Finite Mathematics 0395426936, 9780395426937

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Cleveland Heights. Ohio.44113

CLEVELAND HEIGHTS-UNIVERSITY HEIGHTS CITY SCHOOL DISTRICT

FREE TEXTBOOK RULES Rule l.-This book is the property of the Board of Education and shall not be sold to pupils or anyone else. Rule 2.-This book must not be defaced, marked with pen or pencil, or leaves tom, and it must be kept clean. Rule 3.-If this book is lost or damaged beyond repair the pupil shall pay for it as follows: 100% of cost (a) during the first three years after the date of adoption by the Board of Education (not the date of publication), regardless of the condition of the book when issued; and (b) if, when issued, the book is new or in excellent condition, regardless of the date of adoption. 50% of cost if the book is issued in usable condition more than three years after the date of adoption. Rule 4.-If this book is damaged but repairable the fine shall be fixed at 25 % of the cost of the book when new. Rule 5.-The teacher shall assess all fines. Fines are to be paid in the Main Office.

RECORD OF ISSUE YEAR

BOOK

COND,

PUPIL’S NAME

TEACHER’S NAME

TCM CUKOppCK

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Digitized by the Internet Archive in 2018 with funding from Kahle/Austin Foundation

https://archive.0rg/details/finitemathematicOOOOdelg

John J. Del Grande W. Bisset Edward Barbeau Jr.

Houghton Mifflin Canada Limited 150 Steelcase Road West • Markham, Ontario • L3R 3J9

HOUGHTON MIFFLIN ADVANCED MATHEMATICS

Finite Mathematics John J. Del Grande, Mathematics Consultant, Houghton Mifflin Canada Limited W. Bisset, Head of Mathematics, Earl Haig Secondary School, North York, Ontario Edward Barbeau Jr., Associate Professor of Mathematics, University of Toronto

Reviewers David Andrews, Professor, Dept, of Statistics, University of Toronto Bob McRoberts, Unionville High School, Unionville, Ontario

Canadian Cataloguing in Publication Data Del Grande, J. J., 1922Finite mathematics For use in high schools. Includes index. ISBN 0-395-42693-6 I. Mathematics — 1961- . I. Bisset, William. II. Barbeau, Edward, 1938- . III. Title. QA39.2.D44

1987

512.9

C87-093558-5

All rights reserved. No part of this work may be reproduced or transmitted in any form or by any means, electronic or mechanical, including photocopying and recording, or by any information storage or retrieval system, without permission in writing from the publisher.

Copyright © 1988 by Houghton Mifflin Canada Limited Printed in Canada

345/JD/9210908

CONTENTS

1

Matrices 1-1 1-2 1-3 1-4 1- 5

2

3

4

Mathematical Models What is a Matrix? Addition and Subtraction of Matrices Matrix Multiplication Matrices and Computer Graphics Review Exercises

1 2 6 11 16 24 26

Systems of Equations

31

2- 1 2-2 2-3 2-4 2-5 2-6 2- 7

32 36 41 48 54 59

Simultaneous Events Systems of Equations in Two Variables Solving Systems of m Equations in n Variables Solving Equations Using Matrices Applications Special Matrices Gauss-Jordan Method for the Solution of Systems of Equations Leontief Input-Output Model of an Economy Review Exercises

64 66 69

Linear Programming

73

3- 1 3-2 3-3 3-4 3- 5

74 78 84 89 92 97

Inequalities in One Variable Inequalities in Two Variables Linear Programming Models Simultaneous Inequalities Solving Linear Programming Models Review Exercises

Series

103

4- 1 4-2 4-3 4-4 4-5 4-6

104 110 118 124 130 132 136

Finite Series and Summation Notation Arithmetic Series Geometric Series Infinite Geometric Series Arithmetico-Geometric Series Summing by Differences Review Exercises

111

5

Sets

139

5-1 5-2 5-3 5-4 5-5

140 146 151 157 162 165

Sets: The Basic Concepts Set Diagrams Operations on Sets Counting Elements in Sets The Pigeonhole Principle Review Exercises

6 Combinatorial Analysis 6-1 6-2 6-3 6-4 6-5 6-6 6-7 6-8 6-9

Organized Counting Permutations and Combinations Permutations of Distinct Elements I Permutations of Distinct Elements II Permutations with Repetition Combinations of Distinguishable Elements I Combinations of Distinguishable Elements II Combinations with Some Elements Indistinguishable Additional Problems in Combinations and Permutations Review Exercises

7 The Binomial Theorem 7-1 7-2 7-3 7-4 7-5 7-6

Properties of Pascal’s Triangle The Binomial Theorem for Positive Integer Exponents The General Term of a Binomial Expansion The Binomial Theorem for General Exponents Series Involving Binomial Exponents Proof by Induction Review Exercises

8 Probability 8-1 8-2 8-3 8-4 8-5 8-6

iv

Experimental Probability Theoretical Probability Combinations of Simple Events: “Or” Events Combinations of Simple Events: “And” Events Conditional Probability Genetics and Probability Review Exercises

167 168 174 177 182 186 190 196 199 205 211

215 216 220 225 229 234 238 244

247 248 254 258 265 274 281 284

9

Applications of Probability 9-1 9-2 9-3 9-4 9-5 9-6

Probability Distribution The Binomial Distribution Mathematical Expectation Repeated Trials Without Replacement Markov Chains I Markov Chains II Review Exercises

10 Statistics 10-1 10-2 10-3 10-4 10-5

Representing Data Central Tendency Variability Statistics and Probability The Normal Distribution Review Exercises

289 290 297 306 315 321 326 331

335 336 344 351 362 371 378

Glossary

382

Tables

387

Problems

393

Answers

402

Index

455

V

ACKNOWLEDGEMENTS Editors Frank Eastman David Hamilton Ruth Peckover Richard A. Rader Assembly and Technical Art Dave Hunter Illustrations Tom Sankey Cover Design William Ross The Dragon’s Eye Press Cover Image ©Steven Hunt/The Image Bank Canada Photographs Canadian Tire Corporation, 2 University of Toronto Archives, 3 Chapman Photography, 13, 209, 260 Ontario Ministry of Agriculture and Food, 31 Greenpeace Foundation, 31 Claus Anderson, 35 Ontario Ministry of Transportation and Communications, 54 Bank of Nova Scotia, 103 Smithsonian Institution, 167 Canapress Photo Service, 173, 332, 335, 337, 353 Satellite Data Archives, Environment Canada, 247 Canadian Totalisator Company, 264 General Motors Canada, 289 Econolite Canada Inc., 296 Special thanks to the Metropolitan Toronto Central Library for other photographs.

Photo Research Jill Patrick

vi

Florence Nightingale

The discovery of matrices can be attributed to Arthur Cayley (1821-1895) and James Sylvester (1814-1897), two Englishmen who, as good friends, worked together enthusiastically on many mathematical problems. In his early years, Cayley could not find employment as a mathematician. He elected to study law and became a prominent London lawyer. His interest in mathematics continued, and he wrote many mathematical papers, which eventually led to a position, eagerly accepted, as a professor of mathematics at Cambridge University. Sylvester worked as an insurance actuary and gave evening classes in advanced mathematics. He broke a tradition of his times by allowing a woman to take his course. This person was Florence Nightingale, “the lady with the lamp”, who estab¬ lished battlefield hospitals during the Crimean War. Until about 1920, matrices were of interest only to algebraists. At that time, it was recognized that matrices could be useful tools in the study of other branches of mathematics and science. Since then, applications have been found in such other fields as astronomy, mechanics, electric circuit theory, nuclear physics, and aero¬ dynamics. In the late 1940s, applications for matrices were found in business and economics, when it was shown that a process called linear programming could help to predict how best to minimize costs and maximize profits. The matrix proves to be a powerful concept that can be applied to mathematical modelling.

1

1-1 Mathematical Models Most real-world situations cannot be described in precise mathematical terms, but often can be approximated by a mathematical model. For example, a circle can be used as a model for the wheel of a car. Since the circle is a precise mathematical figure, it does not include any of the imperfections a wheel might have.

The circle is a good model, however, if it predicts the behaviour of the wheel. If we are interested in how far the wheel will roll in 10 revolutions, then the movement of the circle can be studied mathematically. If the circle predicts a distance that is close to what the wheel actually travels, then the circle can be considered to be a good model of the wheel. Sometimes a mathematical model needs to be revised as more information is gath¬ ered about the real-world situation. A revised, or refined, model usually can be expected to yield a more exact prediction.

Sir Isaac Newton’s mathematical model of a free-falling object is usually expressed in the form of an equation, as follows: d = A.9 f

where d is in metres and t is in seconds.

This equation is a good predictor of the distance an object will fall, starting from rest, in t seconds. This model can be used to solve the following problem.

2

EXAMPLE 1

How long will it take a stone dropped from the top of the 400 m high CN Tower to reach the ground?

SOLUTION

Use d = 4.9 where d = 400. 400 = 4.9 400 4.9 +9 s t ^ ±9 =

4000 49 Reject the negative value.

From this, it will take approximately 9 s for a stone to drop 400 m. Another example of a mathematical model is a graph. EXAMPLE 2

Jae Chung’s monthly phone bill for a computer network includes dt basic monthly charge of $9.00 plus $0.20/min for line charges. The monthly bill is a function of the time Jae uses the network and is given by the equation, C = 900 + 20r where C is the monthly charge, in cents, and t is the time, in minutes, the network is used. Draw a graph to represent this function.

SOLUTION

TIME

Minutes

Here, both the graph and the equation are mathematical models of Jae’s monthly network charges. Is the graph a good model? Will it predict Jae’s monthly charges for 70 min on the network in one month? John Synge, an applied mathematician at the University of Toronto, described modelling in the American Mathematical Monthly (Vol. 68, p. 799) as follows: Modelling is a process consisting of three stages: 1. a dive from the world of reality into the world of mathematics, 2. a swim in the world of mathematics, and 3. a climb from the world of mathematics back into the world of reality, carrying a prediction in our teeth.

3

EXERCISE 1-1 A 1. Several possible models of a person are shown below. Study the following situations. Decide which model(s) you could use to describe each situation.

oo

a. A mechanical engineer wishes to estimate how many people can stand in an

elevator with a floor 2 m by 2.5 m.

b. A marine biologist wants to estimate the total skin area of an average person to estimate heat loss of a person when in cold water.

c. A university student must estimate how many of his classmates could be stuffed into a phone booth.

d. An artist wishes to estimate how much gold leaf is required to cover a statue. e. A contractor needs to estimate the surface area of the Statue of Liberty for a cleaning contract.

2. A road map is a graphical model. Given a road map of Alberta, suggest what data might be predicted well and what data might be predicted poorly, or not at all, from the map. 3. Suggest reasons why a small scale model of an aircraft might be tested in a wind tunnel.

B 4. By observing many violent storms such as tornadoes, thunderstorms, hurricanes, and cyclones, meteorologists have found that the same mathematical model, relating the size and the duration of a storm, can be used to make predictions about such storms. = 885 7^

where T is the time in hours and D is the storm’s diameter in kilometres.

a. How long will a cyclone last that is 10 km in diameter?

b. What is the diameter of a thunderstorm that lasts 2 h? c. If a hurricane has a diameter of 200 km, how many days will it last? d. A tornado lasts 12 minutes. What is its diameter? 4

5. One model of a human body is a cylinder whose height is 7 times its diameter. A person 182 cm tall can be approximated by a cylinder 182 cm high and 26 cm in diameter. The volume of the body can then be approximated by irr^h or 96 629 cm^. The human body is about 60% water and has about the same mass as an equal volume of water. Water has a density of 1 g/cm^. If this model is used, a person 182 cm tall would have a mass of about 96.6 kg.

a. Calculate the masses of cylinders of water that are 60, 80, 100, 120, and 140 cm high, using the model described above. b. Plot the results on a graph showing height versus mass. c. Find the height and mass of several classmates and check how well the graph predicts the results. Is this graph a good model for students in your class? 6. A communication authority has the reponsibility of issuing radio broadcast licences. There are four radio frequencies available but 6 cities have applied for licences. A student working at the authority notes that although cities within 180 km of each other must have different frequencies to avoid interference, most cities were a greater distance apart than 180 km. She made a mathematical model of the prob¬ lem in an attempt to resolve the issue. She called the cities A, B, C, D, E, and F and made a chart showing the distances in kilometres between the cities.

A B C D E F

A

B

C

D

E

F

0 250 240 195 170 190

250 0 190 260 200 215

240 190 0 170 175 150

195 260 170 0 160 140

170 200 175 160 0 130

190 215 150 140 130 0

ABC

a. Draw six points to represent 6 cities. b. Join the cities that are within 180 km of each other. c. Can the student assign radio frequencies to the cities so as to avoid interference? If so, how does she do it? 7. Data on the number of yeast cells grown in an experiment are given in the table at the right.

a. Draw a graph of the number of cells versus time in hours. How many cells are there in 11 h? b. The curve rises steeply at first but levels off as the food sup¬ ply is exhausted. Estimate the number of cells in the culture after 30 h. c. Your estimate for (b), in practice, may not be very good as the culture becomes poisoned, and it often happens that the number actually decreases. During what time interval is the rate of growth a maximum?

•

•

•

* D

E

F

Time (h)

Number of cells

0 2 6 8 10 12 14 16 18

6 10 117 234 342 397 428 438 442

5

1-2 What is a Matrix? In the late 1940s, George Dantzig, a mathematician working for the U.S. Air Force, developed the idea of linear programming. It was then discovered that this concept had many business applications. The problems from business often involved many equations with many different variables, and matrices provided an ideal mathemati¬ cal model for them. At about this same time, the computer was being developed, and it was found to be an excellent tool for solving complicated linear programming and matrix problems. A matrix is a rectangular array of numbers such as those found in data tables. For example, cars are often rated in motor magazines for performance as is shown in the following table. uis’/'r

Pontiac Fiero GT Acceleration time 0-100 km/h braking 100-0 (s)

1 RX7

Toyota MK2

.Camaro

7 Ford IROC 7j. Mustang GT

7-3

9.2

'

Stopping distance (m)

49

49

Interior noise (dB)

68

71-

Handling test (km/h)

104

111

112

’iv? - 7:0 " 53

r.

101

74

100'"'-'

In mathematics, a matrix is an array of numbers enclosed by brackets. The size of the matrix is identified by its order, that is, the number of rows and columns in the array. For example:

[1

3

5

5] is a 1 X 4 or one-by-four matrix. This is a row matrix with 1 row and 4 columns.

1 7

5 9

6

8 2] 4

is called a 3 x 2 matrix (3 rows and 2 columns). Its order is 3 x 2.

is a 4 X 1 matrix. This is a column matrix with 4 rows and 1 column.

9 -7

The following terms are frequently used when working with matrices. • A square matrix has an equal number of columns and rows. • The entries or components of a matrix are the numbers in the matrix. • A zero matrix has zero in all entries. Two matrices are equal if they have the same order and if all corresponding entries are equal.

6

EXAMPLE 1

Solve for x and 3^. ' 3 -

7 _

' 3 -l" _ 0 7_

SOLUTION

By inspection x — — 1 and y = 0. Since the matrices are equal the corresponding entries must also be equal.

EXAMPLE 2

Inventory Control A school secretary makes a count of grade

11

and grade 12 books in the stockroom and produces the following table which can be represented by a matrix. Math

Phys

11

15

19

Grade 12

2

0

Grade

Chem

Hist

Eng

0

8

22

13

7

26

a. How many Grade 11 math books are in stock? b. How many Grade 12 books are in stock? c. How many Grade 12 physics books are in stock? SOLUTION

a. From row 1, column 1, we find 15 Grade 11 math books are in stock. b. The sum of the entries in row 2 gives 48 Grade 12 books in stock. c. There are no physics books for Grade 12 in stock.

EXAMPLE 3

Communication or Network Theory Use a network diagram and a matrix to represent the following information.

1. Canada has diplomatic relations with the U.S.A., Cuba, and Mexico. 2. Mexico has diplomatic relations with the U.S.A., and Canada, but not Cuba. 3. Cuba has diplomatic relations with Canada but not Mexico or the U.S.A. 4. The U.S. A. has diplomatic relations with Canada and Mexico, but not Cuba.

SOLUTION

A network diagram shows the countries as points. Diplomatic rela¬ tions are indicated by a line segment joining the appropriate dots.

7

If we use a zero to indicate no diplomatic relations as entries and a to indicate diplomatic relations, the following matrix can be constructed.

1

Canada

U.S.A.

Mexico

0 1 1 1

1 0 1 0

1 1 0 0

Canada U.S.A. Mexico Cuba

Cuba ^

Assume that countries do not have diplomatic relations with themselves

0

Matrices such as those in Example 3 are called communication matrices or network matrices. In a network problem, the important information is not the location of the points and lines, or their lengths, but simply whether or not two points are joined.

