Chapter

Copyright © 2014, 2013, 2010 and 2007 Pearson Education, Inc.

Chapter

Probability

5


Section

Probability Rules

5.1

Copyright © 2013, 2010 and 2007 Pearson Education, Inc.

Objectives1. 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

5-3


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.

5-4


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.

5-5


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.

5-6


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. 5-7

Copyright © 2013, 2010 and 2007 Pearson Education, Inc.5-8

Experiment Example of event Sample Space

Single Birth 1 boy (single event) {g,b}3 births 2 G and 1 B {bbb,bbg,bgb,gbb, not simple event ggg,ggb,gbg,bgg}

2 G and 1 B is not a simple event because it can be broken down into simpler events like ggb, gbg or bgg (simple events)


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)}

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Objective 1

• Apply Rules of Probabilities

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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 spaceS = {e1, e2, …, en}, thenP(e1) + P(e2) + … + P(en) = 1

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 spaceS = {e1, e2, …, en}, thenP(e1) + P(e2) + … + P(en) = 1

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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.

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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.

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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.

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Objective 2

• Compute and Interpret Probabilities Using the Empirical Method

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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.P(E) ≈ relative frequency of E

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.P(E) ≈ relative frequency of E

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frequency of E

number of trials of experiment


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

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

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EXAMPLE Building a Probability Model EXAMPLE Building a Probability Model

(c) Would it be unusual to throw a “Leaning Jowler”?

5-18

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”.


(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

1344

39390.341

5-19



Side with no dot 0.341

Side with dot 0.329

Razorback 0.195

Trotter 0.093

Snouter 0.035


(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”.

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Side with no dot 0.341

Side with dot 0.329

Razorback 0.195

Trotter 0.093

Snouter 0.035


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

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Objective 3

• Compute and Interpret Probabilities Using the Classical Method

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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.

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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


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

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P E number of ways E can occur

number of possible outcomes

m

n



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

where N(E) is the number of outcomes in E, and N(S) is the number of outcomes in the sample space.


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

where N(E) is the number of outcomes in E, and N(S) is the number of outcomes in the sample space.

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P E N E N S


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.

5-26


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

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

(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.

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P(yellow) N (yellow)

N (S)

6

300.2


Objective 4

• Use Simulation to Obtain Data Based on Probabilities

5-28


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.

EXAMPLE Using SimulationEXAMPLE Using Simulation

5-29


Objective 5

• Recognize and Interpret Subjective Probabilities

5-30


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.

5-31


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.

5-32


Section

The Addition Rule and Complements

5.2


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

5-34


Objective 1

• Use the Addition Rule for Disjoint Events

5-35


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

5-36


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.

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Addition Rule for Disjoint Events

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

Addition Rule for Disjoint Events

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

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P E or F P E P F


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

5-40

P E or F or G or ... P E P F P G L


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

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

(a) Verify that this is a probability model.

All probabilities arebetween 0 and 1, inclusive.

0.010 + 0.032 + … + 0.080 = 1

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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


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(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


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(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


Objective 2

• Use the General Addition Rule

5-44


The General Addition Rule

For any two events E and F,

The General Addition Rule

For any two events E and F,

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P E or F P E P F P E and F


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

5-46


P(E) N (E)

N (S)

6

36

1

6

P(F) N (F)

N (S)

10

36

5

18

P(E and F

N (E and F)

N (S)

3

36

1

12

P(E or F) P(E) P(F) P(E and F)

6

36

10

36

3

36

13

36


Objective 3

• Compute the Probability of an Event Using the Complement Rule

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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.

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Complement Rule

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

P(EC) = 1 – P(E)

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Entire region

The area outside the circle represents Ec


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

5-51


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

5-52


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

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

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4895

393,1860.012


(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

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Section

Independence and the Multiplication Rule

5.3


Objectives

1. Identify independent events

2. Use the Multiplication Rule for Independent Events

3. Compute at-least probabilities

5-56


Objective 1

1. Identify Independent Events

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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.

5-58


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.

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Objective 2

• Use the Multiplication Rule for Independent Events

5-60


Multiplication Rule for Independent Events

If E and F are independent events, then

Multiplication Rule for Independent Events

If E and F are independent events, then

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P E and F P E P F


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 thatP(survive) = 0.99186.

5-62

P First survives and second survives P First survives P Second survives (0.99186)(0.99186)

0.9838


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,

5-63

P defective and used P defective P used (0.10)(0.30)

0.03


Multiplication Rule for n Independent Events

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

Multiplication Rule for n Independent Events

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

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P E1 and E2 and E3 and ... and En P E1 P E2 P En


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 EventsEXAMPLE Illustrating the Multiplication Principle for Independent Events

5-65


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

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

5-66


Objective 3

• Compute At-Least Probabilities

5-67


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

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

5-68


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


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

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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).


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).

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


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

5-71



5. If E and F are independent events, then


5. If E and F are independent events, then

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P E and F P E P F


Section

Conditional Probability and the

General Multiplication Rule

5.4


Objectives

1. Compute conditional probabilities

2. Compute probabilities using the General Multiplication Rule

5-74


Objective 1

• Compute Conditional Probabilities

5-75


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.

5-76


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}

Second roll: S = {2, 4, 6}

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P(S) 1

6

P(S) 1

3


Conditional Probability Rule

If E and F are any two events, then

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.

Conditional Probability Rule

If E and F are any two events, then

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.

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P F E P E and F P E

N E and F N E


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 on the next slide.

