SOLUTIONS TO
PROBABILITY AND STATISTICS WITH APPLICATIONS: A PROBLEM SOLVING TEXT SECOND EDITION
Leonard A. Asimow, Ph.D., ASA Mark M. Maxwell, Ph.D., ASA
ACTEX PUBLICATIONS, INC. WINSTED, CONNECTICUT
Copyright 2015 by ACTEX Publications, Inc.
All rights reserved. No portion of this book May be reproduced in any form or by and means Without the prior written permission of the Copyright owner.
Requests for permission should be addressed to ACTEX Learning PO Box 715 New Hartford CT 06057 ISBN: 978-1-62542-838-7
Copyright 2015 by ACTEX Publications, Inc.
All rights reserved. No portion of this book May be reproduced in any form or by and means Without the prior written permission of the Copyright owner.
Requests for permission should be addressed to ACTEX Learning PO Box 715 New Hartford CT 06057 ISBN: 978-1-62542-838-7
TABLE OF CONTENTS Chapter 1: Combinatorial Probability 1 1.1 The Probability Model ..................................... ........................................................ ....................................... ....................................... .........................1 ......1 1.2 Finite Discrete Models with Equally Likely Outcomes ..................................... ...................................................2 ..............2 1.3 Sampling and Distribution ................................... ...................................................... ...................................... ...................................... ......................8 ...8 1.4 More Applications ...................................... ......................................................... ..................................... ..................................... ............................... ............10 1.5 Chapter 1 Sample Examination Examination ...................................... ......................................................... ....................................... ............................. .........13 Chapter 2: General Rules of Probability 15 2.2 Set Theory Survival Kit...................................... .......................................................... ....................................... ...................................... ..................... .. 15 2.3 Conditional Probability Probability ..................................... ........................................................ ....................................... ....................................... ...................... ...23 2.4 Independence ..................................... ........................................................ ....................................... ....................................... ...................................... ...................29 2.5 Bayes’ Theorem ..................................... ........................................................ ....................................... ....................................... ................................. ..............35 2.6 Credibility ...................................... .......................................................... ....................................... ...................................... ....................................... ....................38 2.7 Chapter 2 Sample Examination Examination ...................................... ......................................................... ....................................... ............................. .........40 Chapter 3: Discrete Random Variables 51 3.1 Discrete Random Variables ................................... ...................................................... ...................................... ...................................... ...................51 3.2 Cumulative Probability Distribution ..................................... ....................................................... ..................................... ....................... ....52 3.3 Measures of Central Tendency ...................................... ......................................................... ..................................... .............................. ............ 53 3.4 Measures of Dispersion ...................................... ......................................................... ...................................... ..................................... ...................... ....56 3.5 Conditional Expectation and Variance .................................... ...................................................... ..................................... ..................... .. 59 3.6 Jointly Distributed Random Variables (Round 1) .................................... ....................................................... ...................... ...62 3.7 The Probability Generating Function .................................... ....................................................... ...................................... ...................... ...65 3.8 Chapter 3 Sample Examination Examination .................................... ...................................................... ..................................... ................................ .............66 Chapter 4: Some Discrete Distribution 73 4.1 The Discrete Uniform Distribution ...................................... ......................................................... ....................................... ....................... ...73 4.2 Bernoulli Trials and the Binomial Distribution .................................. ..................................................... ............................ .........76 4.3 The Geometric Distribution ..................................... ........................................................ ....................................... ................................... ............... 78 4.4 The Negative Binomial Binomial Distribution ....................................... .......................................................... ....................................... ....................82 4.5 The Hyper-Geometric Distribution ..................................... ......................................................... ....................................... ....................... ....83 4.6 The Poisson Distribution ..................................... ......................................................... ....................................... ...................................... .....................84 4.7 Summary Comparison of Discrete Distributions ...................................... ........................................................ ..................... ...89 4.8 Chapter 4 Sample Examination Examination .................................... ...................................................... ..................................... ................................ .............93 Chapter 5: Calculus, Probability and Continuous Distributions 97 5.2 Density Functions ..................................... ........................................................ ....................................... ....................................... ............................... ............97 5.3 Great Expectations ....................................... ........................................................... ....................................... ....................................... ........................... .......99 5.4 Mixed Distributions............................. Distributions................................................. ....................................... ....................................... .................................. .............. 104 5.5 Applications to Insurance: Deductibles and Caps Caps ...................................... ........................................................ ..................105 5.6 The Moment Generating Function ..................................... ....................................................... ..................................... ........................ ..... 107 5.7 Chapter 5 Sample Examination .................................... ...................................................... ..................................... .............................. ........... 111