EXERCISE 1-2 A 1. State the order of each of the following matrices.

d. 9

1

.1 "2 1 .2

2 6" 3

b.

1

-

_

7 3 5

8 10“ 5 6 -1 0 1

so _1

a.

'y 9‘

x_

.2

12“ .-10 6_ 9

c. [1

9 4

-10

10]

0 3 5

_

.2

b.

5_

X

3

c. =

4 . 7

-Jd. "-2 3 5 X

4“ 1_

3. Let X be the matrix

■-2 3 . z 7

/ 1_

4-190 3 6 5 -7 4-23

8

a. b. c. d. e.

8

Give the order of X. List the entries in the third row. List the entries in the second column. List the entry in the second row and third column. What is the location of entry - 7?

1

e.

9

X

6

-1“

“l = z 5 _ 1

12

— 5 + -X -f- y + 2z = 3 )

+ z y + 2z -x + y x

69

8. Write an augmented matrix for each of the following systems of equations. c. X = 12 - y a. lx - ly = 12') b. X + y = I )

'}

y = -3^ +

5]

jc + 2y = 4 j d.

3jr - 4y = 0]

-X - y =

e.

>’ = 2x - 7]

X -

- ij

y + 2z=-7"^

3.x - 2y - 7z = 20

f.

2.x -

y + 2z = -2^^

|

3x - 2y +

X - 5y + 2z ^ - I5J

X + 2y +

z = -9 z = 9

i J

9. Solve, using matrices whenever possible. Si.

X + y - 21

b.

X - 3y = S') 2x - 6y = 7j

c.

2x - 5y = 8^ X - 2v = 3J

e.

3x - y -21 X - 9y = 9 1

f.

3x - 4y = 12

jc - y - 2]

d.

jc - 5y = - n 5x -

y = - ij

-X +

4

3>' =

-

10. Find a system of linear equations represented by the following row-reduced echelon matrix. Find a solution for the system. '1 0 0

3 -4 2 5 0-2

8“ -6 4

11. Solve, using row-reduced echelon matrices. a.

d.

g.

y + 2z = -1 ^ b. 3x - 2y - 7z = 20 X - 5y + 2z = - I5J

X -

x+y-t-z = 3^ 2x - 3y 4- z = - 1 I 3x - 4y -f 2z = 2 J

e.

2x - y -h z = 3 ^ X - y - z = - 1 I X - 3y 4- 2z = 7 J

h.

z = 4^ z = 4 I 2z ^ 4j

c.

4x - 3y + 2z = 5^ 2x 4- 3y 4- z = 1 I 6x 4- 6 y - z = 0 J

f.

- 2y - z = 2'1 X + y -H z = 4 I X - 3y -4 2z = 2J

i.

2x 4-

y X 4- 2y X + y

444-

X

2x - y + 2z = -2'^ 3x - 2y 4- z = - 9 I X 4- 2 y + z = 9 J X

-

y - 2z = 3

2x - 3y - 3z = ~ [ X - 2y z lOj -

^

^ _ Q

Find an equation of a circle that passes through the following three points, a. (-4, 1), (3, 0), (5, 4) b. (-1, 1), (0, 2), (3, 1)

13. A plane in a three-dimensional coordinate system can be defined by an equation of the form ax + by + cz = 1. Find an equation of the plane containing the following three points. a. (5, -3, 4), (3, 1, -2), (0, 1, -3) b. (3, - 1, 4), (-2, 3, - 1), (2, 1, 1)

70

-

+ 2y + z = 0 3x - y - 2z = 1 I X - 5y 4- 4z = - 23J X

12. A circle can be defined by an equation of the form x2 -I- y2

=

14. A parabola with its axis parallel to the x-axis is defined by an equation of the form A' = ay~ + by + c where « A 0. Find an equation of a parabola with its axis parallel to the A-axis and passing through the following points. a. (1, -1), (-2, 0), (-1, 3) b. (-2, 2), (-4, 1), (1, -1)

15. A parabola with its axis parallel to the y-axis is defined by an equation of the form y = ax- + bx + c where b 0. Find an equation of the parabola with its axis parallel to the y-axis and passing through the following points. a. (3, -2), (-1, 0), (0, 2) b. (1, 2), (4, 3), (2, 0)

16. In a three-digit number, 100a + lOy + z, the difference between the hundreds and the tens digit is the same as the difference between the tens and the units digit. If the number is divided by the sum of the digits, the quotient is 48. If 198 is subtracted from the number, the remainder has the same digits as the original number but arranged in the opposite order (reversed). Find the number.

17. A grocer makes up three different kinds of fruit baskets that contain apples, oranges, and grapefruit as shown in the following table. basket

apples

oranges

grapefruit

I II III

5 4 3

2 2 3

1 2 2

The grocer has available 68 apples, 35 oranges, and 25 grapefruits. How many of each kind of basket can he make up and use all of the fruit?

18. A fertilizer plant mixes chemical fertilizer to any specifications. A customer wants a fertilizer mixture with 33 tonnes of chemical A and 56 tonnes of che¬ mical B. The plant has a large inventory of two other mixtures with chemical A and chemical B. The manager asks his daughter, who is studying mathematics, if it is possible to use the two mixtures in stock to make up a mixture to the required specifications. The two mixtures are made up as follows. One bag of mixture 1 has 15 kg of chemical A and 25 kg of chemical B. One bag of mixture 2 has 20 kg of chemical A and 35 kg of chemical B. The manager’s daughter used her mathematics to find that it is possible to use the two mixtures in stock to make the new mixture. How many bags of each mixture would have to be used to make up the customer’s order?

19. In a recent election, there were 3 000 000 eligible voters, each of whom favoured one of three parties: Conservatives, Liberals, or Social-Democrats. 1 710 000 eligible voters voted on election day. 50% of the Conservatives voted. 60% of the Liberals voted. 80% of the Social-Democrats voted. The ratio of the number that favoured Liberals to the number that favoured the Social-Democrats was 3:1. Find the number that favoured each party at election time.

20. Show that the following pairs of matrices are inverses. -7“

_1

2.

_ - 1

4_

b.

'l

3“

_o

u

‘3

5‘

'l .0

-3" 1_

00

.0

2_

2 1

21. a. Find the inverse of A =

1_

_

1

L

1

L/1

-3_

by solving the equation

-1

1

H

1

4

7

o

1_

4~

1

1_

“2

1 11

L_

o

Lo ij

1 00

1 d.

7 0]

1

2

1

7“

(N 1_

c.

‘4

1

a.

_0

1_

b. Find the inverse of A using row operations to make the left side of the following matrix into a unit matrix. 1— o 1

_1

o

4 -1

‘2

22. Solve each of the following systems of equations using the inverse of the co¬ efficient matrix. a.

2x + y = 51 4x + 3y = 1)

b.

X -I- 3y = 7 ') 3x + y = 13)

c.

2x + y = -51 -3x + y = 5 )

0

_1

-p^

1 1_

1

o

1

O _0

b.

o

0

o

3

1

0

o

o

1_

o

1

(N 1_

a.

o

23. Show that the following pairs of matrices are inverses.

0-11

-1_

_0

0

2_

4 -11 0 -1 1 Lo 0 ij

24. Use the Gauss-Jordan method to solve the following systems of equations. a.

X + 4y + 6z = 16 ^ 3x + 2y + = 2^ 5x + y + lOz = 39J

b.

i

x - 3y + 3x - 8y + 6x - 5y -iy -

2z 2z 2z 3z

+ w = 1 ^ -(- 5w = 11 / - 4w = -41 - 6w = -9^

25. Given A and B as follows, solve the matrix equation AX = B. a.

72

A =

1 . -2

2" 3_

B =

'll" _ 6_

b.

A =

3 -2 3

0 1 0

3 0 4_

5

_10

3 Linear Programming Recommended Nutrient Intakes 15 year old boy

100%

MP

MILK AND MILK PRODUCTS

BC FV

BREADS AND CEREALS FRUITS AND VEGETABLES MEAT FISH, POULTRY AND ALTERNATES

MA

9 year old girl

100'’u

In 1945, George Stigler, a mathematician working at Columbia University, studied what is now a famous problem. He attempted to find the lowest-cost diet that would satisfy a person’s daily requirements of protein, calcium, iron, vitamin A, thiamine, riboflavin, niacin, and ascorbic acid. After a detailed study of seventy foods, Stigler concluded that the lowest-cost diet was a combination of wheat flour, cabbage, and hog liver. He calculated the approximate amount of each of these foods for a daily diet that would maintain a person in good health. In 1945 dollars, the cost of this diet would have been $59.88 a year. The cost in 1987 would be about a dollar a day. Stigler was limited in developing his original solution because the number of calcula¬ tions involved was great. With present-day computers, the task can be more easily managed and additonal variables can be included. One result of a new study would most certainly be a more appetizing diet. Stigler’s problem set the stage for the development of a technique in mathematics called linecir programming. A linear function is used to build a mathematical model of a problem. This function is then maximized or minimized as one of the steps in finding a solution. The problem can be a real-world situation that is then more care¬ fully studied and understood. 73

3-1 Inequalities in One Variable Mathematical statements called inequalities use symbols such as > , < , > , < , and ^. jc > 10 means x is greater than 10. X > 10 means x is greater than or equal to 10. X

< 10 means

x

is less than or equal to 10.

0 :< X < 10 means 0 is less than or equal to x and x is less than or equal to 10. In other words, x is between 0 and 10, inclusive. Inequalities can be transformed using most of the same rules that are used for trans¬ forming equations.

The same number can be added to both sides of an inequality. If fl > ^, then a + c > b + c.

EXAMPLE 1

Since 10 > 7, then 10 -f- 3 > 7 + 3.

Each side of an inequality can be multiplied (or divided) by the same positive number. If fl > ^, then ac > be when c > 0.

EXAMPLE 2

Since 10 > 7, then 10 x 2 > 7 x 2.

The rules for transforming equations and the rules for transforming inequalities dif¬ fer when both sides of an inequality are multiplied or divided by a negative number. In the following example, the inequality is multiplied by a negative number and the direction of the inequality symbol must be reversed in order to maintain a true statement.

If fl > ^, then ac < be when c < 0.

EXAMPLE 3

Since 10 > 7, then 10(-2) < 7(-2); that is, -20 < -14.

Two inequalities can be added just as a pair of equations can be added. If a < ^ and c < d, then a + c < b + d.

EXAMPLE 4

74

Since 3 < 5 and - 1 < 10, then 3 - 1 < 5 + 10.

There is no rule for the subtraction of inequalities. This can be illustrated by two instances. 20 > 17 is true. and 10 > 6 is true. But 20 ~ 10 > 17 - 6 is false.

EXAMPLE 5

20 > 17 is true, and 9 > 8 is true. 20 - 9 > 17 - 8 is true.

Solve 2x - 3 > 5x + 12, x e R, and graph the solution on a number line.

SOLUTION

2x - 3 > 5jc + 12 2x - 5x > 12 -f- 3 - 3x > 15 Multiply each side of the inequality by - ^. X < -5 The solution set of the original inequality is all of the real numbers less than -5. The graph of the solution set is shown as follows. An open circle is drawn at -5 since -5 is not a solution.

2

Number-line graphs are useful to show all of the solutions of an inequality. Study the following examples. x< -2,xeR

^-0-1-1-1-^-\-^-► -3-2-101234

The open circle indicates that - 2 is not included in the solution. X > 3, X e R

-1-1-1-%-10 12 3 4 The solid circle indicates that 3 is included in the solution.

-2 ^ X < 3,x e R

--♦—I-1-1-1-0-1-► 2-10

1

-

x< -2 or X > 3,

X e

R

^

'I -5

I-1 -4

-3

234

0-1-1-1-1-® -2

-1

0

1

2

3

I

^

4

5

An inequality can be used to model the following situation. Suppose a person has $55 to buy a Walkman radio. The amount, A, that the person can pay for the radio can be expressed by this inequality. 0 < A < 55 If a tape costing $6 is included in the purchase, the inequality containing A must be modified. 0 < A + 6 < 55 When purchasing items such as radios and tapes, a person generally uses a process of elimination to make a choice. Consideration must be given to whether or not the radio is monaural or stereo, and what length of time the tape will play since these factors affect the cost of each item. Final decisions may take into account the limita¬ tions such as those expressed by the above inequality. EXAMPLE 6

For the inequality 2 < jc < 20, ;r is an integer, list the numbers that satisfy all of the following conditions. (a) (b) (c) (d)

SOLUTION

The The The The

number number number number

is is is is

not a multiple of 5. not a factor of 24. not prime. even.

Condition (a) eliminates the numbers 5, 10, 15, 20. Condition (b) eliminates the numbers 2, 3, 4, 6, 8, 12. Condition (c) eliminates the numbers 2, 3, 5, 7, 11, 13, 17, 19. Condition (d) eliminates the numbers 3, 5, 7, 9, 11, 13, 15, 17, 19. The numbers satisfying all of the conditions are 14, 16, and 18.

EXERCISE 3-1 A

1. Write an inequality that represents each of the following statements. a. y is more than 7. d. y is at most 7.

b. y is less than 7. e. y is greater than 7.

c. y is at least 7. f. y is between 3 and 10 inclusive.

2. Given that -5=!

- 1) = 19^

10. a. Verify the identity. 1 X -

b. Find the sum of the following series. 2

\

X

+

i

15

1

—

x+1

r = 91 r - r

+

1

-

3

+

12

1

-

4*

30

11. Prove that for each positive integer n, n^ - {n - \)- + (n-2)^ - 0? - 3)^

12. Solve for n. 11

S'

p=n

134

P

2' + s 10

p= = n

P

4-

■ • ■ ± 2^ + U = 1 + 2 + 3

4-

• ■

4-

/?.

■ +

1680

The Snowflake Curve The snowflake curve is approximated by beginning with an equilateral triangle. At the start, each side of the triangle is divided into three equal segments, the middle of which is swung out to admit a fourth segment of the same length.

The second approximation is the twelve-sided star-shaped figure shown below.

To obtain each approximation from the preceding one, follow the same process of splitting each side into three equal segments, and then swinging the middle one outwards to allow a fourth equal segment to be inserted. The next two approximations follow.

If continued indefinitely, these curves get closer and closer to a limit curve. This curve will have an infinite perimeter but will enclose a finite area. 1. Show that the perimeter of each approximating curve is equal to | the perimeter of the preceding one.

2. Show that if a given approximation has N sides, the interior of the next one contains the given one plus N additional equilateral triangles. 3. Let A be the area of the original triangle.

a. What additional area is enclosed by the first approximation? The second one? The third one?

b. What area is enclosed by the limit curve? 135

REVIEW

UNIT 4

1. Identify the arithmetic sequences and the geometric sequences. a. 4.0, 6.5, 9.0, 11.5, 14.0 b. 3, 12, 48, 192, 768

c. j, 1, 3, 9, 27, 81

d. 36, 18, 9, 4.5, 2.25

f.

e. 9, 7, 5, 3, 1, - 1

- 1, 2, -4, 8, - 16

2. Find the first four terms of a sequence whose ^th term is given. Identify the se¬ quence as arithmetic or geometric. a.

= 4/7 - I

b.

= 5(2)''-'

c.

= 3"

2/7 — 3 ^-

d.

3. In an arithmetic sequence, the first term is -21 and the seventh term is -45. Find the common difference. 4. Find the unknown variable for each arithmetic sequence. a. If

= 17 and rjy = 97, find

c. If^jg = 13 and (7 =

b. If r, = | and 0.

(a + by

The expansion of (a + by has the following characteristics: 1. The number of terms in the expansion is n + 1. 2. The terms are usually arranged in descending powers of a, and the term number of any term is one more than the exponent of b. 3. In any term the exponent of b is the same as the lower number in the combina¬ torial symbol.

4. The coefficients of terms equidistant from the ends of the expansion are equal.

EXAMPLE 1

Use the Binomial Theorem to expand (2x - 3y)^.

SOLUTION

Make the substitutions a = 2x, b = -3y, and n = 6. (2x - 3y)« = (2xf + (^6^ (2x)5(-33') +

{2x)H-3yf + (f) (2x)(-3>-)5 +

(9

(-3#

= 64x6 - 6(32x5)(3y) -t- 15(16x4)(9y2) - 20(8x3)(27y3) 4- 15(4x2)(81/) - 6(2x)(243y5) + 729y6 = 64x6 _ 576x5y + 2160x4y2 - 4320x3y3 + 4860x2/ - 2916xy5 -h 729y6

EXAMPLE 2

Find the 5th term in the expansion of (x -f- y)20.

SOLUTION

In the 5th term, the exponent of y is 4. The 5th term is of the form x26/. The coefficient is

^ since 30 is the binomial exponent and

4 is the exponent of y. Thus, the 5th term is

222

x26/.

