statisticssite.iugaza.edu.ps/mriffi/files/2018/02/ch05.pdf · 2018. 2. 6. · 2 of 2)

STATISTICSINFORMED DECISIONS USING DATAFifth Edition

Chapter 5

Probability

Copyright © 2017, 2013, 2010 Pearson Education, Inc. All Rights Reserved


5.1 Probability RulesLearning Objectives

1. Apply the rules of probabilities

2. Compute and interpret probabilities using the empirical method

3. Compute and interpret probabilities using the classical method

4. Use simulation to obtain data based on probabilities

5. Recognize and interpret subjective probabilities


5.1 Probability RulesIntroduction to Probability (1 of 5)

Probability is a measure of the likelihood of a random phenomenon or chance behavior occurring. 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.


5.1 Probability Rules Introduction to Probability (2 of 5)

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

The long-term proportion in 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.



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 event 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.



EXAMPLE Identifying Events and the Sample Space of a Probability Experiment

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”.

(a) e1 = boy, boy, e2 = boy, girl, e3 = girl, boy, e4 = girl, girl

(b) S = {(boy, boy), (boy, girl), (girl, boy), (girl, girl)}

(c) E = {(boy, girl), (girl, boy)}


5.1 Probability Rules5.1.1 Apply the Rules of Probabilities (1 of 4)

Rules of probabilities

1. The probability of any event E, P(E),must be greater than or equal to 0 and less than or equal to 1.

That is, 0 ≤ P(E) ≤ 1.

2. The sum of the probabilities of all outcomes must equal 1.

That is, if the sample space

S = {e1, e2, …, en}, then P(e1) + P(e2) + … + P(en) = 1



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 ModelIn 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



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

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

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


5.1 Probability Rules5.1.2 Compute and Interpret Probabilities Using the Empirical Method (1 of 6)

Approximating Probabilities Using the Empirical Approach

The probability of an event E is approximately the number of times event E is observed divided by the number of repetitions of the experiment.



EXAMPLE Building a Probability Model

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.

Outcome Frequency

Side with no dot 1344

Side with dot 1294

Razorback 767

Trotter 365

Snouter 137

Leaning Jowler 32



EXAMPLE Building a Probability Model

(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”?

Outcome Frequency

Side with no dot 1344

Side with dot 1294

Razorback 767

Trotter 365

Snouter 137

Leaning Jowler 32



Outcome Probability

Side with no dot 0.341

Side with dot 0.329

Razorback 0.195

Trotter 0.093

Snouter 0.035

Leaning Jowler 0.008

(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”.



Outcome Probability

Side with no dot 0.341

Side with dot 0.329

Razorback 0.195

Trotter 0.093

Snouter 0.035

Leaning Jowler 0.008

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


5.1 Probability Rules5.1.3 Compute and Interpret Probabilities Using the Classical Method (1 of 5)

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.



Computing Probability Using the Classical Method

If an experiment has n equally likely outcomes and if the number of ways that an event E can occur is m, then the probability of E, P(E) is



Computing Probability Using the Classical Method

So, if S is the sample space of this experiment,



EXAMPLE 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.



EXAMPLE Computing Probabilities Using the Classical Method

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


5.1 Probability Rules5.1.4 Use Simulation to Obtain Data Based on Probabilities

EXAMPLE Using Simulation

Use the probability applet 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.


5.1 Probability Rules5.1.5 Recognize and Interpret Subjective Probabilities (1 of 2)

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.


5.1 Probability Rules5.1.5 Recognize and Interpret Subjective Probabilities (2 of 2)

EXAMPLE Empirical, Classical, or 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?

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.


5.2 The Addition Rule and ComplementsLearning Objectives

1. Use the Addition Rule for Disjoint Events

2. Use the General Addition Rule

3. Compute the probability of an event using the Complement Rule


5.2 The Addition Rule and Complements5.2.1 Use the Addition Rule for Disjoint Events (1 of 8)

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.



Addition Rule for Disjoint Events

If E and F are disjoint (or mutually exclusive) events, then



The Addition Rule for Disjoint Events can be extended to more than two disjoint events.

In general, if E, F, G, . . . each have no outcomes in common (they are pairwise disjoint), then



EXAMPLE The Addition Rule for Disjoint Events

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

(a) Verify that this is a probability model.

All probabilities are between 0 and 1, inclusive.

0.010 + 0.032 + … + 0.080 = 1

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




(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


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




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

P(one or two or three)

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

= 0.010 + 0.032 + 0.093

= 0.135


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


5.2 The Addition Rule and Complements5.2.2 Use the General Addition Rule (1 of 3)

The General Addition Rule

For any two events E and F,



EXAMPLE Illustrating the General Addition Rule

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.