iv
TABLE OF CONTENTS
Chapter 6: Some Continuous Distributions 117 6.1 Uniform Random Variables ...................................... .......................................................... ....................................... ............................... ............ 117 6.2 The Exponential Distribution ...................................... .......................................................... ....................................... ............................. ..........118 6.3 The Normal Distribution ..................................... ........................................................ ....................................... ...................................... .................. 122 6.4 The Law of Averages and the Central Limit Theorem................................... Theorem................................................. ..............124 124 6.5 The Gamma Gamma Distribution ...................................... ......................................................... ....................................... ..................................... .................127 6.6 The Beta Family of Distributions ...................................... .......................................................... ....................................... ....................... ....129 6.7 More Continuous Distributions ....................................... ........................................................... ....................................... ......................... ...... 132 6.8 Chapter 6 Sample Examination Examination ...................................... ......................................................... ....................................... ........................... .......134 Chapter 7: Multivariate Distributions 139 7.1 Joint Distributions for Discrete Random Variables ................................... ..................................................... ..................139 7.2 Conditional Distributions – The Discrete Case .................................... ...................................................... ........................ ......141 7.3 Independence – Discrete Case...................................... ......................................................... ....................................... ............................. ......... 143 7.4 Covariance and Correlation ...................................... .......................................................... ....................................... ................................ .............143 7.5 Joint Distributions for Continuous Random Variables ................................... ................................................ .............147 147 7.6 Conditional Distributions – The Continuous Case ..................................... ....................................................... ..................151 151 7.7 Independence and Covariance in the Continuous Case................................... ................................................ .............153 7.9 Bivariate Normal Distribution ....................................... .......................................................... ....................................... ............................ ........ 158 7.10 Moment Generating Function for a Joint Distribution .................................... ................................................. .............159 7.11 Chapter 7 Sample Examination ....................................... ........................................................... ........................................ ......................... .....159 Chapter 8: A Probability Potpourri 167 8.1 The Distribution of a Transformed Random Variable ..................................... ................................................. ............167 8.2 The Moment-Generating Function Method ...................................... .......................................................... ........................... .......182 8.3 Covariance Formulas...................................... Formulas......................................................... ...................................... ....................................... ........................ ....183 8.4 The Conditioning Formulas ....................................... .......................................................... ....................................... ............................... ........... 183 8.5 Poisson Processes Revisited ....................................... .......................................................... ....................................... ............................... ........... 187 8.6 Chapter 8 Sample Examination Examination ...................................... ......................................................... ....................................... ........................... .......191 Chapter 9: Statistical Distributions and Estimation 197 9.1 The Sample Mean as an Estimator ....................................... ........................................................... ....................................... .....................197 9.2 Estimating the Population Variance ...................................... ......................................................... ...................................... .....................198 9.3 The Student t -Distribution -Distribution ..................................... ........................................................ ....................................... .................................... ................200 9.4 The F -Distribution................................................. -Distribution.................................................................... ....................................... .................................... ................201 9.5 Estimating Proportions ..................................... ........................................................ ....................................... ....................................... ..................... .. 203 9.6 Estimating the Difference Between Between Means ....................................... ........................................................... ........................... .......204 9.7 Estimating the Sample Size .................................... ........................................................ ....................................... .................................. ............... 205 9.8 Chapter 9 Sample Examination Examination ...................................... ......................................................... ....................................... ........................... .......206 Chapter 10: Hypothesis Testing 211 10.1 Hypothesis Testing Framework Framework ..................... ........................................ ...................................... ...................................... ........................ ..... 211 10.2 Hypothesis Testing for Population Means ...................................... .......................................................... ............................. ......... 211 10.3 Hypothesis Testing for Population Variance .................................... ....................................................... ............................ .........213 10.4 Hypothesis Testing for Proportions ..................................... ........................................................ ...................................... ...................... ...213 10.5 Hypothesis Testing for Differences in Population Means..................................... ............................................ .......214 214 10.6 Chi-Square Tests .............................. ................................................ ..................................... ...................................... ....................................... ....................215 10.7 Chapter 10 Sample Examination ....................................... .......................................................... ...................................... ........................ ..... 216
TABLE OF CONTENTS
v
Chapter 11: Theory of Estimation Estimation and Hypothesis Testing Testing 221 11.1 The Bias of an Estimator .................................... ....................................................... ..................................... ..................................... ..................... .. 221 11.2 Building Estimators ....................................... ........................................................... ....................................... ...................................... ........................ .....221 11.3 Properties of Estimators ...................................... .......................................................... ....................................... ..................................... .................. 228 11.4 Hypothesis Testing Test ing Theory .................................... ...................................................... ..................................... .................................... .................232 11.5 More General Likelihood Ratio Tests ...................................... .......................................................... .................................... ................ 235 11.6 Bayesian Estimation ....................................... ........................................................... ....................................... ....................................... ....................... ...237 11.7 Chapter 11 Sample Examination ....................................... .......................................................... ....................................... ........................ ....244 Chapter 12: A Statistics Potpourri 251 12.1 Simple Linear Regression: Basic Formulas ..................................... ........................................................ ............................ .........251 12.2 Simple Linear Regression: Estimation of Parameters ...................................... .................................................. ............256 12.3 Comparing Means Using ANOVA ...................................... ......................................................... ...................................... ..................... .. 258 12.4 Non-Parametric Tests T ests .................................... ........................................................ ....................................... ....................................... ........................ ....259 12.6 Chapter 12 Sample Examination ....................................... .......................................................... ....................................... ........................ ....260
PREFACE
TO
SOLUTION MANUAL
Our solution manual contains detailed solutions to the questions in our Probability and Statistics with Applications textbook. Frequently, multiple solution methods are illustrated. Our goal is to help you learn and utilize calculus-based probability. Your level of understanding will reflect the level of thought and the connections that you make while solving problems (practicing), not just the solutions written by others (reading). When you are attempting to solve problems, we recommend that you be careful and take your time. Always define variables that you introduce. If you do not know how to start, rewrite the question. Then write down facts that you know that might help you solve the problem. Then think. The time invested in thinking through solutions to problems may seem excessive, especially to busy students juggling many activities. However, the investment will pay strong dividends down the road in terms of your comprehension and your ability to make those critical connections to by now familiar routines. Look at the written solution as your last resort. When looking at a solution, try to minimize your reliance on this crutch. Specifically, once you have seen enough to get unstuck, try to finish your solution on your own. Finally, go back and read the written solution carefully to see if there are insights or shortcuts that will help you in the future. The most ineffective method to attempt to understand any mathematics is by initially looking at a written solution. Learning mathematics is about your journey, not the punch-line. Every year we hear students say “I understood in class and I understood while doing homework, but I did not understand on the exam.” What they mean to say is “I followed when the teacher solved a problem in class and I followed reading a written solution, but the first time I was responsible to solve a problem on my own, I could not.” For students wishing to master this material, a useful exercise is to think about ways to modify a question that you solved into a new question and/or to think of alternate solutions. MMM
SOLUTIONS TO PROBABILITY AND STATISTICS WITH APPLICATIONS A PROBLEM SOLVING TEXT
CHAPTER 1: COMBINATORIAL PROBABILITY Section 1.1 The Probability Model
1-1
(a) We systematically list the sample space. Main Course
Side
Beverage
hamburger hamburger hamburger hamburger hamburger hamburger chicken sandwich chicken sandwich chicken sandwich chicken sandwich chicken sandwich chicken sandwich
potato chips potato chips potato chips coleslaw coleslaw coleslaw potato chips potato chips potato chips coleslaw coleslaw coleslaw
soda water milk soda water milk soda water milk soda water milk
(b)
{hamburger & potato chips & soda, hamburger & coleslaw & soda, chicken sandwich & potato chips & soda, and chicken sandwich & coleslaw & soda}.