EXERCISE 7-2 A

1. Explain how each of the following terms is formed from the four binomial fac¬ tors of the product (a + b)(a 4- b)(a -I- b){a -I- b). a.

b. a^b

c. a^b^

d. aP

e. M

2. a. Explain how the term a^^b^ is formed in the expansion of {a -i- b)^^. b. Eind the combinatorial expression for the coefficient of a^^b^ in the expan¬ sion of (a -f- b)^^. 3. Consider the binomial (a 4- b)^^. a. How many binomial factors are there? b. What is the degree of each term in the expansion? c. How many terms are there in the expansion? d. Compare the number of terms in the expansion with the binomial exponent 30.

4. In a binomial expansion of {a +

by, a term including a^b^^ occurs.

a. What is the value of nl b. What is the coefficient of c. How many terms are there in the expansion ? 5. Expand {a +

by and {b -i- a)^. Compare the results.

6. Explain why the coefficient of the expansion of (a 4- by^.

is the same as the coefficient of a^b^^ in

7. In the binomial expansion of (x +

there is a term in

a. Eind the value of k. b. Eind the coefficient of the term. B

8. In the expansion of (1 + Jc)^, find all the terms with coefficient 84. 9. Eind the middle term in the expansion of (« + y)^.

10. Use the binomial theorem to find the first three terms of each expansion, a. (u 4- v)’6

b. {x 4- 2yy

c. (3x -

d. (x +

e. (x^y + yh)^

f.

(j + f)

h. (x + ly

i.

(i +

g. (1 +

11.

abf^

Express each of the following in the form (fl -Ia. 10^ -h 4

X

103

4-

6

X

102 -h 4

X

jj

by.

iqi + 10^

b. 10^ 4- 7 X 10^ 4- 21 X 10^ + 35 X 10"^ + 35 x 10^ -f 21 x 10^ -t- 7 x 10' -h lO'^

223

12. a. Expand (1 + x)^, (1 - x)^, and(l - x^)^. b. Use the expansions in part (a) to verify that (1 - x^)^ = (1 + x)^(l - x)^.

13.

Verify that (1 + x)^ - (1 - x)^ = 2x(3 + x^) in the following two ways: (i) by expanding (1 + x)^ and (1 - x)^, (ii) by factoring (1 + x)^ - (1 - x)^ as a difference of cubes.

14. Verify that: (x + y)4 + (x^ + y^) = 2(x^ + xy +

15. Verify that:

^

a. (x + y)3 - (x^ + y^) = 3xy(x + y) b. (x 4- y)^ - (x^ + y^) = 5xy(x + yXx^ + xy + y^) c. (x + y)^ - (x^ + y^) = 7xy(x + y)(x^ + xy + y^X

16. Find the coefficient of x^ in the expansion of (1 - 2x^(2 + x)^. C 17. Show that the coefficient of x'^ in the expansion of (1 + x)^'^ is equal to twice the coefficient of x'^ in (1 + x)^'^ “ ^.

18. Lety = x + x“f a. Show that x^ + x~^ = y^ - 2. b. Showthatx^ + x~^ = y^ - 3y. c. Express x^ + x“^andx^ + x“^ as polynomials in y. Hint: Expand (x + x~^y and (x + x~*)^.

19. a. Trinomial expressions can be expanded with repeated use of the binomial theorem. Expand (a + b + cX by first writing it as ((a + b) + cX and treating (a -h b + c) as a binomial in (a + b) and c. Expand each of the powers of (a + b) that occur. b. Check the result in part (a) directly by expanding the product of three tri¬ nomials. The product (a + b + c)(a + b -h cXa + b + c) can be written as the sum of 27 = 3^ terms, each a product of three letters, one chosen from each factor. For example, a^b occurs three times with the b coming from one of the three factors and the a from the other two. To obtain terms in abc, there are three ways of designating the trinomial from which a comes, there are two trinomials left from which to take b, and finally c must come from the remaining trinomial. c. Check that the expansion obtained for (a + b -h cX yields the correct result under the substitution a = b = c = 1.

20. Expand (x^ - 6x

-I- 9)^.

21. Expand (fl + /? V3)^ + (a - b\f3X.

224

7-3 The General Term Of A Binomial Expansion The binomial theorem shows that the expansion of {a + by consists of the sum of (n + \) terms of the form

where r is an integer between 0 and n

inclusive. This is called the general term of the expansion. Sometimes, in order to efficiently identify a term in a binomial expansion, it is helpful to assume it has the form of the general term and then try to identify the values of the variables n, a, b, and r.

EXAMPLE 1

Given the expansion of (1 + z)^^, a. find the coefficient of b. find the coefficient of

SOLUTION

a. The general term for (1 + z)^"^ is The coefficient of z^ is b. The coefficient of z^^ is

EXAMPLE 2

In the expansion of (x^ - x a. find the coefficient of b. find the coefficient of x“ c. find the constant term, if it exists.

SOLUTION

In an expansion of (a + by, where a = x^, n = 5, the general term of this expansion is

b = -{x

and

(J) a^-V = = (- IX

{t\

= (-ir (^) a. The term in

corresponds to the value of r for which

10 - 5r = 5. Thus, r = 1. Substituting this value for r in the general term shows that the coefficient of x^ is (-1)'

-5.

225

b. For the general term to represent a term in ;c Make the substitution r = 4. The coefficient of is (-1)“^

10

=5.

c. The only powers of jc occurring in the expansion are of the form ^10-5r^ where 0 < r < 5. There is a non-zero constant term only if there is a term inx^. Therefore, 10 - 5 r = 0. r = 2 Therefore, the constant term is

= 10.

An alternative solution using the Law of Exponents provides a more convenient working form. (x^ - x~^)^ =

- 1)]^ =

- 1)^

a. The coefficient of

in the expansion jc~ - 1)^ is the same as the coefficient of x^^ in the expansion of - 1)^, which can be calculated to be - 5.

b. The coefficient of is the same as the coefficient of (x^ - 1)^, which can be calculated to be 5.

in the expansion of

c. The constant term is the same as the coefficient of x^^ in the expansion of (x^ - 1)^, which can be calculated to be 10.

EXAMPLE 3

Find the coefficients of

SOLUTION

The binomial theorem can be applied with a - c, b = -2^ n = 1.

and

in (c -f- -2)^. 2

The (r -I- l)th term in the expansion is To find the term containing c^, 7 - 3r = 3 is needed. This does not occur for any positive integer value of r. Therefore, there is no term in and the coefficient of is 0. To find the term containing

1 - 2r = 4 So that The term is

r = 1 and the required coefficient is 14.

EXAMPLE 4

Three consecutive terms in the binomial expansion of (1 + x)" have coefficients 136, 680, and 2380. Find n.

SOLUTION

Suppose that the consecutive coefficients are

(,!l) = 136. («) = 680,and(^”J = 2380

226

n r-\

It follows that

in - r)\r\

(r - !)!(« -^+1)1 jr -

l)!(/t - r + 1)!

{n - r)!r! in - r + \) r 680 136 = 5 n - r + \ = 5r n = 6r - I

(

"

1

\r/

Vr+l/

©

^-1_ r 1)!

=_

(r + l)l{n

-

-

n — r r + I 2380 680 7

2 2n - 2r = Ir + 1 2n = 9r + 1 Combining © and

@

we have: 2(6r - 1) = 12r - 2 = 3r = r =

9r + 7 9r + 7 9 3

Substituting in ©, we have; n = 6(3) - 1 :.n = 17

EXERCISE 7-3 A

1. Write the general term for the expansion of the binomial. a. (1 +

d. (1 2. A term independent of 3: has a zero exponent for v and is called a constant. Deter¬ mine if there are terms independent ofv in each expansion. Find the value of the terms. a.

(2v-)

b. i4x^ + 3x 0*^

227

3. Find the term in

for the indicated value of k in the expansion.

a. (1 + x)'2

k = 1

b. (1 - x)35

fc = 18

C. (1 + x2)3

k = 4

d. (6x - 1)'3

^ = 11

e. (x + x^)^

k = 2

f.

(3x + 2x^)^

^: = 13

4. Find the 10th term in each of the following:

b. (x - 2)100

a. (1 + B

5. Write the term in x^o in its simplest form in each expansion. a. (1 + x)i^

b. (x +

d. (x - x^)^

e. (x + x^)i

2)100

c. (x +

6. Determine the value of n in the expansion of (a + by in descending powers of a, if it is found that two adjacent coefficients are each equal to 126. 7. If (4 - 2zy is expanded in ascending powers of z, find the fourth term. 8. Write every fifth term in the expansion of each binomial. a. (a + b. (a + ^)i^ C

9. Three consecutive terms in the expansion of (1 + x)'^ have coefficients 120, 210, and 252. Find n.

10. In the expansion of (1 + x)'^ in ascending powers of x, it is found that one of the coefficients is equal to twice the preceding coefficient. a. Show that this happens if and only \f n + 1 is a multiple of 3. b. Give the four smallest positive integers, n, for which this occurs.

11. Find all the values of n for which 35 is a coefficient in the expansion of (1 + x)'^ 12. If x^ occurs with a non-zero coefficient in the expansion of (x + x i)*^, show that r - s must be an even number.

13. Let « be a given positive integer. Identify those values of k for which x^ has a non-zero coefficient in the expansion of (x^ - x“ l)'^

14. For some value of k, the expansion of (3x^y“^ -I- 2x“^y)^ has a term of the form c(xy)^. Find it.

15. Show that the general term of the expansion of (a + Z? + c)" is given by , , , aPWc’', wherep + q + r = n. plqlri

16. Expand (1 -i- x -f- x^y.

228

7-4 The Binomial Theorem For General Exponents The mathematician and physicist Sir Isaac Newton (1642-1727) is credited with investigating the binomial expansion of (1 + xY when n is not a positive integer. He inferred the result from observed patterns but he did not give a formal proof. The process Newton used can be shown by considering the first few expansions of (1 + xY for positive integer values of n. (1 + xY = I (1 + xY = 1 + JC

(1 +

= 1 + 2.x +

x'^

(1 + = 1 + 3x + 3x^ + (1 + x)^ = 1 + 4x + 6x^ + 4x^ + jY (1 + x)^ = 1 + 5x 4- lOx^ + lOx^ + 5x^ + x^ The coefficients of the terms in each expansion are, of course, found in Pascal’s Triangle. To expand (1 + x)'^ where ^ is a negative integer, extend Pascal’s Triangle upwards into negative rows, while maintaining the relationship each row has to the one above. The first step is to start with Pascal’s Triangle and fill out the rows with zeros to the right. Upper, or negative rows, can now be added one at a time by using the same rule employed in the original Pascal’s Triangle. A number in a row below a pair to its right and left in the row above it, is the sum of the pair above. The fol¬ lowing triangle shows the result of adding three upper or negative rows.

1

Row - 3

1

Row -1

1

Row 0

1

Row 1

2

Row 3

33

3

- 10

-4

-1 I -I 0 0 0

1

Row 2

6

-2

1

Row -2

-3

0 1

0

1

0

1

..

1-1 0 0 0 0 0

0 0

-2

3

-4

5

r'yyv''/\

5

0

0

...

y/yyy\

... 0...

0

0

11 ...

000000...

The triangle can be extended upwards for as many rows as desired. With the use of the extended Pascal’s Triangle, the following expansions can be conjectured as being true: From Row -1 From Row -2 From Row - 3

(1 + x)“' = 1 - x -f- x^ - x^ -f- x^ (1 + x)~^ = I - 2x + 3x^ - 4x^ -i- 5x^ ■ (l+x)“^ = l-3x + 6x^ - lOx^ ■

The expansion for (1 4- x)“i is an infinite geometric series, with ratio -x, whose sum can be found using the formula from Chapter 6, provided that \x\ < 1.

= 1 - X + x^ - x"^ -h

229

Although the above procedure does not constitute a mathematical proof, it appears that under certain conditions the extended Pascal’s Triangle gives the coefficients for the terms in the expansion of a binomial with a negative exponent. The next concern is to investigate the possibility of a formula for the binomial expansion in cases where the exponent n is not only a negative value but is also a fraction. The meaning of

can be generalized to the case in which n is not a positive integer.

In so doing, its meaning as the number of ways of doing something must be abandoned for a purely algebraic definition. The algebraic definition will be equivalent to the expression for

when « is a positive integer. Consider the equation /^\ ^ n{n-\){n-2) ■ ■ {n-r + 1) \r) r\

The left side of the equation is a combinatorial symbol and has meaning only for non-negative integers n and r. The right side, however, has meaning for any real number n. Accordingly, the right-hand expression can be used to define the left-hand expression for any real number n, where r is a positive integer. In evaluating

note that the number of factors in the numerator equals r and the

denominator is r!. For example, f ^ j =

EXAMPLE 1

Compute

{n){n - \){n - 2) 3!

a.

SOLUTION

-56 (81)(24) 7 243

The binomial theorem can be formulated for rational number values of the exponent n. When n is not a non-negative integer, none of the coefficients of a power of x vanish, because no factor in the numerator is zero. Thus, the result is an infinite series. (1 -I-

230

= 1 + nx +

n(n - l)x^

2!

-r

n(n - l){n - 2)x' 3!

The infinite series expansion can be used for all binomial exponents and, in particular, for exponents that are fractions. EXAMPLE 2

Find the binomial expansions of 1 a. (1 + xY b. (1 + ;c)“4

SOLUTION

a. Use the binomial expansion for n

1 2

'

(1 +

(1 +

+ jX +

xf

+

+

2!

+

3!

+

+

1

+

{^]x

—

(ib)-^^

b. (1 + X)- = 1 - 4x .

2!

V 128

\x^

, (-4)(-5)(-6)x3 3!

(-4)(-5)(-6)(-7)x4 4! = I - 4x + \0x^ - 20x^ + 35.^4 Since the series (1 + does not always terminate, this raises a question of its validity for numerical substitutions for x. When - 1 < x < 1, and as more terms are added, the sum on the right side of the equation gets closer and closer to some value. This means that if x is small enough, a reasonably close approximation to the value of (1 + x)" can be obtained by taking the first few terms on the right side of the equation. EXAMPLE 3

Evaluate the sum of the first four terms in the expansion of 1 (1 + x)^ when: a. X = 0.1.

b. X = 8.

c. Compare the results with those obtained by using a calculator. i

SOLUTION

a.

i

= (1 + O.l)^

= 1 + yo.l) - |(0.01) + -|L(0.001) = 1 + 0.05 - 0.00125 + 0.000 062 5 = 1.048 81 . . .

-1

b. When x = 8, the sum of the first four terms is: 1 + i(8) - i(64) + i(512)

=1+4-8 + 32 = 29

c. A calculator will provide the following result: (1.1)^ = 1.048 809 . . . A calculator is not needed to see that:

(1

+

8)2

=

V9

= 3 When comparing the results of parts (a) and (b) with part (c), it can be seen that the binomial expansion gives good approximate values for powers of numbers that can be expressed in the form (1 + xY,\f the absolute value of x is small compared to 1. In such a case, the values of higher powers of x , such diS , and so on, approach 0 very quickly. In part (b), the expansion gives a poor result because 8 is not small compared to 1. In order to work with the more general expression {a + by, first write it in the form /

a^\ \ +

by

expand this form using the general expression.

2

EXAMPLE 4

a. Find the first four terms in the expansion of (8 4- x)^. b. Using the result of part (a), estimate (^8. 3)^ to two decimal places.

SOLUTION

a. (8 + xj = 8^(1 + |)^

= 4[1 + f(|)

=

4[1 +

= 4 +

2!

JC

12

3

(9)(64) X" 144

+ '

3! 4x3 (81)(512)

10 368

+

b. Since 8 + x = 8.3 when x = 0.3, substitute x = 0.3. (VO)2 =

4 +

0.3

= 4 + 0.1 = 4.10

0.09 144

+

0.027 10 368

0.0006

Note that only three terms were needed to obtain the accuracy required.

232

A

EXERCISE 7-4 1. Write, in unsimplified form, the first four terms in the expansion of the binomial. i

a. (1 +

b. (1 -

2

xy

3xy

c. (1

2. Each of the following is a term in a binomial expansion. Give its term number in the expansion, the exponent in the binomial, and the binomial. (-3)(-4)(-5)(-x)3

(i)(_i)(_|)(_|)(2,)4

3! B

4!

3. Write the first four terms of the binomial expansion. -3

a.

(1

+

b.

(1 + x)

e.

(1

^

c.

(1

+ r)2-4

f.

(1

+ u)~

h. (1 + /)7-2

1.

(1

+ 1)7-

c.

(1

+ 2x)'-3

—

X)4

_ o .j

d. (1

—

^

X) a

g-

(1

+ uf

•

Find the general term of the binomial expansion of 1

b.

