint

● Because A1 and A2 cover all of S, E ∩ A1

and E ∩ A2 cover all of E

● This means that we have divided E into two disjoint pieces

E = (E ∩ A1) U (E ∩ A2)


● Because E ∩ A1 and E ∩ A2 are disjoint, we can use the Addition Rule

P(E) = P(E ∩ A1) + P(E ∩ A2)

● We now use the General Multiplication Rule on each of the P(E ∩ A1) and P(E ∩ A2) terms

P(E) = P(A1) • P(E | A1) + P(A2) • P(E | A2)


P(E) = P(A1) • P(E | A1) + P(A2) • P(E | A2)

● This is the Rule of Total Probability (for a partition into two sets A1 and A2)

● It is useful when we want to compute a probability (P(E)) but we know only pieces of it (such as P(E | A1))

● The Rule of Total Probability tells us how to put the probabilities together



● The general Rule of Total Probability assumes that we have a partition (the general definition) of S into n different subsets A1, A2, …, An

Each subset is non-empty None of the subsets overlap S is covered completely by the union of the subsets

● This is like the partition before, just that S is broken up into many pieces, instead of just two pieces



● In a particular town 30% of the voters are Republican 30% of the voters are Democrats 40% of the voters are independents

● This is a partition of the voters into three sets There are no voters that are in two sets (disjoint) All voters are in one of the sets (covers all of S)

EXAMPLE The Rule of Total ProbabilityEXAMPLE The Rule of Total Probability

● For a particular issue 90% of the Republicans favor it 60% of the Democrats favor it 70% of the independents favor it

● These are the conditional probabilities E = {favor the issue} The above probabilities are P(E | political party)


● The total proportion of votes who favor the issue

0.3 • 0.9 + 0.3 • 0.6 + 0.4 • 0.7 = 0.73

● So 73% of the voters favor this issue



• In our male / female and business / non-business majors examples before, we used the rule of total probability to answer the question

What percent of studentsare business majors?

• We solved this problem by analyzing male students and female students separately


• We could turn this problem around• We were told the percent of female students

who are business majors• We could also ask

What percent of business majorsare female?

• This is the situation for Bayes’s Rule


● For this example We first choose a random business student (event E) What is the probability that this student is female?

(partition element A1)

● This question is asking for the value of P(A1 | E)

● Before, we were working with P(E | A1) instead The probability (15%) that a female student is a

business major


• The Rule of Total Probability– Know P(Ai) and P(E | Ai)

– Solve for P(E)

• Bayes’s Rule– Know P(E) and P(E | Ai)

– Solve for P(Ai | E)


• Bayes’ Rule, for a partition into two sets U1 and U2, is

• This rule is very useful when P(U1|B) is difficult to compute, but P(B|U1) is easier

)U|B(P)U(P)U|B(P)U(P)U|B(P)U(P

)B|U(P2211

111




The business majors example from before

Business

Not Business

Business

Not Business

Female

Male

0.55

0.45

0.55•0.15

0.55•0.85

0.45•0.20

0.45•0.80

0.0825

0.0900

Total = 0.1725

0.4675

0.3600

EXAMPLE Bayes’s RuleEXAMPLE Bayes’s Rule

● If we chose a random business major, what is the probability that this student is female? A1 = Female student

A2 = Male student

E = business major

● We want to find P(A1 | E), the probability that the student is female (A1) given that this is a business major (E)

EXAMPLE Bayes’s RuleEXAMPLE Bayes’s Rule


● Do it in a straight way first We know that 8.25% of the students are female

business majors We know that 9% of the students are male business

majors Choosing a business major at random is choosing

one of the 17.25%

● The probability that this student is female is8.25% / 17.25% = 47.83%

EXAMPLE Bayes’s Rule (continued)EXAMPLE Bayes’s Rule (continued)


)A|E(P)A(P)A|E(P)A(P)A|E(P)A(P

)E|A(P i2211

11

● Now do it using Bayes’s Rule – it’s the same calculation

● Bayes’s Rule for a partition into two sets (n = 2)

P(A1) = .55, P(A2) = .45

P(E | A1) = .15, P(E | A2) = .20

We know all of the numbers we need



4783

090008250825

204515551555

2211

11

.

...

......

)A|E(P)A(P)A|E(P)A(P)A|E(P)A(P



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