5.2 The Addition Rule and Complements5.2.3 Compute the Probability of an Event Using the Complement Rule (1 of 6)

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)



EXAMPLE Illustrating the Complement Rule

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



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?

Travel Time Frequency

Less than 5 minutes 24,358

5 to 9 minutes 39,112

10 to 14 minutes 62,124

15 to 19 minutes 72,854

20 to 24 minutes 74,386

25 to 29 minutes 30,099

30 to 34 minutes 45,043

35 to 39 minutes 11,169


45 to 59 minutes 15,650


90 or more minutes 4,895

Source: United States Census Bureau



Travel Time Frequency

Less than 5 minutes 24,358


10 to 14 minutes 62,124

15 to 19 minutes 72,854

20 to 24 minutes 74,386

25 to 29 minutes 30,099

30 to 34 minutes 45,043

35 to 39 minutes 11,169


45 to 59 minutes 15,650


90 or more minutes 4,895

Source: United States Census Bureau

There are a total of 24,358 + 39,112 + … + 4,895 = 393,186 residents in Hartford County.

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


5.3 Independence and the Multiplication RuleLearning Objectives

1. Identify independent events

2. Use the Multiplication Rule for Independent Events

3. Compute at-least probabilities


5.3 Independence and the Multiplication Rule5.3.1 Identify independent events (1 of 2)

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.


5.3 Independence and the Multiplication Rule5.3.1 Identify independent events (2 of 2)

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.


5.3 Independence and the Multiplication Rule5.3.2 Use the Multiplication Rule for Independent Events (1 of 6)

Multiplication Rule for Independent Events

If E and F are independent events, then



EXAMPLE Computing Probabilities of 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 two randomly selected 60 year old females will survive the year?

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



EXAMPLE Computing Probabilities of Independent Events

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?

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



Multiplication Rule for n Independent Events

If E1, E2, E3, … and En are independent events, then



EXAMPLE 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 four randomly selected 60 year old females will survive the year?



EXAMPLE Illustrating the Multiplication Principle for Independent Events

P(all 4 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


5.3 Independence and the Multiplication Rule5.3.3 Compute At-Least Probabilities (1 of 4)

EXAMPLE Computing “at least” Probabilities

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 − (0.99186)500

= 0.9832



Summary: Rules of Probability

1. The probability of any event must be between 0 and 1. If we let E denote any event, then 0 ≤ P(E) ≤ 1.

2. The sum of the probabilities of all outcomes in the sample space must equal 1.

That is, if the sample space S = {e1, e2, …, en}, then P(e1) + P(e2) + … + P(en) = 1




3. If E and F are disjoint events, then

P(E or F) = P(E) + P(F).

If E and F are not disjoint events, then

P(E or F) = P(E) + P(F) − P(E and F).

4. If E represents any event and EC represents the complement of E, then P(EC) = 1 − P(E).




5. If E and F are independent events, then


5.4 Conditional Probability and the General Multiplication RuleLearning Objectives

1. Compute conditional probabilities

2. Compute probabilities using the General Multiplication Rule


5.4 Conditional Probability and the General Multiplication Rule5.4.1 Compute Conditional Probabilities (1 of 8)

Conditional Probability

The notation P(F|E) is read “the probability of event F given event E”. It is the probability that the event F occurs given that event E has occurred.



EXAMPLE 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?



The probability of event F occurring, given the occurrence of event E, is found by dividing the probability of E and F by the probability of E, or by dividing the number of outcomes in E and F by the number of outcomes in E.



EXAMPLE 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 on the next slide.



blank 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?







East 204 36 15

Midwest 212 29 13

South 219 26 9

West 152 76 26



EXAMPLE Murder Victims

In 2005, 19.1% of all murder victims were between the ages of 20 and 24 years old. Also in 2005, 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?


5.4 Conditional Probability and the General Multiplication Rule5.4.2 Compute Probabilities Using the General Multiplication Rule (1 of 2)

In words, the probability of E and F is the probability of event Eoccurring times the probability of event F occurring, given the occurrence of event E.


5.4 Conditional Probability and the General Multiplication Rule5.4.2 Compute Probabilities Using the General Multiplication Rule (2 of 2)

EXAMPLE Murder Victims

In 2005, 19.1% of all murder victims were between the ages of 20 and 24 years old. Also in 2005, 86.9% of murder victims were male given that the victim was 20 − 24 years old. What is the probability that a randomly selected murder victim in 2005 was a 20 − 24 year old male?