+

1

(1

(1

1

a.

5. a. Find the value of (1.1)^ to three decimal places, using a binomial expansion. 1

b. Find the value of (8.16)^ to three decimal places, using a binomial expansion. 6.

Expand and simplify the first four terms in the expansion of the binomial. a. (2 - x)

b. (2 - f)-2

c. (8 +

d. (9 - 2zy

e. (8 + 12y)

f.

7. In the expansion of (1 + ax) Find the value of a.

2ay

{x -

5x^

the fourth term is ^

8. Use the binomial expansion to find approximate values to three decimal places of a. (123)^

b. (123)^

c. (123^

d. (123)-i

^ . 2x^ 9. In the expansion of (

The numbers of the Fibonacci sequence can be calculated from this formula by using n = 1, 2, 3, and so on. n = \, n = 2,

F, = 1 = F2 = \ =

1 2V5 1

[(1 +

V5)

[(1

+

V5)2

-

(1

-

V5)2]

[(1

+

V5)3

-

(1

-

V5)3]

- (1 -

V5)]

4V5

n = 3,

F3 = 2 =

n = 4,

F4 = 3

=

1

8V5 _i_

-

16V5

[(1

+

'Jsf

-

(1

-

VS)'*]

Use the binomial expansion on the right side to check these equations. Then, find F^ and F^ and verify that the formula is correct.

238

7-6 Proof by Mathematical Induction Consider the problem of finding the sum of the series I’l! + 2*2! + 3*3! 4- 4’4! +

+ ri’nl.

To see what the sum might be, calculate a few special cases for small values of n. For n = 2,

1-1! + 2-2! = 5 = 3! - 1

For ^ = 3,

1-1! + 2*2! + 3-3! =1+4+18 = 23 = 4! - 1

Forn = 4,

M! + 2*2! + 3-3! + 4-4! = 1+4 + 18 + 96 = 119 = 5! - 1

There is a pattern in the results of the calculations. On the basis of what has been seen so far, it is reasonable to guess that 1-1! + 2-2! + 3-3! + 4-4! + 5-5! = 6! - 1. This, in fact, turns out to be true when the two sides of the equation are evaluated. Perhaps there is some way of using the information previously obtained to derive new equations. The left side of the new equation is the same as that of the previous equation with one additional term. Given that: M! + 2-2! + 3-3! + 4*4! - 5! - 1, can the sum with one more term be found? M! + 2*2! + 3*3! + 4-4! + 5-5! = = = = =

(M! + 2-2! + 3-3! + 4-4!) + 5-5! (5! - 1) + 5-5! (1 + 5)5! - 1 6-5! - 1 6! - 1 , since 6-5! = 6!.

Note that this procedure not only saves a fair amount of direct computation, but it also provides the basis for developing new numerical patterns. Repeating the process of adding one more term yields the sum with the next term included. 1-1! + 2*2! + 3-3! + 4-4! + 5-5! + 6-6! = = = =

(11! + 2*2! + •• + 5-5!) + 6-6! (6! - 1) + 6-6! 7-6! - 1 7! - 1.

The following conjecture can be made based on the pattern that has emerged so far. For every positive integer n, M! + 2-2! + 3-3! + 4*4! +

• + n'n\ = (n + 1)! - 1.

239

This conjecture has been verified for n = 1, 2, 3, 4, 5, and 6. The next step is to establish the conjecture in general. The procedure used so far will verify it to any selected number, say 100, by working through all of the integers between 6 and 100. If a computer were used to check these results, it would be useful to have a subroutine to apply at each step. Suppose that the equation has been established for n = k. 1-1! + 2-2! + 3-3! +

• + k'k\ = {k + 1)! - 1.

The left side for the case n ^ k + 1 is then 1-1! + 2*2! + 3-3! + ••• + k'k\ + (/: + l)'(k + 1)! = (k + ly. - \ + (k + 1) (k + 1)

= (I + k + l)(k + ly - I = (k + 2) {k + 1)! - 1 = (k + 2)! - 1. This is the right side for the case n = k + 1.

The final member of this equation is exactly what is specified in the conjecture when the substitution n = k + 1 is made. Thus, this step can be applied for successive values of k, 6, 7, 8, and so on, up to any specified number. On the basis of this argument, it can now be asserted that the above conjecture is true for all values of n. This process of proving mathematical statements can be formalized in the Principle of Mathematical Induction. The symbol P(n) is used to denote a statement that involves an integer n. For example, in the previous discussion, P{n) is: 1-1! + 2-2! + 3-3! + P(l) is P{1) is

• + n^nl = {n + 1)! - 1.

M! = (1 + 1)! - 1. M! + 2-2! = (2 + 1)! - 1.

The Principle of Mathematical Induction asserts the following: Suppose that: 1 there is an integer r for which P{r) is true; and 2. if P{k) is assumed to be true, P{k + 1) can be shown to be true. Then the statement P{n) is true for each integer n > r.

.

Applying the induction principle is like climbing a ladder. Step one says that a bottom rung can be found, and step two says that it is possible to go from one rung to the next. Note that step two is a hypothetical statement; it does not assert the truth of the origi¬ nal statement, but only relates the truth of one particular case to the truth of another case. Thus, it is necessary to establish the statement for one particular value of n in order to get started.

240

Another analogy for induction is the domino effect. In Japan, there is an annual competition in which the competitors set up an extended pattern of upright dominoes arranged in rows. When the first domino is tipped it hits and topples the second domino, which in turn topples a third, and so on. The winner is the competitor who can arrange the largest chain of toppling dominoes. In an induction proof, once the first domino is tipped, the domino effect will continue forever. The steps in an induction proof are as follows: 1. If not already done, formulate a result as a statement which depends on an integer «. 2. Prove the result for an initial value of n (usually n = Oovn = 1, the bottom rung). 3. Prove that if the result is true for n = k, then it is true for n = k + 1. 4. Conclude on the basis of steps 2 and 3, using the Principle of Mathematical Induc¬ tion, that the result is true for all values of n from the initial value on. EXAMPLE 1

Prove by induction that , +

12 -f- 22 + 32 + 42 -h SOLUTION

Let P(n) be

12

+ 22 -1-

32

-f-

^

n(n

+

42

+

P(l) is

^ 6 (1)(2)(3), which is true.

P(k) is

+ 2^ + 3^ + ■■■+ k^ =

P{k + \) is

l)(2n + 1) ^ f^ for

1 o q = 1, 2, 3, ... .

n{n -f l)(2/i -1- 1)

=

k{k + 1)(2^ -h 1)

12 + 22 + ■ ■ ■ + (^ + 1)2 =

(k + 1)(^ + 2){2k + 3)

Now suppose that P{k) is true. It must be shown that P{k -I- 1) is true. Add {k -\- 1)2 to each side of P{k). The left side is U -h 22 + • ■ ■

- /c2 -f- (/c + 1)2

4

= (I2

-

4

22

- • •

4

+ /c2)

- (/: 4- 1)2.

4

Making use of P{k), the right side is:

I k{k + \)(2k + 1) + (/t + \f = (k + 1)( — (k 4- 1)

k{2k 4- 1)

+ {k + 1))

2^2 + Ik + 6

(k + i) is a factor.

(k + l){k + 2){2k + 3)

This is of the form ^ n{n -I- \)(2n 4- 1) when n = k + 1. Hence, P(k) implies P{k + 1). Since P(n) is true for « = 1, and if P(k) is true then P{k + 1) is true, then by the Principle of Mathematical Induction Pin) is true for n = 1,2,3,.... 241

EXERCISE 7-6 A 1. Check the following statements; 1 2

1 2

1

1 + 6

1 2

1

1 + 1 + ± 12 2 6 1 + 1 + J_ 2

6

12

1 5

a. b. c. d.

What is the pattern by which the denominators 2, 6, 12, and 20 are formed? Discover a pattern for the above numerical equations. Write and check that the next two equations follow the pattern. Write a general statement, depending on the positive integer n, that has the above equations as particular cases. e. Prove this general statement by mathematical induction.

2. What is wrong with the following argument? Statement: Any two consecutive integers are equal. Let P(n) be the statement n = n -h 1. P(k) is k = k -h 1. Assume this to be true. P(k + 1) is k + 1 = k + 2. Prove that this is true. Adding 1 to each side of k = k + 1 yields k + \ = k + 2. But this is P(k + 1). Hence, P(k) implies P(k + 1). Therefore, any two consecutive integers are equal. 3. Consider the function/(«) = rP + n + 41. Verify that/(O) = 41,/(I) = 43,/(2) ^ 47,/(3) = 53,/(4) = 61. a. Calculate/(5),/(6), and/(7). b. Does the evidence you have uncovered so far support this conjecture: f{n) is a prime number for each non-negative integer n? c. Have you actually proved the conjecture? d. Find a counterexample, that is, a value of n for which the conjecture is false.

4. Consider the following argument. Is it a proof by induction? Give reasons for your answer. Proposition: For any positive integer n, l+2 + --

-\-n=^ n{n -i- 1).

Proof: Write down the sum, S, twice. S' = \ + 2 +■■■ + (n-2) + {n — \) + n. S = n {n — l)+ '-i3 -f2 -fl. Adding vertically and then summing across, we find that 2S is equal to (1 + A?) + [2 + (/7 - 1)] + ■ + [(« - 1) -r 2] + (/? + 1) = n{n + 1), from which the result follows.

242

5. Use mathematical induction to prove the proposition in question 4. 6. Prove that:

a. 5 + 9 + 13 + 17 + • • • + {An + 1) = n{2n + 3) b. 7 + 15 + 23 + 31 + . . . + (8^z — 1) = n{An + 3) c. 1 + 3 + 9 + 27 + • ■ • + S"-' = —■ ~ * B

7. Write out and check the first three cases of each statement. Give a proof by mathematical induction that the statement is true for each positive integer n. Ill 1-3 3-5 5-7

+

1 n " + (2n - \){2n + \) ~ 2n + I

b. The sum U + 3^ + 5^ + • •

-I- {2n - 1)^ of the first n odd squares is

equal to ^(2/i - l)2n{2n + 1). 8. Prove, using mathematical induction.

a. 1*2 + 2*3 + 3*4 + ••• + n{n + 1) = ^ n{n + \){n + 2)

,

b. 1-2-3 4- 2-3-4 + 3'4*5 + ••• + n{n + \){n + 2) = ^ n{n + l){n + 2){n + 3) c. Look at the results of parts (a) and (b) and guess a formula for the sum 1-2-3-4 + 2'3*4*5 + 3'4-5*6 + •■• + n{n + l){n + 2){n + 3). Check your formula for n = 1, 2, 3. Give a general proof by induction that the formula is true.

d. Can you guess an even more general result? 9. Prove a.

n

_0

b_

'l .0

k

0 '

_0

1

o'

c

a

n

1_

10. a. Prove that

b. Prove that

c. Prove that

'l .0

kn 1 _

n{n 4 1)

2

'

n{n 4 l){2n 4 1)

6 ^n{n 4 l)y

11. Prove for each positive integer n that - (n - \)^ + (n - 2)^ + ■■■ + (- If "’(1)^ = n + {n-\)+

-

+2+\.

243

C 12. Show by induction that, for each positive integer n:

r = k-\

13. Consider the integers 1, 2, 3, 4. The set of all possible products of two of the integers is 2, 3, 4, 6, 8, 12, and the sum of these numbers is 35. a. Find the sum of all possible products b. How many pairs of the integers 1, 2, c. Show that the sum of the products of 1,2, 3, . . . , « is given by ^ +

of two of the integers 1, 2, 3, 4, 5. 3, . . . , n are there? all possible pairs of the integers \)(n - l)(3n + 2).

14. Show that the sum of all possible products of three of the integers 1,2,3, . . . , n is given by ^+ 1)2^^ _ _ 2). 15. Prove that + n , for n = 1, 2, 3, 2{rP- + n + 1)

St +

r=

16. a. Show that A U B = (A\ B) U B expresses any union of two sets as a union of two disjoint sets. b. State a generalization of part (a) that displays the union of n sets A^, A2, . . . , as a union of n pairwise disjoint sets. c. Prove your assertion in part (b) by induction.

17. Prove by induction a. 1+2+4 + 8 + 16+

- + 2'^-i =2^-1

n- 1

b.

2j

a + kd = na + \ n(n - l)d.

k=0

^

18. A straight line divides the plane into two regions. If we draw a second straight line not parallel to the first, the plane is divided into four regions. A third straight line, not parallel to either of the first two lines and not passing their point of intersection, will result in a division of the plane into seven regions (one of which is a triangle). Show that n lines drawn in the plane, no pair of which are parallel, will divide the plane into 1 + n + 19. a. Verify that ^

.

b. The Fibonacci sequence

is defined by

F, = F2= I {n > 2) Prove that for each n. Fn
)

x

P{B\A), for events A and B.

16. Show that A and B are independent events if and only if P{A\B) = P{A\B).

280

8-6 Genetics and Probability The Austrian botanist, Gregor Mendel (1822-1884) developed the theory of genetics in an experiment concerned with the colour of seeds in pea plants. Mendel found that characteristics, such as colour, are passed on from one generation to another by “characteristic carriers” called genes. Some of these genes are called dominant, denoted by A, and some are called recessive, denoted by a. Each living thing has a genotype or gene pair of the form AA, Aa, or aa for some particular trait or characteristic. If a person inherits a dominant gene from either parent (that is, his or her genotype is AA or Aa), he or she will demonstrate the dominant trait — for example, right-handedness. A person will demonstrate the reces¬ sive trait, such as left-handedness, only if he or she inherits that gene from both parents (that is, his or her genotype is aa). Each parent passes on one of his/her genes to the offspring to make up the offspring’s gene pair.

EXERCISE 8-6 A 1. Assuming that a person has equal probability of inheriting either gene, the proba¬ bilities of various genotypes can be calculated. Find the following probabilities: a.

b.

P(child is AAjboth parents are AA) P(child is aa both parents are aa)

AA AA A-

•

•

A-

•

•

Parent 1 AA AA —I-1-

c.

d. e.

f. g-

h.

P(child is AA|both parents are Aa) P(child is Aa|both parents are Aa) P(child is aa|both parents are Aa)

P(child is AA|one parent is AA, one is aa) P(child is Aa|one parent is AA, one is aa) P(child is aa|one parent is AA, one is aa)

A

A

A -

•

•

a -

•

•

}

I

A

a

Parent 1

Parent 2

"

Parent 2

a Parent 1 a -I-(-

•

1.

•

Jk.

A

P(child is AA|one parent is AA, one is Aa) P(child is Aa one parent is AA, one is Aa) P(child is aa|one parent is AA, one is Aa)

A

Parent 2

a Parent 1 A “1-1

A

1. P(child is AA one parent is aa, one is Aa) m. P(child is Aa|one parent is aa, one is Aa) n. P(child is aa|one parent is aa, one is Aa)

A

Parent 2

a Parent 1 A n-1

a

a

Parent 2

281

o.

B

Copy and complete the following Summary Chart.

2. Can the probabilities of a particular trait be predicted when these characteristics are not spread evenly over the whole population? To solve this problem, make the following assumptions: • The probabilities that a person has a particular genotype are: P(AA) = X, P(Aa) = y, iP(aa) = z, where x + y + z = 1. • These probabilities are the same for males and females. Father’s Genotype

Child’s Genotype

Probability that each branch occurs. G)G)(1) - jr2

W(>’)(^) =

^xy

a. Copy the tree diagram and enter the probabilities from question 1 in the appro¬ priate places. b. Write an algebraic expression for each genotype combination.

282

3. Since branches of the tree represent mutually exclusive events, we can write an expression for the probability of a child’s being a specific genotype. For example:

P(AA) = P(AA|AA, AA) + P(AA|AA, Aa) + P(AA|Aa, AA) + P(AAiAa, Aa) Show that:

a. P(AA)

=

(x

+

b. P(Aa) = 2(x + ^y){z + ^y) c. P(aa) = (z + d. P(AA) + P(Aa) + P(aa) = 1 4. Suppose a characteristic that determines earlobe shape exists in a population in the proportion:

P(AA) = 0.30

P(Aa) = 0.60

P(aa) = 0.10

a. Find the probability of each of the genotypes AA, Aa, and aa in the next gen¬ eration.

b. Using the results of part (a) as the next parent population, find the probability of each of these genotypes two and three generations from now. c. Compare the answers to parts (a) and (b), and comment on the question, ‘ ‘Will a particular rare recessive characteristic die out?” d. Using P(AA) = x, P(Aa) = y, and P(aa) = z, find the probability of each of these genotypes two and three generations from now. This result is called the Hardy-Weinberg principle which states that, if there is random mating, then the population of genotypes should stabilize over a period of time. 5. Show that if equilibrium.