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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

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(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



East 204 36 15

Midwest 212 29 13

South 219 26 9

West 152 76 26

5-81

204

204 36 150.8

P believes in God lives in the east

N believe in God and live in the east N live in the east



Believe in God



East 204 36 15

Midwest 212 29 13

South 219 26 9

West 152 76 26

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204

204 212 219 1520.26

P lives in the east believes in God

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


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?

EXAMPLE Murder VictimsEXAMPLE Murder Victims

5-83

P male 20 24 P male and 20 24 P 20 24

0.166

0.1910.869109 0.869 86.9%


Objective 2

• Compute Probabilities Using the General Multiplication Rule

5-84



The probability that two events E and F both occur (E and F are dependent events) is

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


The probability that two events E and F both occur (E and F are dependent events) is

In words, the probability of E and F is the probability of event E occurring times the probability of event F occurring, given the occurrence of event E.5-85

P E and F P E P F E


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?

EXAMPLE Murder VictimsEXAMPLE Murder Victims

5-86

P male and 20 24 P 20 24 P male 20 24

0.8690.191 0.165979 0.166


Section

Counting Techniques

5.5


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

5-88


Objective 1

• Solve Counting Problems Using the Multiplication Rule

5-89


Multiplication Rule of Counting

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

different ways.

Multiplication Rule of Counting

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

different ways.

5-90

pqr


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.

EXAMPLE Counting the Number of Possible MealsEXAMPLE Counting the Number of Possible Meals

5-91


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

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Objective 2

• Solve Counting Problems Using Permutations

5-93


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

5-94


Number of permutations of n Distinct Objects Taken r at a Time The number of arrangements of r objects chosen from n objects, in which1. the n objects are distinct,2. repetition of objects is not allowed, and3. order is important, is given by the formula

Number of permutations of n Distinct Objects Taken r at a Time The number of arrangements of r objects chosen from n objects, in which1. the n objects are distinct,2. repetition of objects is not allowed, and3. order is important, is given by the formula

5-95

n Pr n!

n r !


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

EXAMPLE Betting on the TrifectaEXAMPLE Betting on the Trifecta

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

5-96

10 P3 10!

10 3 ! 10!

7!

10987!

7!1098 720 ways


Objective 3

• Solve Counting Problems Using Combinations

5-97


A combination is a collection, without regard to order, of n distinct objects without repetition. The symbol nCr represents the number of combinations of n distinct objects taken r at a time.

5-98


Number of Combinations of n Distinct Objects Taken r at a Time The number of different arrangements of r objects chosen from n objects, in which1. the n objects are distinct,2. repetition of objects is not allowed, and3. order is not important, is given by the formula

Number of Combinations of n Distinct Objects Taken r at a Time The number of different arrangements of r objects chosen from n objects, in which1. the n objects are distinct,2. repetition of objects is not allowed, and3. order is not important, is given by the formula

5-99

n Cr n!

r! n r !


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

EXAMPLE Simple Random SamplesEXAMPLE Simple Random Samples

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.Use Formula (2) with n = 20 and r = 4:

20 C4 20!

4! 20 4 ! 20!

4!16!

2019181716!

4 32116!

116,280

244,845

There are 4,845 different simple random samples of size 4 from a population whose size is 20.

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Objective 4

• Solve Counting Problems Involving Permutations with Nondistinct Items

5-101


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

where n = n1 + n2 + … + nk.

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

where n = n1 + n2 + … + nk.

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n!

n1!n2 !L nk !


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

EXAMPLE Arranging FlagsEXAMPLE Arranging Flags

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

differentverticalarrangements

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10!

5!3!2!

1098765!

5!3!2!2,520


Objective 5

• Compute Probabilities Involving Permutations and Combinations

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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?

EXAMPLE Winning the LotteryEXAMPLE Winning the Lottery

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EXAMPLE Winning the LotteryEXAMPLE 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.

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The size of the sample space isEXAMPLE Winning the LotteryEXAMPLE Winning the Lottery

Each ticket has two sets of 6 numbers, so a player has two chances of winning for each $1. If E is the event “winning ticket,” then

There is about a 1 in 10,000,000 chance of winning the Illinois Lottery!

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P E 2

20,358,5200.000000098

N S 52 C6 52!

6! 52 6 !

52515049484746!

6!46!20,358,520


Section

Putting It Together: Which Method Do I Use?

5.6


Objectives

1. Determine the appropriate probability rule to use

2. Determine the appropriate counting technique to use

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Objective 1

• Determine the Appropriate Probability Rule to Use

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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:

EXAMPLE Probability: Which Rule Do I Use?EXAMPLE Probability: Which Rule Do I Use?

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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.


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P E N E N S

7

260.269


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.


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P E N E N S

7

260.269


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?


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P one or more tatoos and pierced ears P E P F | E 0.14 0.65 0.091


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(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.

P E and F P E P F | E 0.14 0.65 0.091

So, the chance of selecting an adult American at random who has one or more tattoos and pierced ears is 9.1%.


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Objective 2

• Determine the appropriate counting technique to use

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The Hazelwood city council consists of 5 men and 3 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)

EXAMPLE Counting: Which Technique Do I Use?EXAMPLE Counting: Which Technique Do I Use?

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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.

N(subcommittees) = 10 • 3 = 30. There are 30 possible subcommittees that contain 3 men and 2 women.

To select the women, we must consider the number of arrangements of 3 women taken 2 at a time. Since the order of selection does not matter, we use the combination formula again.

N ways to pick 3 women 3C2 3!

2!1!3


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N ways to pick 3 men 5 C3 5!

3!2!10


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?


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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 = 2,961,840We 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

N top four 43 P4 43!

43 4 ! 43424140 2,961,840


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