5.1 Counting TechniquesLearning Objectives

1. Solve counting problems using the Multiplication rule

2. Solve counting problems using permutations

3. Solve counting problems using combinations

4. Solve counting problems involving permutations with nondistinct items

5. Compute probabilities involving permutations and combinations


5.5 Counting Techniques5.5.1 Solve Counting Problems Using the Multiplication Rule (1 of 4)

Multiplication Rule of Counting

If a task consists of a sequence of choices in which there are pselections for the first choice, q selections for the second choice, rselections for the third choice, and so on, then the task of making these selections can be done in



EXAMPLE Counting the Number of Possible Meals

The fixed-price dinner at Mabenka Restaurant provides the following choices:

Appetizer: soup or salad

Entrée: baked chicken, broiled beef patty, baby beef liver, or roast beef au jus

Dessert: ice cream or cheesecake

How many different meals can be ordered?



EXAMPLE Counting the Number of Possible Meals

Ordering such a meal requires three separate decisions. Choose an appetizer. For each choice of appetizer, we have 4 choices of entrée, and for each of these 2 • 4 = 8 parings, there are 2 choices for dessert. A total of

2 • 4 • 2 = 16

different meals can be ordered.



If n ≥ 0 is an integer, the factorial symbol, n!, is defined as follows:

n! = n(n − 1) • • • • • 3 • 2 • 1

0! = 1

1! = 1


5.5 Counting Techniques5.5.2 Solve Counting Problems using Permutations (1 of 3)

A permutation is an ordered arrangement in which r objects are chosen from n distinct (different) objects so that r ≤ n and repetition is not allowed. The symbol nPr represents the number of permutations of r objects selected from n objects.



Number of Permutations of n Distinct Objects Taken r at a Time

The number of arrangements of r objects chosen from n objects, in which

1. the n objects are distinct,

2. repetition of objects is not allowed, and

3. order is important, is given by the formula



EXAMPLE Betting on the Trifecta

In how many ways can horses in a 10-horse race finish first, second, and third?

The 10 horses are distinct. Once a horse crosses the finish line, that horse will not cross the finish line again, and, in a race, order is important. We have a permutation of 10 objects taken 3 at a time.

The top three horses can finish a 10-horse race in


5.5 Counting Techniques5.5.3 Solve Counting Problems Using Combinations (1 of 3)

A combination is a collection, without regard to order, in which r objects are chosen from n distinct objects with r ≤ nwithout repetition. The symbol nCr represents the number of combinations of n distinct objects taken r at a time.



Number of Combinations of n Distinct Objects Taken r at a Time

The number of different arrangements of r objects chosen from nobjects, in which

1. the n objects are distinct,

2. repetition of objects is not allowed, and

3. order is not important, is given by the formula



EXAMPLE Simple Random Samples

How many different simple random samples of size 4 can be obtained from a population whose size is 20?

The 20 individuals in the population are distinct. In addition, the order in which individuals are selected is unimportant. Thus, the number of simple random samples of size 4 from a population of size 20 is a combination of 20 objects taken 4 at a time.


5.5 Counting Techniques5.5.4 Solve Counting Problems Involving Permutations with Nondistinct Items (1 of 2)

Permutations with Nondistinct Items

The number of permutations of n objects of which n1 are of one kind, n2 are of a second kind, . . . , and nk are of a kth kind is given by


5.5 Counting Techniques5.5.4 Solve Counting Problems Involving Permutations with Nondistinct Items (2 of 2)

EXAMPLE Arranging Flags

How many different vertical arrangements are there of 10 flags if 5 are white, 3 are blue, and 2 are red?

We seek the number of permutations of 10 objects, of which 5 are of one kind (white), 3 are of a second kind (blue), and 2 are of a third kind (red). Using Formula (3), we find that there are


5.5 Counting Techniques5.5.5 Compute Probabilities Involving Permutations and Combinations (1 of 3)

EXAMPLE Winning the Lottery

In the Illinois Lottery, an urn contains balls numbered 1 to 52. From this urn, six balls are randomly chosen without replacement. For a $1 bet, a player chooses two sets of six numbers. To win, all six numbers must match those chosen from the urn. The order in which the balls are selected does not matter. What is the probability of winning the lottery?




The probability of winning is given by the number of ways a ticket could win divided by the size of the sample space. Each ticket has two sets of six numbers, so there are two chances of winning for each ticket. The sample space S is the number of ways that 6 objects can be selected from 52 objects without replacement and without regard to order, so N(S) = 52C6.


5.6 Putting It Together: Which Method Do I Use?Learning Objectives

1. Determine the appropriate probability rule to use

2. Determine the appropriate counting technique to use


5.6 Putting It Together: Which Method Do I Use?5.6.1 Determine the Appropriate Probability Rule to Use (1 of 8)



EXAMPLE Probability: Which Rule Do I Use?