= 4jcz, then the distribution of genotypes is unchanged, or in

That is, prove:

a. (x -I-

=

X

b. 2(x + ^){z + 2^ = y c. (z +

= z

6. The Hardy-Weinberg principle can be used to predict the probability that a person is a carrier of a rare recessive trait. Suppose one person in 30 000 in a popu¬ lation in equilibrium is a dwarf.

a. What is the probability that a randomly selected person is a carrier of the trait “dwarfism” (a non-dwarf carrying a recessive gene)?

b. What is the probability that the child of two carriers is a dwarf. Hint: Use the relationships from question 2.

283

REVIEW

UNIT 8

1. A pair of dice is thrown. What is the probability of throwing: a. exactly one three? c. a total of 7 or 11?

b. at least one three? d. a pair or a total of 8?

2. A card is drawn from a deck of 52 cards. a. What is the probability of drawing an honour card {A, K, Q, J, 10}? b. What are the odds in favour of drawing an honour card? 3. Using the two spinners shown below, what is the probability of spinning:

5. a. A bag contains 5 blue chips, 4 red chips, and 3 green chips. Three chips are drawn randomly without replacement. Draw the tree diagram corresponding to the above situation. b. Calculate the probability of drawing 3 red chips. c. Calculate the probability of drawing exactly 2 red chips. 6. Three dice are thrown. What is the probability of throwing a. a triple (3 dice the same)? b. a total of 6? c. exactly one three? d. at least one three?

7. Two cards are drawn without replacement from a deck of 52 cards. a. What is the probability of drawing two aces? b. What are the odds in favour of drawing two honour cards? 8. Using the two spinners below, what is the probability of spinning

a. exactly one 3? b. at least one 3? c. a pair or a total of 8?

284

9. The probability that Hugo makes his own lunch is

If he makes it today,

then the probability that he makes it again tomorrow is y. a. Draw a tree diagram illustrating Hugo’s lunch-making habits over a 2-day period. b. Find the probability that Hugo will not make his lunch tomorrow.

10. Find the probability that: a. the total of two dice is 7 given that the total is odd. b. the total is odd given that the total is 7.

11. For any two events A and B, prove that: a. If PiA) ^ P(B), thenP(/I|B) ^ P{B\A). b. P[(A U 5)|B] = P[A\{A 0 5)] = 1.

12. Two tetrahedral dice are coloured so that one die has 3 coloured faces and 1 white face and the other die has 4 coloured faces. If the two dice are placed in a hat and one is selected randomly and placed on a table so that all sides show coloured, what is the probability the other side is white?

13. In a population of students, it was observed that 50% used radios with head¬ phones and 50% never used headphones. Health officials reported that 35% of the students have severe hearing problems, 40% have moderate hearing problems, and the remainder have no hearing problems. It was also observed that 50% of students using headphones have severe hearing problems and 30% of students using headphones have moderate hearing problems. a. Are the events use headphones and have a hearing problem independent? b. Find the probability that a person has no hearing problem knowing that this person does not use a radio with headphones. c. Find the probability that a person doesn’t use headphones knowing that this person does not have a hearing problem.

14. Two members of a 3-person jury each arrive independently at a correct decision with probability P. The third member flips a fair coin. The decision of the majority rules. Find the probability of a correct decision by the jury.

15. A and B are mutually exclusive events such that P{A) = 0.3 and P{B) = 0.2. a. Illustrate these events using a Venn diagram. b. Find P{A)

c. Find P{B)

d. Find P{A Pi B).

e. Are the events A and B independent?

285

16. Many students can’t walk and chew gum at the same time. It has been observed that 40% of students are not walking while 70% of students chew gum. a. What is the probability that a student is walking and chewing gum? b. If a student is seen to dance in the halls, what is the probability the student is chewing gum?

17. In a Stanley Cup final, the winning team is the first one to win 4 games. Mon¬ treal is playing Edmonton and in any game P(Montreal wins) = 0.4. What is the probability that the series is over in just 4 games?

18. A woman has triplets. Assume that P(boy) = P(girl). a. What is the probability that the triplets are all girls? b. What is the probability that at least one of the triplets is a girl? c. What are the odds in favour of there being at least one child of different sex?

19. A 30-year-old Canadian male marries a 25-year-old Canadian female. Using the Life Tables on page 390, find the probability that the couple will celebrate their golden wedding anniversary, assuming they remain married.

20. A team T must play 3 other teams A, B, C in the season, and will reach the playoffs if it wins at least 2 of the 3 games. The probabilities of P’s winning over A, B, and C are P(A) = |, P{B) =

and P(Q =

Tied games are not

possible. What is the probability that T will reach the playoffs?

21. Four people are starting a game of bridge. Each player picks one card from a full pack of 52 cards to decide upon the dealer. What is the probability that all four players pick an ace?

22. The Football Players Union is concerned about the issue of drug testing and arranges for an independent organization to run an experiment to determine the accuracy of the tests. A test group of 50 players took the drug, while a control group took a placebo. The results are summarized below.

Player

Drug

Placebo

Tested Positive Tested Negative

38 12

4 46

Totals

50

50

a. Estimate the probability that a player who did not take the drug tested positive.

b. Estimate the probability that the test gives a correct result. c. Estimate the probability that the test makes an error.

286

23. How accurate are lie detectors? An experiment was conducted in which a group of suspects was instructed to lie or tell the truth to a set of questions, and a group of polygraph experts along with the polygraph (lie detector) judged whether the suspect was telling the truth or not. The results are tabulated below.

\

Experts’ \ Suspeets’ Answers \ Judgement \ Truth Lie Truth Lie

93 7

11 89

Totals

100

100

a. Estimate the probability that a suspect who is telling the truth is found to be honest by the experts. b. Estimate the probability that a suspect is a liar and gets away with the lie. c. Estimate the probability that the experts’ judgement is correct. d. Estimate the probability that there will be a miscarriage of justice.

24. A town is served by two hospitals, the large downtown hospital and the small county hospital. It is a well-known fact that about 50% of births are girl babies and 50% are boy babies. One day this year, 45 babies were born in the large hospital and 15 babies were born in the county hospital. A doctor in one of the hospitals observed that 60% of the babies born were girls. Which hospital was that doctor more likely to work at? To determine whether 9 girls in 15 babies is a more likely occurrence than 27 girls in 45 babies, a simulated experiment can be performed using random num¬ ber tables (or tossing coins.) Experiment I Select one sample of 15 random digits. Assign the even digits to the event a girl baby. Assign the odd digits to the event a boy baby. If this sample has exactly 9 girls, mark it with a /; otherwise, mark it with an x. Repeat for 100 samples. How many / were recorded? Experiment II Select one sample of 45 random digits. If this sample has exactly 27 girls, mark it with a /; otherwise, mark it with an x. Repeat for 100 samples. How many / were recorded? Can you answer the question now?

25. The Spider and the Fly Problem A spider and a fly are sitting at opposite corners of a 5 x 5 square grid. The spider moves randomly east or north and the fly moves randomly west or south at the same rate. Eind the probability that they meet for lunch.

s

26. The Birthday Problem a. If two people are selected at random, what is the probability that their birth¬ days are different? (Assume there are no leap-year birthdays.)

b. A third person appears. What is the probability that the three birthdays are all different?

c. What is the probability that at least two have the same birthday? d. Find a formula that will generate the probability that in a group of n ran¬ domly selected people at least two have the same birthday. Can you write a computer program that would generate these values?

e. Using the formula from part (d), complete the following table of values.

f.

Draw a graph plotting n against the probability of at least one matching birthday and join the points by a smooth curve.

g. In a room containing thirty people, show that the proba¬ bility of at least two people having the same birthday is greater than 60%.

p j

q

q

_

^

h. How many people must be in the room before you are absolutely certain that two will have the same birthday?

lO

27. What is the probability that in a group of four randomly selected people at least two: a. were born on the same day of the week? b. have the same sign of the zodiac? c. have the same birth month?

28. Johnny Follet, a gambler, made this statement: “On this road we’ve been meet¬ ing about 20 cars an hour. I’ll give you even money that in the next hour, two cars will have the same last two digits in the licence number.”(i^ (This is known as a Sucker Bet.) a. What is the probability that Johnny Follet will win? b. What odds should he give? The Spoilers, Desmond Bagley, Fontana Books, Pub. Wm. Collins, 1969.

29. Two red cards and two black cards are selected from a deck, shuffled, and placed face down on a table. Two of these cards are selected at random. What is the probability that there is one of each colour?

288

9

Applications Of Probability

“The best laid schemes o’ mice and men gang aft a-gley.” With this statement Robert Burns reminds the reader that chance events abound in everyday activities. Chance is a factor in determining the sex of a child, and a chance encounter may help determine one’s future course in life. Chance or random events occur in industry. A machine may turn out components with precise dimensions, but now and then it inexplicably turns out a defective part. People on production lines build superb cars but occasionally, for no apparent reason, a lemon is created. The random production of defects is undesirable but unavoidable. The telephone system is designed on the theory that calls are made randomly. The system becomes jammed when telephone calls cease to be randomly distributed dur¬ ing a day, as in power emergencies when many people try to contact the power com¬ pany at the same time. The occurrence of random events does not imply chaos or the absence of order. Al¬ though we cannot be absolutely certain that the next roll of a pair of dice will turn up a total of seven, we know that in 6000 trials the total seven will turn up about one sixth of the time. In the course of repetition, order emerges from chaos. This order is described by the rules of probability. The purpose of this chapter is to extend these rules and apply them to the everyday world. The mathematician, by applying both probability and statistics, attempts to make sensible predictions about seemingly unpredictable events. Laws of probability, like laws of motion, help in prediction of future events.

289

9-1 Probability Distributions One mathematical idea that is important to the development of the laws of probability is that of a set. Another fundamental concept, the function, will now be considered to further develop the subject of probability. The following problem illustrates the use of a function. A supermarket manager is concerned about the length of time a customer may wait in the check-out line. A survey is made in which the length of time each customer waits is recorded to the nearest minute. Each customer is assigned the value of a function, r, representing waiting time. T may take on the values 0, 1, 2, 3, . . . .To each customer check-out (c) a whole number of minutes (r) is assigned. T\ check-out (c)

time (r).

Since the value that Ttakes on for a particular customer is unpredictable, Tis called a random variable. A random variable is a function that assigns a real number to each outcome of an experiment. Suppose that the record for one of the check-out counters from 10:00 to 11:00 on a Saturday was as follows:

Time

Frequency

1 2 3 4 5 6

2 3 2 4 3 1

The number of times that the function T takes on the value 5 is 3. This means there are 3 customers who wait 5 min. Abbreviate Ttakes on the value 6>/5 as T = 5. In the experiment, the number of outcomes for which T = 5 is 3. The idea of a function can be used by assigning a probability number to each value of the random variable. In the check-out counter experiment, the set of possible values of the random variable, along with the corresponding relative frequencies, is given below.

T

290

Frequency

Relative Frequency

1 2 3 4 5 6

2 3 2 4 3 1

0.13 0.20 0.13 0.27 0.20 0.07

Total

15

1.00

The relative frequencies can be considered as probabilities. The probability of the outcome for which T = 4 is 0,27. The notation P{T = 4) = 0.27 is often used to express this fact. Thus, we also have:

P{T P{T P{T P{T P{T

= = = = =

1) 2) 3) 5) 6)

= = = = =

0.13 0.20 0.13 0.20 0.07

There is just one probability for each value of the random variable. The set of all possible values of a random variable such as T, along with the corres¬ ponding probabilities, is called a probability distribution. {(1, 0.13), (2, 0.20), (3, 0.13), (4, 0.27), (5, 0.20), (6, 0.07)} EXAMPLE 1

The probability distribution in the experiment of finding the num¬ ber of heads {H) in the tossing of two coins is given as follows:

Value of H Probability

0

1

2

1

1

1

4

2

4

Find P{H = 0), P{H = 1), and P{H = 2). SOLUTION

P{H = 0) is the probability of H having a value of zero (no heads), P(H = 0) = \ = 0.25 Similarly,

P(H = 1) = ^ = 0.5 P(H = 2) = \ = 0.25

The preceding example illustrates that a probability distribution defines a function called a probability distribution function or simply a probability function. EXAMPLE 2

For the experiment of rolling two dice, let the random variable X represent the sum of the up faces.

a. Make a probability distribution table. b. Find P{X = x) for x e |2, 3, 4, 5, . . . , 12}. c. Draw the graph of the probability function. 291

SOLUTION

a.

X

Frequency

2

1

3

2

4

3

5

Relative frequency or Probability 1 = 0.028 36

2 36 3 36 4 36 5 36 6 36 5 36 4 36 3 36

4

6

5

7

6

8

5

9

4

10

3

11

2

12

1

Total

= 0.056 = 0.083 =

=

0.167

= 0.139 =

0.111

= 0.083 0.056 =

0.028 1.000

36

2

_ _

P{X = 3) = P{X = 11) = 36

PiX = 4) = P{X = 10) = X 36 4

_ ^

P{X = 5) - P{X = 9)

= 36

P(X = 6) = P(X = 8)

= 36

P(X = 7)

2 1

♦

♦

♦

i

♦ •

♦ t ? ^ • 4 • i 4.• ■ #.♦#

♦4 #

-#■

0.111

= 0.139

36 36

5 4

• • # •

12

2 1

6

3

b. P{X = 2) = P{X = 12) = 4 36

36

c. The probability graph follows.

In the example of rolling 2 dice, there were exactly 11 outcomes. In this example, there are a finite number of outcomes and X is called a discrete random variable. The corresponding probability function is called a discrete function.

292

2_ 36

_

p(X=n)

3

4

5

6

A continuous random variable can be a real number within some interval. For example, the waiting time in the supermarket could have been a continuous random variable if it were not rounded to the nearest minute. The process of assigning probability numbers to a continuous variable becomes diffi¬ cult if an attempt is made to apply the same process used for discrete variables. This problem can be overcome if an area graph is used instead of a point graph. If we use the example of rolling a die, the probability distribution can be graphed as follows: V

P(X = x) . ‘

Area Graph

•

Point Graph •••••

X

1

1

1

1

2

•

1

3

4

1

5

6

123455'

• The probability of an outcome v is represented by the area of the rectangle centred on outcomes. For discrete random variables, each outcome is assigned the central position of the base of the rectangle. • The total area of the rectangles is 1.

1.5

2

2.5

P{x = 2) = area of rectangle centred on 2 is 1 x f. D

The area model can be adapted to continuous random variables. EXAMPLE 3

A long-distance phone call is scheduled to come to your home between 6:00 p.m. and 8:00 p.m. In returning home, you miss your bus and do not arrive until 6:25 p.m.

a. Draw a graph showing the probability distribution of the time of arrival of the call. b. What is the probability that you receive the call? SOLUTION

Assuming that the call is no more likely to come at one time than any other, we assign equal probabilities to each equal interval of time. Let X be the number of minutes after 6:00 p.m., where X is a con¬ tinuous random variable. The graph of the distribution is a rectan¬ gle whose area is 1. This is usually referred to as a uniform or rectangular distribution. The width of the rectangle is 120 (time interval of 120 min.). Thus, the height is j ^q (area = 1). 1 120

j X —► 0

25

120

The probability that you receive the call is equal to the area of the unshaded region.

P(X > 25) = (120 - 25)

X jIq = ^ = 0.79

Note: the probability of getting a call at exactly 7:00 p.m. is zero. Why? The preceding problem could have been answered more easily by comparing the interval of time that you were available with the whole interval of time. There is value, as will be shown, in using the same graphical model for the probability distri¬ butions of both discrete and continuous random variables.

EXERCISE 9-1 A 1. Identify the following random variables as being discrete or continuous:

a. b. c. d. e.

your birthdate, your mass at birth, your time of arrival at school, the total outcome in rolling 3 fair dice, the outcome of a roulette wheel.

2. For each of the following uniform probability distributions, establish the vertical scale. Devise an experiment to fit each graph.

a.

____

b.

-—I—-1——r—-- -I—*—

0

c.

12

0.1

3

—-0

^

0.2

0.3

0.4

0.5

0.6

0.7

d.

60

20

^

40

3. A time study was commissioned by an office manager to analyse the time people spent copying materials at the copy machine. The results are shown in the table below. Time is rounded to the nearest minute.

a. Define the random variable X. b. Using the frequency distribution table, find P{X = x) for X e {1, 2, 3, . . . , 10}.

294

0.8

4. At City Bank, transaction times with the tellers during a one-hour period are recorded. Time, in the table showing the results, is rounded to the nearest minute.

Time

Frequency

1 2 3 4 5 6 or more

18 14 10 6 2 0

a. Define the random variable X. b. Using the frequency distribution table, find P{X = x) forx e {I, 2, 3, 4, 5,6]. 5. A pair of dice are rolled and the sum of the up faces is recorded as follows:

Sum of Up Faces 4 3 6 3 00

6 5 9 11 11

4 7 11 4 8

7 7 7 6 9

4 4 6 8 7

6 2 5 8 11

7 12 6 8 7

7 10 6 8 10

10 9 9 6 7

3 10 9 11 9

a. Define the random variable X. b. Make a frequency distribution table. c. Find P(Z = x) for.v e {2, 3, 4, 5, ... , 12}. d. Draw an area probability graph for the experiment. 6. a. Draw an area probability graph for the experiment of tossing a coin n times, where the random variable X represents the number of heads for each of the following cases:

(i) n = 2,

(ii) n = 3,

(iii) n = 4.

b. Using the graphs, show the regions associated with each of the following events: (i) there are exactly 2 heads,

(ii) there are less than 2 heads.

7. a. Draw an area probability graph for the experiment of rolling two dice . Let the random variable X represent the total outcome of the two dice, b. Illustrate and calculate the probability of the event toss a 7 or an 11.

8. A drawer contains 4 red and 3 blue socks. Four socks are drawn, without re¬ placement.

a. Draw an area probability graph where the random variable X represents the number of red socks. b. Calculate the probability of “at least one pair” and illustrate this event on the graph.

9. A stop watch with a sweep second hand is stopped randomly.

a. What is the probability that the hand has stopped between 40 s and 50 s? b. Illustrate your answer using an area probability graph. c. The watch is started again and stopped randomly. What is the probability that it stops in the same interval? d. The watch is started and stopped twice randomly. What is the probability that it stops between 15 s and 20 s both times?

10. Traffic light A at a busy intersection is red for 20 s, amber for 5 s, and green for 35 s in the direction you are driving. Passing through this intersection, you make a right turn and are confronted with a second, unconnected traffic signal B whose timing pattern is red for 25 s, amber for 5 s, and green for 40 s.

a. If it is legal to enter an intersection on an amber signal and the traffic signals operate independently, what is the probability that a law-abiding driver (i) will not be stopped? (ii) will be stopped twice? b. Show the probability distribution for each traffic signal using an area graph.

11. The graph shown represents an area probability graph for the arrival time of a commuter at her office.

a. Give an interpretation of the horizontal scale and decide upon a valid variable. b. Establish an appropriate vertical scale. c. What is the probability that the commuter is there on time, assuming that r = 0 is the beginning of the workday? d. What is the probability that the commuter will be on time for a whole week? What did you assume in this calculation? C 12. A rope 20 m long is cut into two segments randomly and each part is used to form the boundary of a square.

a. Draw the probability graph in which the random variable L is the length of the longer part. b. Find the probability that the larger square has an area greater than 9 m^. c. Find the probability that the total area of the two squares is greater than 20.5 m2.

296

9-2 The Binomial Distribution In Chapter 8, tree diagrams were used as models for events that occurred more than once. Tree diagrams, however, become cumbersome when an experiment is repeated many times. For example, a young couple may be planning to have a family of four children. What is the probability of having exactly two boys and two girls? What is the probability of having exactly four girls? These questions can be answered by making use of the tree diagram shown or by making use of counting methods.

A second approach involves counting the number of subsets that have the desired characteristic and comparing that number with the total number of possible events. The number of ways of having 2B or 2G is

=

6.

There are 2"^ = 16 possible

families. We assume the probabilities of having a boy or a girl are equal. Thus, each possible family is equally likely. Then, the probability of having 2 boys and 2 girls is-|or|. An idea introduced earlier in the book makes use of the same counting procedure. The binomial expansion of {b + g)^ is

The binomial expansion can be used to calculate probabilities that apply to succes¬ sive events. If the probability of having a boy baby is b, and the probability of having a girl baby is g, then the probability of having exactly 2 boys and 2 girls is

This expression is the third term of the binomial expansion for {b -f g)^:

(b + gf = Assume that/? = g =

+ (3)^'^’ + (4)^'*The probability of having 2 boys and 2 girls is

297

The binomial model can be used to find the probability of having four girls. 1

The probability of having 4 girls is

The binomial model used above applies to situations that are equivalent to drawing from a hat with replacement. These are called Bernoulli Trials.

Bernoulli trials have the following characteristics: • They have exactly 2 outcomes, success or failure. • Each trial is independent. • The probability of each outcome is the same for each trial of the experiment.

The probabilities for various events determined by a Bernoulli experiment can be found easily by using the binomial expansion. {q + pY =

+

+

4- . . . +

+

. . ■

In this expansion, p = probability of success on any single trial, q = probability of failure, and p + q = 1. The general term {^q^~^p^ represents the probability of having exactly r successes in n trials if the probability of success is p. EXAMPLE 1

What is the probability of having at least 2 girls in a family of 4 children?

SOLUTION

Let b = probability of a boy and g = probability of a girl, where b = g = ^. There are 4 repetitions of the experiment having a

child. {b + gf = M The probability of having at least 2 girls is the sum of the probabilities of having 2, 3, or 4 girls.

+

(f)bg^

+

= 6 X

+ 4 X

(0

+

16

An alternative approach considers the complementary event. The probability of having at least two girls is the difference between 1 and the probabilities of having 4 or 3 boys.

U 16

298

0

+

A sequence of Bernoulli trials has a binomial distribution. If the random variable X represents the number of successes in n trials, then PiX = x) = Here p = the probability of success on any single trial of the experiment and

p + q = I, or P{X

EXAMPLE 2

a. Write a probability function for the experiment of tossing a coin i. twice, ii. three times, iii. four times, where the random variable X represents the number of heads, b. Draw the area probability graph of each distribution.

SOLUTION

Let t represent the probability of a tail and h the probability of a head.

a. i. (t + hf- =

+ 2ht +

PiX = x)

Assume h = t = \ 2 -X

P{X = x) =

1

PiX = 0) =

4

P(X = 1) = (i

2

1

4

2

P(X = 2) = Q

1

Number of heads

4 PiX =

ii. {t + /z)3

= ^3 +

3^2/^

4.

3^/^2

X)

4.

The probability function is given by 3

-X

P(X = x) =

Selects: = 0, 1, 2, 3 and calculate

P{X = 0) = ^,

P{X = 1) = 8

P{X = 2) =

PiX = 3) = i

Number of heads

’

299

iii. (r + hf =

+ At^h + 6tV + Ath^ +

The probability function is given by

Select X = 0, 1, 2, 3, 4 and calculate

P(X = 0) = P(X = 4) = V, P(X = 1) = P{X = 3) = i P{X = 2) = |. P{X =

X)

1 2 3 8 4 J_

8

0

2

3

4

The graph of a distribution often shows characteristics of an experiment that specific formulas may obscure. To illustrate this point, consider the following example of a binomial distribution in which the probability of success is not equal to the probabil¬ ity of failure. EXAMPLE 3

A bag contains 100 marbles of which 20 are yellow and 80 are blue. The bag is shaken, a marble is drawn and returned, and the bag is shaken again. This process is repeated three more times.

a. Write the probability function, where B = the number of blue marbles. b. Draw an area probability graph of this function. c. Find the probability of drawing i. exactly 3 blue marbles, ii. at least 3 blue marbles. SOLUTION

a. In the binomial expansion (y + Let b =

letn = 4.

^ be the probability of drawing a blue marble

and y = ^ the probability of drawing a yellow marble. (y + by =

+ Ay^b -I- 6y^b^ -I- Ayb^ + b^

The probability function for x blue marbles is

300

4

b. P{B - 0)

P{B =

X)

625 16 625

P{B = 1) P{B - 2)

1

96 625

P(B = 3) = III P(B = 4) = Iff

B = number of blue marbles

c. i. PiB = 3) = Iff The probability of drawing exactly 3 blue marbles is

9 S^

ii. P(B > 3) = P(B = 3) + P(B = 4) = Iff The probability of drawing at least 3 blue marbles is

5 1 2

The binomial distribution can be used to illustrate an episode from the early history of probability. One of the problems sent by Le Chevalier de Mere to Pascal was the following. De Mere knew the odds favoured at least one six in four rolls of a die. Example 4 will show these odds to be 671 : 625. What would the odds be in favour of a double six occurring in 24 tosses of two dice? De Mere’s thinking was as follows. Knowing that two dice had 36 possible outcomes (6 times that of a single die), de Mere felt that if 6 times as many tosses (24) were carried out with a pair of dice, the double six would be favoured. A method of solving this problem will be established by considering the simpler problem of tossing one die.

EXAMPLE 4

Find the odds in favour of tossing at least one six in four rolls of a die.

SOLUTION

This is a Bernoulli trial. Consider the complementary event to roll

a number other than a six. P(at least one six) = I - P (no sixes)

625 1296 _

671 1296

The odds in favour are 671 : 625. You will be asked to solve de Mere’s problem in Exercise 9-2, Problem 17.

301

Bernoulli trials can be applied to solve problems in which the experiment can be repeated indefinitely. EXAMPLE 5

Two noblemen have challenged each other to a duel. Nobleman A has a shooting accuracy of 70% and Nobleman B has a shooting accuracy of 40%. Nobleman A, being a gentleman, gives B the first shot and they take turns after that. What is the probability that

B wins the duel? SOLUTION

The progress of the duel is shown in the following tree diagram.

To find the probability that B wins the duel, find the probability of reaching each end of a branch marked B wins.

P(B wins on 1st shot) = 0.4 wins on 2nd shot) = 0.6 x 0.3 x 0.4 = 0.18 x 0.4 P(i5 wins on 3rd shot) = (0.18)^ x 0.4 and so on P(5 wins) = 0.4 + 0.18 X 0.4 + (0.18)^ x 0.4 + . . . = 0.4(1 + 0.18 + (0.18)2 + . . .) 1 + 0.18 + (0.18)2 + . . . is an infinite geometric series and

a Soo

=

.■.P(5wins) =

1 - r

, |r| < 1

0.4 1 - 0.18

20 41

An alternative solution can be attained by considering the following argument: At the end of round 1, P{B wins) = 0.4, P{A wins) = 0.6 x 0.7 = 0.42. Thus, the odds in favour of B winning are 0.4 : 0.42 or 20 : 21; therefore, P{B wins) = |y.

302

The binomial distribution can also be used to explain some common fallacies or mis¬ conceptions in probability. In the book The Bridge Over the River Kwai the following excerpt occurs with regard to paratroopers jumping into the jungle: . if they do only one jump, you know, there’s a fifty percent chance of an injury. Two jumps, it’s eighty percent. The third time, it’s dead certain they won’t get off scot free. You see? It’s not a question of training, but the law of averages.” EXAMPLE 6

If a paratrooper has a 50% chance of injury on any one jump, what is the probability of not being injured after

a. two jumps? b. three jumps? SOLUTION

a. This is a binomial situation in which n = 2. {q + pf =

+ 2qp + p^,

where q = ^ (the probability of an injury) and P = ^ (the probability of no injury). The probability of not being injured after two jumps = p^ = ^.

b. This is a binomial situation in which n = 3 and p = q ^ (q

+

p)^

=

q^

+

3q^p

4-

3qp^

+

p^

The probability of no injury after three jumps =

EXERCISE 9-2 A

1. The following terms come from different binomial distributions. Describe each distribution by finding i. n, the number of repetitions, ii. p, the probability of success on any single trial and iii. the number of successes.

303

3. Describe situations for which the binomial expansions in question 2 could deter¬ mine the probabilities. B

4. A die is tossed eight times and two occurs 4 times.

a. In a binomial model, the terms of the expansion of (r a 7 occurs, in which case the player loses.

a. Find the probability of winning on the first roll. b. Find the probability of losing on the first roll. c. Find the probability that a player rolls the following points and then subse¬ quently wins. i. four ii. ten iii. six d. Find the probability that the player wins at dice.

19. Yahtzee In the game of Yahtzee, a player has three tries at rolling some or all of a set of five dice, attempting to achieve such results as four of a kind or two pairs. If the player rolls a pair of fours on the first toss, and rolls only the non-fours showing on subsequent tosses. Find the probability of rolling five 4s.

9-3 Mathematical Expectation In a game of chance, the pay-off can be associated with the probability of success in that game and can be determined by the formula Expectation = Pay-off x P (winning). Suppose there is a prize of $10. To win this prize, a person must get either 3 heads or 3 tails in three tosses of a coin.

P (winning) = |

|

Mathematical expectation = ^ x $10 = $2.50. Mathematical expectation does not mean a person will win $2.50 on any particular attempt since one can either win or lose, thereby receiving either $10 or $0. Mathe¬ matical expectation means that in the long run a person will win $2.50 per game. If 100 games are played the winnings will likely be close to $250, but unlikely to be exactly $250. Mathematical expectation should govern how much to invest or bet. In this example, if it costs $3 for each attempt at winning $10, a player would lose 50C per game in the long run. In many government lotteries the mathematical expectation is that players will get back about 50 cents for each dollar bet. Expectation is a way of combining results when there are many outcomes. It is the total or sum of the products of outcomes and their corresponding probabilities.

EXAMPLE 1

Ten cards are marked as follows: $1 is written on 4 cards, $2 is written on 3 cards, $3 is written on 2 cards, and $5 is written on 1 card. The cards are shuffled and one card is drawn. The prize is the amount that appears on the card drawn.

a. Calculate the mathematical expectation of the game. b. If the game operator charges $2 to play the game, does a player win or lose in the long run?

SOLUTION

Let P{X = x) = probability of drawing a card worth $x.

P(X = I) = -^

P(X = 3) = -|

P{X = 2) = X

P(X = 5) = X

a. Mathematical expectation = $(-^xl+-j^x2+-^x3 + -jLx5

= $0^) = $2.10 b. In the long run, players will win IOC per game.

306

r-pi

..

The average winnings of the group

4x$l+3x $2 + 2

This can be rewritten ^ x $1 + ^ x $2 + ^ x $3 +

X

$3 + 1

X

-

$5

x $5

and is the same as the expectation of each person in the game. If 10 people play the game and each person receives one of the 10 cards, the prize money would be divided such that 4 3 2 1

people people people person

win $1, win $2, win $3, and wins $5.

EXAMPLE 2

A raffle is being held by the local hockey league. The first prize is $1000, there are two second prizes of $500 each, and 10 prizes of $20 each. If 5000 tickets are sold at $1 each, find the expected winnings of a person buying one ticket.

SOLUTION

A ticket qualifies for first prize if it is drawn first. A ticket qualifies for a second prize if it comes up on the second or third draw. A ticket qualifies for a third prize if it comes up on any of the fourth through the thirteenth draws. Consider the probabilities in turn. The probability that a ticket is drawn first is The probability that a ticket is drawn second is: 4999 5000

1 ^ 1 4999 5000’

4999

since it must not be drawn on the first draw (with probability Jq^)Similarly, the probability that a ticket comes up on the third draw is also

_J_ 5000-

4999 y 4998 ^ 1 ^ 1 5000 4999 4998 5000

The probability that a ticket comes up on any draw is This result can be determined using symmetry. Since there is no bias as to the position at which a ticket is drawn, it is equally likely that it can occur on any draw, meaning one chance out of 5000. From the preceding discussion we know that: P(first prize) -

First Prize $1000

^,

5

^(second prize) =

^ +

5

and since the outcomes are mutually exclusive,

Second Prize $500 Third Prize $20

P(third prize) =

307

Hence, the expected winnings are $1000 X ^ + $500 X

+ $20 X 5^ = $0.44.

Consequently, in the long run, a person would lose 560 each time a raffle ticket was purchased. Note that the average winnings of a ticket purchaser _ total amount of prize money number of purchasers $(1000 + 2 X 500 + 10 X 20) 5000 ■ ^

,,

This is equal to the mathematical expectation.

Mathematical expectation can be applied to many business and economic situations in which chance can affect the outcome. A useful model for these situations is a decision tree. The following example illustrates the use of a decision tree.

EXAMPLE 3

A business person is negotiating a deal worth $5000. She assumes that there is a 0.7 probability of concluding the deal if she makes a trip and negotiates in person. The trip will cost her $1500 in transportation, meals, and hotel expenses. If she does not make negotiations in person, the probability of making the deal is only 0.4, and the cost of phone calls and miscellaneous expenses is $500. Should she make the trip and negotiate personally or not?

SOLUTION

The decision tree is shown below. success

failure success

failure

The mathematical expectation if the deal is arranged personally is $(0.7 X 5000 -f- 0.3 X 0) = $3500. Therefore, the expected net gain is $3500 - $1500 = $2000. The mathematical expectation if the deal is not arranged person¬ ally is $(0.4 X 5000 + 0.6 x 0) = $2000. Therefore, the net expectation is $2000 - $500 = $1500. Therefore, the person should make the trip and negotiate the deal personally.