In the game show Deal or No Deal?, a contestant is presented with 26 suitcases that contain amounts ranging from $0.01 to $1,000,000. The contestant must pick an initial case that is set aside as the game progresses. The amounts are randomly distributed among the suitcases prior to the game. Consider the following breakdown:

Prize Number of Suitcases

$0.01−$100 8

$200−$1000 6

$5,000−$50,000 5

$100,000−$1,000,000 7




The probability of this event is not compound. Decide among the empirical, classical, or subjective approaches. Each prize amount is randomly assigned to one of the 26 suitcases, so the outcomes are equally likely. From the table we see that 7 of the cases contain at least $100,000. Letting E = “worth at least $100,000,”we compute P(E) using the classical approach.

Prize Number of Suitcases

$0.01−$100 8

$200−$1000 6

$5,000−$50,000 5

$100,000−$1,000,000 7



The chance the contestant selects a suitcase worth at least $100,000 is 26.9%. In 100 different games, we would expect about 27 games to result in a contestant choosing a suitcase worth at least $100,000.




According to a Harris poll in January 2008, 14% of adult Americans have one or more tattoos, 50% have pierced ears, and 65% of those with one or more tattoos also have pierced ears. What is the probability that a randomly selected adult American has one or more tattoos and pierced ears?




The probability of a compound event involving ‘AND’. Letting E = “one or more tattoos” and F = “ears pierced,” we are asked to find P(E and F). The problem statement tells us that P(E) = 0.14, P(F) = 0.50 and P(F|E) = 0.65. Because P(F) ≠ P(F|E), the two events are not independent. We can find P(E and F) using the General Multiplication Rule.


5.6 Putting It Together: Which Method Do I Use?5.6.2 Determine the Appropriate Counting Technique to Use (1 of 5)



EXAMPLE Counting: Which Technique Do I Use?

The Hazelwood city council consists of 5 men and 4 women. How many different subcommittees can be formed that consist of 3 men and 2 women?

Sequence of events to consider: select the men, then select the women. Since the number of choices at each stage is independent of previous choices, we use the Multiplication Rule of Counting to obtain

N(subcommittees) = N(ways to pick 3 men) • N(ways to pick 2 women)




To select the men, we must consider the number of arrangements of 5 men taken 3 at a time. Since the order of selection does not matter, we use the combination formula.




On February 17, 2008, the Daytona International Speedway hosted the 50th running of the Daytona 500. Touted by many to be the most anticipated event in racing history, the race carried a record purse of almost $18.7 million. With 43 drivers in the race, in how many different ways could the top four finishers (1st, 2nd, 3rd, and 4th place) occur?




The number of choices at each stage is independent of previous choices, so we can use the Multiplication Rule of Counting. The number of ways the top four finishers can occur is

N(top four) = 43 • 42 • 41 • 40 = 2,961,840

We could also approach this problem as an arrangement of units. Since each race position is distinguishable, order matters in the arrangements. We are arranging the 43 drivers taken 4 at a time, so we are only considering a subset of r = 4 distinct drivers in each arrangement. Using our permutation formula, we get


5.7 Bayes’s Rule (on CD)Objectives

1. Use the Rule of Total Probabilities

2. Use Bayes’s Rule to compute probabilities


5.7 Bayes’s Rule (on CD)5.7.1 Introduction (1 of 2)

Addition Rule for Disjoint Events

If E and F are disjoint (mutually exclusive) events, then


5.7 Bayes’s Rule (on CD)5.7.1 Introduction (2 of 2)

General Multiplication Rule

The probability that two events E and F both occur is


5.7 Bayes’s Rule (on CD)5.7.2 Use the Rule of Total Probabilities (1 of 18)

EXAMPLE Introduction to the Rule of Total Probability

At a university

55% of the students are female and 45% are male

15% of the female students are business majors

20% 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 female

‒ 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%, or0.45 • 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

Branch 1 − 55% are Female

Branch 2 − 45% are Male




Branch 3 – 15% of Female is business.

Branch 4 – 85% of Female is not business.

Branch 5 – 20% of Male is business.

Branch 6 – 80% of Male is not business.




Multiply out the branches.




Add the two business branches.



• 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 two non-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 and E ∩ 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 ndifferent 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



Rule of Total Probability

Let E be an event that is a subset of a sample space S. Let A1, A2, A3, …, An be partitions of a sample space S. Then,



EXAMPLE The Rule of Total Probability

• 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 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


5.7 Bayes’s Rule (on CD)5.7.3 Use Bayes’s Rule to Compute Probabilities (1 of 11)

• 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 students are 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 majors are 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)



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



Bayes’s Rule

Let A1, A2, A3, …, An be a partition of a sample space S. Then, for any event E for which P(E) > 0, the probability of event Ai for i = 1, 2, …, n given the event E, is



EXAMPLE Bayes’s Rule

The business majors example from before




• 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)




• 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 is

8.25% / 17.25% = 47.83%




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

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

statisticssite.iugaza.edu.ps/mriffi/files/2018/02/ch05.pdf · 2018. 2. 6. · 2 of 2)

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