308

EXAMPLE 4

A basketball player is in the process of shooting two foul shots. If her probability of success on each shot is 0.7, what is the number of points she is expected to score?

SOLUTION

Method I Construct a tree diagram as shown.

s

M

S

S = Successful Shot

M

M = Missed Shot

The expected number of points = 2 X (0.7)2 + I X (0.7)(0.3) + 1 X (0.3)(0.7) + 0 = 1.4.

Method II Let the random variable X represent the number of successful foul shots scored. Let E{X) be the expected number of points. Multiply the probability by the point outcome. 2 X P(X = 2) = 2 X (0.7)2 ^ 0.98 1 X P(X - 1) = 1 X 2 X (0.7)(0.3) = 0.42 0 X P(X = 0) = 0

E{X) = 0.98 + 0.42 = 1.4

The number of points that the player is expected to score in this situation can be thought of as the average number of points the player will score in the long run. Using the probability distribution with random variable X, the mean of the distribution E{X) can be defined. E{X) can be thought of as a weighted average of the values assumed by X. In general, if there are n outcomes n

The mean, or expected value, of a distribution may not be a possible outcome of the experiment. The mean represents the expected or average result if the experiment were repeated over and over.

309

The mean of a distribution is an important characteristic of a distribution, and it is worth investigating whether or not this value can be found easily. EXAMPLE 5

Find the mean for a binomial distribution with probability of suc¬ cess p in the case

b. n = 3.

a. n = 2, and SOLUTION

Each success is assigned the value 1 and each failure the value 0. Consider the random variable X, which measures the number of successes in n trials.

a. n = 2,

(q + p)^ =

+ 2qp + p^

P(X = 0) = q^ = {\ - pf P{X = \) =2qp P{X = 2) = p^ Therefore,

E(X) = 0

X

q^ + I X 2qp -h 2 x p^

= '^piq + P) = 2p, since q + p = 1.

b. n = 3,

(q + pf’ = q^ + 3q^p + 3qp^

p^

P(X = 0) = q^ P{X = 1) = 3q^p P(X = 2) = 3qp^ PiX = 3) = p^ Therefore,

E(X) = = = =

0 X q^ + I X 3q^p 3p{q^ + 2qp + p^) 3p{q pf 3p, since q + p = 1.

2 x 3qp^ + 3 x p^

In general it can be shown that the mean of a binomial distribution with n repetitions and probability of success p on any single trial is np. A proof of this statement follows.

(q + pY = q^ +

{^q''~^P^ +

+

. +

p\

where q + p = \. 'p' + 2x (^^q" ^p^ + . . . + rx (^^q" 7^' + . . . + np'^

E(X) = Oxq" + lx fj

—

I

= np[q"^^ + 2 X —-—q"~^p^ + ... + p"~^]

310

After removing the factor np, the general term becomes:

r(n - 1)! (n - r)!r!

(n - 1)! (n — r)l(r - 1)!

V

This is the general term of (q + pY ~ ^. Therefore,

E{X) = np{q + pY ^ = np, since q + p = \.

A gambler should be aware of the expected return on an investment in the many situations that can occur both legally and illegally. Such organizations as provincial lotteries or casinos allow the expected loss to be calculated. If a game has expectation 0 it is called di fair game. Lotteries and casinos do not operate fair games.

EXERCISE 9-3 B

1. The following table gives the four cash prizes and the probability of winning each prize for a lottery held to aid a school soccer team. If each ticket costs $5, what is the net expectation for a player?

Winning

$1000

$100

Probability

0.001

0.005 0.01 0.02

$50

$10

2. A time study is done at the local supermarket. Three check-out clerks are observed and the length of time customers wait in line is observed and categorized. Based upon the survey the probability that a customer waits for each of the given time intervals is estimated.

0 to 2 min 2 to 4 min 4 to 6 min 6 to 8 min 8 to 10 min Clerk A

0.1

0.2

0.25

0.35

0.1

Clerk B

0.05

0.3

0.3

0.25

0.1

Clerk C

0.1

0.3

0.3

0.15

0.15

In the long run, which clerk is the fastest? (Use the mid value of the time interval as the representative value for that interval.) 3. A businesswoman is contemplating a trip to try to arrange a business deal worth $5000. The trip will cost $500, but if she goes, there is an 80% chance of con¬ cluding the deal successfully. If she does not take the trip, she can try to arrange the deal by phone at an estimated cost of $200, or by mail at a smaller cost of $50. There is a 60% chance of making the deal by phone but only a 40% chance of making the deal by mail. What should she do and why?

311

4. The Scotch Mist umbrella store finds that on rainy days its expected profit is $500 per day, on cloudy days it is $300 per day, and on sunny days its expected loss is $200 per day.

a. For the month of October, the weather office predicts the following proba¬ bilities: P (rain) = 0.3, P (cloud) = 0.5, and P (sun) = 0.2. Calculate the expected daily profit for the store in October. b. For the month of July, the weather office predicts P (rain) =0.1, P (cloud) = 0.4, and P (sun) = 0.5. Calculate the expected daily profit for the store in July. 5. A bowl contains 10 red marbles and 10 blue marbles. A game operator lets the player draw two marbles and makes the following bet. If both marbles are the same, the player wins $1 and if they are different, the operator wins $2.

a. Find the probability that the player wins. b. Find the mathematical expectation of the player. 6. What is the expected outcome of one roll of a single die? 7. In a one and bonus situation, a basketball player is allowed a second free throw only if she makes the first. If she is successful in 70% of her free throws, what is the expected number of points she will make in each one and bonus situation?

8. A fortune-teller advertises, with a money-back guarantee, that for only $10 she can correctly predict the sex of an unborn child. What is her expected profit on each prediction? C 9. A con artist takes 8 coins out of his pocket and asks the customer to guess the number of heads that will turn up. The con artist gives 2:1 odds in favour of the customer. What is the expected profit of the con artist if the customer bets $1?

10. Overbooking

\

Fly-By-Night Airlines has observed that 5 % of the people who buy airline tickets do not show up for flights. As a consequence FBN sells 105 tickets for each plane having 100 seats.

a. What is the probability that a flight is overbooked? b. In 1000 flights, how many flights are expected to be overbooked? 11. Expectation and Life Insurance The break-even premium charged by an insurance company is the expected amount to be paid if an insured person dies within one year. For example, the probability that a 17-year-old male does not survive one year is The break-even premium on a $100 000 life policy is $127.36. See the life tables on page 390.

x $100 000 or

Find the break-even premium on a $100 000 policy for each of the following:

a. a 20-year-old female c. a 40-year-old female

312

b. a 20-year-old male d. a 40-year-old male

12. Boyd’s of Boston, the insurance conglomerate, offers flight insurance for FlyBy-Night’s Space Shuttle. The company’s statistics show that FBN has a failure rate of only 1 flight in 2500. What should Boyd’s charge for a $10 000 policy if they wish to have a profit margin of 20%?

13. Roulette A European roulette wheel has 37 numbered slots 0, 1, 2, . . . , 36. There are 18 black (the even numbers) and 18 red (the odd numbers) and 0 is coloured white. An American roulette wheel has the same 36 coloured slots, but instead of a white 0 it has a green 0 and another green 00. In both European and Ameri¬ can casinos, betting black is an even money bet.

a. Calculate the probability of winning by betting black on each kind of wheel. b. Calculate the net expected return on each dollar bet on each kind of wheel. (When you bet $1, the casino is obliged to match your dollar.) European Roulette

American Roulette

14. A property developer wants to develop a condominium village near a mountain that has great potential for a ski resort. The success of the development lies in discovering a suitable water supply. A drilling company does a feasibility study. Based upon local experience, it costs $2000 to drill a well, and the probability of hitting a reservoir of water is 0.65. The drilling company submits a price of $200 000. They are paid only if water is found.

a. What is their expected profit if they drill just one well? b. If two wells are drilled, what is the probability that at least one strikes water? What is the expected profit? c. What number of wells should be drilled to maximize the expected profit?

15. Chuck-a-Luck Chuck-a-Luck is a game played in many small carnivals around North America. The operator has a cage containing 3 dice. The dice are rolled and a player can bet on the occurrence of any single number from one to six. If a player bets $1, and one of her numbers occurs, the operator pays her $ 1. If two of her numbers occur, the operator pays her $2. If three of her numbers occur, the operator pays her $3. Suppose six friends each bet $1 on a different number. What would the operator’s profit be

a. b. c. d.

if the 3 dice showed 3 different values? if a pair and a singleton occurred? if all three dice showed the same outcome? Find the operator’s expected profit on each $600 bet.

313

16. The St. Petersburg Paradox The rules of the game are as follows. A player (P) tosses a coin and the house agrees to pay the player $x if P gets a head on the first toss, $x^ if P gets a head on the second toss, $jc^ if P gets a head on the third toss, and so on. Find the mathematical expectation of P when a. jc = 1,

h. X = 1.99,

c. jc = 2.

The paradox occurs in the case when x = 2. Here the mathematical expectation of P is infinite! For further reading and discussion of this paradox see the follow¬ ing books. Lady Luck by Warren Weaver and Probability with Statistical Appli¬ cations by Mosteller, Rourke, and Thomas. 17. The One-Armed Bandit Present-day slot machines have 3 wheels, each with 20 slots on them. Each slot contains one or two symbols (cherries, oranges, etc.) The number of arrange¬ ments of the slots that can appear on the pay line is 20 x 20 x 20 = 8000. The number of times each symbol appears on a wheel is shown in the following table: Reel 1 Cherries Oranges Plums Bells Bars Melons 7s

Reel 2

2 5 7 1 3 2 1

6 5 3 5 2 2 1

Reel 3 0 5 4 9 1 3 1

In some cases, two symbols appear in one slot. For instance, if a melon and an orange appear in one slot, then either may be used. a. The following chart indicates the winning arrangements and the pay-off. Com¬ plete the table, find the total number of ways of winning, and find the total pay-off in 8000 plays.

Winning Arrangement 1 2 3 3 3 3 3 3

cherry on reel 1 only Cherries Oranges Plums Bells Melons (Jackpot) Bars (Jackpot) Sevens (Double Jackpot)

Reel 1

Reel 2

Reel 3

Pay-off Ways

2

14

20

560

Total in coins Pay-off

b. Show that the take of the casino is 5.55% in the 8000 plays of the game.

314

2 5 10 14 18 100 100 200

1120

9-4 Repeated Trials Without Replacement The binomial distribution provides a model for experiments that have two outcomes and are repeated under identical circumstances. For example, a card is drawn from a well-shuffled deck and then is replaced. The deck is shuffled again and a second card is drawn. If the card is not replaced before the second draw, there are still two possi¬ ble outcomes at each repetition of the experiment; however, the probabilities are not the same for each trial. Problems of this nature are examples of hypergeometric distribution and have already been studied in Chapter 8. This section will further develop this topic. Drawing objects one at a time without replacement is equivalent to drawing a set of objects once from a group of n distinct objects. The first way of considering the problem can be modelled using tree diagrams; the second makes use of counting subsets.

EXAMPLE 1

Three marbles are drawn from a bag that contains five blue and four yellow marbles.

a. Draw a probability graph showing the distribution of the num¬ ber of blue marbles that can be drawn and their probabilities.

b. Find the mean of the distribution. SOLUTION

a. Let X represent the number of blue marbles. The number of subsets with 3 marbles is f ^ j • The number of subsets with

and P(X = 3)

exactly 3 blue marbles is

42

-

10

Similarly, P(X = 2) =

21

’

PiX = x)

and

b. 0 X P(X = 0) = 0 1 X P(X = 1) = -^

2

X

X

P(X = 2) =

3 X PCX = 3) = ^ £(X) = 0 +

^

5 3

-

2

Therefore, the mean is 1^^.

315

These probabilities can also be found by using a tree diagram, but that procedure is much longer. A hypergeometric distribution involves a set of n objects that can be divided into two groups of 5 successes and n - s failures. From this set, m objects are drawn one at a time without replacement. If X represents the number of successes in each draw.

, where

^ j = the number of subsets of size m, selected from the whole group of = the number of subsets of size x, selected from the set of successes 5, and

f n - s \ = thg number of subsets of size m - x, selected from the set of n - 5 failures. \m - X/ EXAMPLE 2

A manager has a short list of candidates consisting of 6 men and 4 women from his staff. The manager selects 3 people randomly from the list to form a promotion committee.

a. Establish a probability distribution for the random variable W = the number of women. Illustrate your answer with an area probability graph.

b. Find the probability that there is at least one woman on the committee. SOLUTION

The number of subsets of size 3 selected from 10 people is

PiW)

2

3

6

0

The area probability graph is shown.

b. P(W > 1) = 1 - P{W = 0) = 1 - i 6

316

5

6

12

3

Many problems associated with such card games as poker or bridge may be solved using these methods. EXAMPLE 3

In the game of poker, 5 cards are dealt from a 52-card deck to each player. A deck of cards contains 4 suits each having 13 cards. A flush is a hand in which all 5 cards are the same suit and do not form a straight, which is a sequence of 5 consecutive cards. Find the probability of a flush.

SOLUTION

The total number of possible poker hands is In any one suit there are

•

flushes.

Since there are 4 suits in the deck of cards, there are 4 flushes. This includes, however, 40 straight flushes (that is, A 2 3 4 5, 23456, . . . ,10JQKAin the 4 suits) that must be subtracted.

P (any flush)

The hypergeometric distribution can be extended to situations that involve more than two categories, such as those that occur in the game of bridge. EXAMPLE 4

Find the probability that a bridge hand of 13 cards will contain 5 spades, 4 hearts, 3 clubs, and 1 diamond.

SOLUTION

The total number of possible bridge hands is There are

and

52 ^

ways of selecting 5 spades, ( V)

selecting 4 hearts,

(3^)

selecting 3 clubs,

(\^)

selecting 1 diamond.

Therefore, the probability that a bridge hand will have the required

distribution is

317

The combinatorial symbol

becomes difficult to calculate when n is large and r is

not close to n or 0. For example,

^

but it is very time consuming to evaluate ^

^

relatively easy to calculate,

^ . It can be shown that the binomial

distribution gives a reasonable approximation to the hypergeometric distribution for large values oi n. To illustrate this point, consider the following example: EXAMPLE 5

In ajar of 100 Jellybeans there are 80 green and 20 yellow jellybeans. A handful of 5 jellybeans is selected at random. What is the proba¬ bility that there are exactly 3 green jellybeans in this selection?

SOLUTION

There are

^ 5^^ possible subsets of 5 jellybeans.

• Use the hypergeometric distribution, and

P(G = 3)

= 0.2073.

• Now, use the binomial distribution and assume that each jelly¬ bean is returned after being selected, even though this is not a correct procedure for the problem. P(G = 3) = ('5j(0.8)3(0.2)2 = 0.2048 Even for such a relatively small value of n, the approximation using the binomial distribution is quite good, and it can be shown that the approximation improves as n becomes larger. The length of time needed to calculate the result using binomial methods is much faster than using hypergeometric formulas. A useful application for this approximation is sampling in statistics.

EXERCISE 9-4 A

1. The following expressions are from different hypergeometric distributions. In each case identify i. n, the number of objects in the set, ii. m, the size of the subset, iii. 5, the number of successes in the given set, and iv. jc, the number of successes in the subset.

318

2. Each of the following expressions is missing some information. Write the terms needed to form complete hypergeometric expressions.

B 3. A bag contains 10 identically shaped blocks, 5 blue and 5 green, from which 3 blocks are selected.

a. Draw a probability graph showing the distribution of blue blocks. b. Calculate the probability of drawing at least 2 blue blocks. c. Find the mean of the distribution. 4. Repeat question 3 using a bag that contains 6 blue and 4 green blocks. 5. A jury of 12 people is being selected for a sexual discrimination trial. The defence lawyers know the list of potential Jurors contains 20 women and 15 men. What is the probability that there will be a majority of women on the jury if it is selected randomly? 6. On a roster of 20 astronauts, 8 are women and 12 are men. A crew of 4 is selected randomly for a simulated flight. Find the probability that

a. there are exactly 2 women in the crew, b. the crew is all female, c. there are at least 2 women in the crew. 7. When forming a parliamentary committee of 5 people, the cabinet minister has 11 rural members and 16 urban members from which to choose. If the commit¬ tee is selected randomly, find the probability that

a. there are 3 rural members on the committee, b. there are at least 3 rural members on the committee. 8. Poker Find the probabilities of each of the following poker hands:

a. 3 aces and 2 kings, b. 3 deuces and 2 jacks, c. a full house (3 cards of one kind and a pair). 9. Bridge Find the probability of each of the following bridge hands:

a. b. c. d. e. f.

7 hearts, 6 spades, 4 clubs, 3 diamonds, 3 hearts, 3 spades, 4 aces, 4 kings, at least 1 ace, no aces, no kings, no queens, no jacks, no tens (a Yarborough). 4 clubs, 5 diamonds, 3 hearts, 2 spades

319

In questions 10-12 use a binomial distribution to estimate the hypergeometric situation.

10. Quality Control The quality control inspector at the Popsicle Factory estimates that 2 % of the Popsicles shipped by the factory are broken.

a. In a box of 20 Popsicles find the probability that i. there are 0 broken Popsicles, ii. there is 1 broken Popsicle, iii. there are 2 broken Popsicles, iv. there is at least 1 broken Popsicle. b. In a shipment of 1000 boxes, how many boxes would you expect to contain at least one broken Popsicle?

11. Sampling From a population that contains 60% Whig voters and 40% Tory voters, a sam¬ ple of 20 people is selected randomly.

a. Find the probability that 10 people will support the Whigs. b. Find the probability that at least 10 people will support the Whigs. c. What is the expected number of Whig supporters in the sample?

12. In Toronto, 40% of the households have V.C.R.s. In a random sample of 10 households what is the probability that

a. exactly 4 households have V.C.R.s? b. exactly 3 or 5 households have V.C.R.s? 13. Find the probability of the following distributions in the game of bridge: a. 4-3-3-3

b. 7-6

c. 5-5-2-1

14. Poker Hands a. How many different poker hands (5 cards) can be dealt from a deck of 52 b. c.

d.

e.

cards? How many different poker hands containing 4 aces can be dealt? What is the probability of being dealt 4 aces? A straight flush is a run of 5 cards of the same suit. How many different straight flush poker hands can be dealt? An ace can be used for a high or a low straight: lOJQKA or A2345. Which is more likely to occur, a straight flush or 4 aces?

15. Calculate the probability of each type of poker hand. See Chapter 6, page 209. 16. A four flusher is a term derived from the game of poker. In the game of 5 card stud a player is dealt one card down and four cards up. If the four cards dealt up are the same suit, then the player may have a true flush (5 cards of the same suit) or be a four flusher. What is the probability that a person showing 4 cards of the same suit is a four flusher?

320

9-5 Markov Chains I In many experiments such as tossing a coin, each toss of a coin is regarded as an independent trial. In other kinds of experiments, such as flicking a light switch, the outcome depends upon the previous state of the light. A Markov chain is an experi¬ ment in which each outcome depends only on the outcome of the preceding trial. EXAMPLE 1

A “random watchdog” home lighting system is designed so that once each hour, a light switch will be activated with probability If the lights are “on” when the owner turns on the system, find the probability that the lights will be “on” at the end of the second hour.

SOLUTION

A tree diagram can illustrate the possibilities.

P(“On” after hour 2) = P(light is on for 2 h) + P(light changed twice)

The outcomes, such as “on” or “off’ ’, are called states of the system and the proba bilities | (of changing state) and ^ (of maintaining state) are called transition probabilities. Transition probabilities can be summarized in a transition arrow diagram. 2 3

Markov chains have applications in such diverse areas as weather prediction, genetics, and economics.

321

In the “battle of the colas”, market researchers have found that the probability that a person who buys “brand C” now will buy it again tomorrow is 0.8. The probability that he/she will buy “brand P’ ’ tomorrow is 0.2. They have also found that the probability of a person who buys “brand P” now will buy “brand P” tomorrow is 0.7, and the probability that he/she will switch to “brand C” is

EXAMPLE 2

0.3. a. Draw a transition arrow diagram illustrating these probabilities. b. Find the probability that a person who buys brand C today will also buy brand C two days from now.

b. We have the following cases (starting at C): Today

Day 1

Day 2 Probability

C

0.8

C

0.8

C

0.8

X

0.8 = 0.64

C

0.2

P

0.3

c

0.2

X

0.3 = 0.06 0.70

Therefore, the probability that a person who buys “brand C” today will buy it two days from now is 0.70. Matrices can also be used to summarize transition data. A transition matrix consists of columns that are probability vectors whose components are transition probabilities. In the “battle of the colas”, we have the following matrix: From C

P

C

0.8

0.3

P

0.2

0.7

To

This element gives the probability of going from state P to state C.

One advantage of using a matrix model is that multiplication of matrices can be used to determine the transition probabilities several stages from now. For example, using the “battle of the colas” transition matrix, 0.8 0.2

0.3 0.7

we can evaluate and interpret T^;

322

0.8 0.2

0.3 0.7_

0.7 0.3

0.45" 0.55

0.8 .0.2

0.3 0.7

Consider the element 0.7. As we have seen, 0.7 = (0.8)(0.8) + (0.2)(0.3) is the probability that a purchaser of “brand C” will buy “brand C” two days from now. This is the element in the first row and first column of T^. From C P

cro.7 0.45

To

pLo.3

0.55

= T2

This matrix shows that not only is the result from example 2 easily found, but also that all other possible transition probabilities for two days from now are illustrated. It can be shown that higher powers of T, such as and will determine transition probabilities three and four days from now. Transition matrices are especially useful when there are more than two components in the probability matrix. EXAMPLE 3

In the mini-van field. General Motors (GM), Ford (F), and Toyota (T) dominate the market. Over the past few years, market research indicates that the following transition arrow diagram describes annual buyer preference.

GM 0.8

a. Write an annual transition matrix in the format: From GM F T To

GM r F T

b. Evaluate T^. c. Find the probability that a person who buys a Toyota now will buy another Toyota two years from now. SOLUTION

a.

To

GM

From F

T

GM ‘0.80 0.05 F T _0.15

0.05 0.70 0.25

0.20 0.20 0.60

323

b.

"0.80 0.05 _0.15

T2 =

0.05 0.70 0.25

0.20“ 0.20 0.60_

0.125 0.5425 0.3325

0.6725 0.105 0.2225

0.80 0.05 _0.15

0.05 0.70 0.25

0.20" 0.20 0.60_

0.29 0.27 0.44

c. The probability that a person buys a Toyota mini-van two years from now given that he/she bought one today is 0.44, as shown in row 3, column 3.

EXERCISE 9-5 A 1. For each of the following transition arrow diagrams find the corresponding transition matrices.

4^ I Ui

b.

2. For each of the following transition matrices form a transition arrow diagram.

b.

Q

1

Li

C

F

0.9

0.4

X

1 1

0.6_

Y

1 A

Z

1 1

_1

Q

O

Q Q

C

J

F .0.1

1^

c.

X

Y 2 4

4

Z 5 10

2-“ 0.4

B 3. The transition arrow diagram illustrates the purchase pattern for laundry detergents X and Y. For example, the probability that a person buying brand X now will buy brand X next time is 0.8 and buy brand Y next time is 0.2.

0.2

a. What is the probability that a person who buys brand Y today will switch brands next time? b. Write a transition matrix T in the format

From X Y To

c. Evaluate

X Y

and T^.

d. What is the probability that a person who buys brand X now will buy brand X (i) two times from now? (ii) three times from now?

324

4. Suppose the probability that a car will start easily on a given day depends on whether it started easily on the previous day. The probabilities are summarized in the following transition matrix (S = start easily, N = does not start easily). Now Next Day

S [0.95 0.05

S N

N 0.9' 0.1_

a. What is the probability that the car will start easily on Wednesday if it started easily on Monday? b. What is the probability that the car will start easily on Friday if it did not start easily on Wednesday? 5. In the city of Erewhon, which has a fixed population, it has been observed that in each 5-year period there has been a popu¬ lation shift involving the city core(C) , the suburbs(S), and rural areas(R) according to the transition arrow diagram shown at the right.

0.7

a. Write a population transition matrix T in the format: From C S R

c To

C

r

S R

b. FindT^. c. What is the probability that, in ten years, a person who lives in the city now will be living (i) in the city? (ii) on a farm?

6. The working population of Erewhon is categorized as management(M), white collar(W), or blue collar(B). Data on the jobs held by children of the working population shows that children of the management group are distributed among the jobs with 40% being in management, 40% being white-collar workers, and 20% being blue-collar workers. For the children of the white-collar workers, 60% are white-collar workers and the rest are evenly divided between manage¬ ment and blue-collar jobs. For the children of blue-collar workers, 40% are in the same type of jobs as their parents, while the rest are twice as likely to be whitecollar workers as working in management. From M W B

a. Form an arrow diagram illustrating the transition data. b. Form a transition matrix T in the format shown. To c. Evaluate T^. d. Find the probability that the grandchildren of blue-collar workers are: (i) blue-collar workers, (ii) in management.

M ' W B

9-6 Markov Chains II Markov chain transition matrices can be used to store transition data and determine transition probabilities two and three stages from now, as has been shown. They can also be used to determine long-term effects on a given population. To illustrate this, the market research data on the consumer buying patterns of two headache remedies, Asperol and Tyrin, is given by the transition arrow diagram: 0.20

From A T and the transition matrix: To

A [0.6 T LO.4

0.2 0.8

Suppose further that a test is performed on a sample of people 50% of whom now use Asperol and 50% of whom now use Tyrin. This data can be written as a column matrix and is called a probability vector.

Consider the matrix product:

rl V =

0.6 .0.4

0.2 0.8_

0.5 .0.5_

0.4 ~

0.6_

_

This means, that after one purchase, 40% of the group studied use Asperol and 60% use Tyrin.

'0.6

0.2' o bo 1_

HV =

o

If we denote HV as V', we can form another matrix product. '0.4' .0.6_

0.36 ” _0.64_ This means, that after a second purchase, 36% of the original test group now use Asperol and 64% use Tyrin. It should also be noted that the matrix product: HV = H(HV) , since HV = V The product H(HV) can be shown to be the same as (HH)V or H^V.

326

This agrees with our result from the previous section, that is, if a situation can be represented by a Markov chain transition matrix T, and the Markov process were applied twice, the matrix would give the transition matrix data. In general, T" will represent the transition probabilities n stages from now. To study long-term effects, we need to find large powers of the transition matrix. EXAMPLE 1

Use the headache remedy transition matrix A A T

H

To.6 0.4

T

0.2 0.8

and the initial market vector V

0.5 0.5

a. Find: (i) H2,

(ii) H4,

(iii) H8.

b. Predict the market share of Asperol after 8 purchases.

c. Does the market share seem to be stabilizing? SOLUTION

a. (i)

"0.6 _0.4

0.2" 1

o bo

_

"0.6

(iii)

H8

K

H*V =

0.44 0.56

0.28 0.72

0.44 0.56

0.28 0.72

0.28“ 0.72_

"0.44 _0.56

0.3504 0.6496

0.3248 0.6752

0.3504 0.6496

0.3248 0.6752

0.3338 0.6662

0.3331 0.6669_

‘0.3338 .0.6662

0.2“ 0.8_

0.3504 0.6496

0.3331' 0.6669.

0.3248 0.6752

"0.5“ _0.5_

0.3334

0.6666_ c. The probability column vectors of

are both about the same and both approximately equal to the market share vector after 8 purchases.

327

Using this somewhat artificial model, which assumes • the population remains fixed, and • the buying habits of Asperol and Tyrin purchasers maintain the market prefer¬ ence as predicted by the original transition arrow diagram, we arrive at the conclusion that after many purchases the market share stabilizes or reaches an equilibrium state. Can this state be found without finding high powers of the transition matrix?

Let the equilibrium vector This vector must have the following characteristics: • Vj + V2 = 1 It is a probability vector. • HVg = Vg

EXAMPLE 2

Since it is the equilibrium state and applying H does not change it.

Find the equilibrium vector for the headache transition matrix

^

SOLUTION

0.6 I0.4

~

0.2 0.8

Since HV^ =

• «

•

0.6 .0.4

0.2' 0.8_

^1 LV2J

^1 LV2J

0.6vj -1- 0.2v2 0.4vj -t- 0.8v2_

LV2J

This matrix equation gives rise to two linear equations:

0.6v, +

0.2v2

= Vj

(1)

0.4vj + O.8V2 = V2

(2)

Each of these equations simplifies to: 2V|

= Vj

Combining this with Vj -I- V2 = 1, we have

1

^ and V2 =

2

the equilibrium state is given by In the long run, if the buying pattern continues, ^ of the market will belong to Asperol and ^ of the market will belong to Tyrin.

328

Not all transition matrices have the characteristic of reaching an equilibrium state. A

0 1

simple example of a transition matrix that does not is:

I 0

If a transition matrix does produce an equilibrium state, it is called regular.

EXERCISE 9-6

_0.2

o

c.

_1

(ii)

1

(i) o oo

o

1 b.

4 3 4^

o

2 1 ^2

1

1

o

1

o '■O

r

a.

1__1

1. For each of the following transition matrices T, find:

1

A

0.6_

2. For each of the transition matrices in question 1, find the probability vector V such that TV = V. B

3. Researchers have found that ^ of the offspring of white mice are white, and ^ piebald. Only ^ of the offspring of piebald mice are white, the rest being piebald. Deduce the eventual proportion of piebald mice in a population of white and piebald mice. 4. In a study on population migration, it is found that 2 % move annually from rural to urban areas, while 1 % move from urban to rural areas. Predict the eventual population distribution between rural and urban areas, if the trend continues. 5. In training a rat in a maze, it was found that if the rat found the right path to the food it remembered it ^ of the time. If it did not find the food, the probability of the rat’s finding it next time was only On what proportion of the rat’s attempts would you expect it to find the food? 6. Three dogs are all fighting for the same bone. When A has it, there is a probability of ^ that he will keep it for 5 min, and a probability of ^ that he will lose it to B, and similarly for C. The other probabilities are given in the matrix: From

ABC A To

B C

1 4 3 4

0

0 3 4 ^

If A has the bone initially, find which dog is most likely to have the bone after a.

10 min,

b. 20 min,

c. eventually.

329

7. Long-term observation of a mathematics teacher who gives surprise quizzes shows the following pattern: • If she gives a quiz one day, she won’t quiz the next. • If she doesn’t give a quiz one day, she quizzes the next day ^ of the time. We see that the probability that a quiz is given on a particular day depends on what happened the previous day. Predict the long-term pattern of quizzes for this teacher.

C

8. In genetics, animals possessing gene combination AA or Aa (where A repre¬ sents a dominant gene and a represents the corresponding recessive gene) will exhibit the dominant trait (attached earlobes, for example) while only those pos¬ sessing the aa combination will exhibit the recessive trait (detached earlobes). A A is a dominant combination, Aa hybrid, and aa recessive. Suppose hybrid males are mated with 1024 dominant females, 512 hybrid females, and 512 recessive females. Hybrid-dominant matings will produce ^ dominant, ^ hybrid offspring. Hybrid-hybrid matings will produce ^ dominant, ^ recessive, and

hybrid offspring. Hybrid-recessive matings will produce ^ hybrid and ^

recessive offspring. The transition matrix for the process is: From D H R

To

D

^

H

1 1 1

R

0

2

2

^

0

4^

2

= T

2

1 1 4

a. Find: (i) T2,

2^

(ii) Tl

b. Find the population vector after (i) 1 transition, (ii) 2 transitions,

(iii) 3 transitions.

c. Predict the “long-run” probability vector. d. Find the long-run population distribution of 2048 females. 9. See question 5 on page 325 for the transition diagram for the city of Erewhon. If the city population is 10 000, the suburban population is 20 000, and the rural population is 1000, find the population in each region:

a. in five years

330

b. in ten years

c. eventually.

REVIEW

UNIT 9

1. a. Annie Oaklie is in a rifle competition. Her probability of hitting the target is 0.95. If she shoots 15 times, what is the probability that she hits the target more than 12 times?

b. If Annie’s probability of getting a bull’s eye is 0.20, what is the probability that she gets more than 4 bull’s-eyes in 15 shots? 2. Twenty coins are poured from a sack onto a table. What is the probability that

a. exactly 10 coins come up with heads? b. more than 10 coins come up with heads? 3. If 60% of the voters in California favour Clint Westwood for mayor, what is the probability that, in a random sample of 25 voters, the majority favour him? 4. A census of a Canadian city of 40 000 families shows that 80% of the families have refrigerators. Ten families are selected randomly for a sociological and economic survey. What is the probability that 7 or fewer of these families have refrigerators?

5. a. If 5 % of all children are left-handed, what is the probability that there is at least one left-handed student in a Grade 10 class of 20 students selected randomly?

b. If the caretaker provides just one chair with a left-handed writing surface (the other 19 are right-handed) in a classroom, what is the probability that all students using this classroom will have an appropriate chair? 6. Five cards are taken from a deck of cards and each card is replaced before the next card is selected. What is the probability of picking

a. 5 spades?

b. 3 red cards?

c. 4 tens?

7. Repeat question 6 but draw the cards without replacement.

. a. Draw an area probability graph for a binomial distribution with=

8

and (i) n = 4,

(ii) n = 5,

3

,