(c)
Pr(soda with meal)
number of meals with a soda 4 . total number of possible meals 12
Note:
Reducing
4 1 has the interpretation that one of the three beverages is a 12 3
soda.
1-2
(a)
The event that doubles are rolled is the set of outcomes 11, 22,33, 44,55,66 . There are 6 ways to roll doubles.
(b)
Pr(sum is prime) Pr(sum equals 2, 3, 5, 7, or 11)
15 . 36 1
2
1-3
SOLUTIONS TO EXERCISES – CHAPTER 1
(a) Legend
P 1
First plain M&M® package
P 2
Second plain M&M package
P 3
Third plain M&M package
1
N 2 B D A
First peanut M&M package Second peanut M&M package Peanut butter M&M package Dark chocolate M&M package Almond M&M package
There are two ways to list the sample space. If you grab two packages at the same time, then the outcome of DA represents one package of dark M&M’s and one package of almond M&M’s. In this case DA is the same outcome as AD (order does not matter), and therefore only needs to be listed once in the sample space. Then there are 28 outcomes: { P 1 P 2 , P 1 P 3 , P 1 N 1 , P1 N2, P1 B, P1 D, P1 A, P 2 P 3 , P 2 1 , P 2 2 , P2 B, P2 D, P2 A, P 3 1 , P 3 2 , P3 B, P3 D, P3 A, 1 N 2 , N 1 B, N 1 D, 1 A, 2 B, N 2 D, 2 A, BD, BA, DA}. If you envision the experiment as grabbing one package and then grabbing a second package (in other words, order matters), there will be 56 elements in the sample space. (b) If order doesn’t matter the answer is
15 . If order matters then the answer is 28
30 15 30 . Of course , so both perspectives lead to the same answer so long 56 28 56 as you are consistent in your treatment of the sample space. (c)
18 36 . 28 56
Section 1.2 Finite Discrete Models with Equally Likely Outcomes
1-4
There are 3 2 6 36 current outfits
3
2
6
Shoes
Pants
Shirts
Adding shoes will allow 4 2 6 48 different outfits, adding a shirt will allow 3 2 7 42. You increase the total number of outfits most by adding to the item that you have the least. The problem mentions shoes or shirts only, so the correct answer is “shoes.”
PROBABILITY AND STATISTICS WITH APPLICATIONS: A PROBLEM SOLVING TEXT
1-5
(a)
3
63
6
6
Red White (b)
6 Blue
There are six choices for the blue die. 1 1 6 1 (c) Pr(first die is 3 and second die is 4) . 36 63 (d) No, tree diagrams are not so useful if tracking more than 50 or so outcomes. 6
1-6
109
1-7
10 23 26 23 10 10 10
1-8
One could try to use a tree diagram to systematically list all possible way to rob four banks. The thief could rob any bank first. He has only five choices of bank to rob second (he does not rob the bank he robbed first). Similarly, the thief has five choices of bank to rob third (it just cannot be the bank he robbed second, but it could be the bank he robbed first). The total is 6 5 5 5 750.
1-9
8 10 10 9 10 10 10 10 10 10
1-10
(a)
(b) (c)
7! 7! 7! (7 6)! 1! 13! 13 12 11 10 9 8 7 6 5 4 3 2 13! 1! 7! 7 6 5 4 840 3!
(d)
20
(e)
4! 24
(f)
Undefined
(g)
23 22 21
4
SOLUTIONS TO EXERCISES – CHAPTER 1
1-11
1-12
(h)
3! 1 3!
(i)
5! 120
(j)
10
(k)
7! 765 35 3! 4! 3 2 1
(l)
1
(a)
17!
(b)
16!
(c)
16! 1 17! 17
Kendra gets her own exam, the other 16 exams can be redistributed to any of the remaining 16 students.
1 8 P 3
654
1-13
6 P 3
1-14
12 P 4
1-15
r 23. Solve 1
1-16
r 4. Solve 1
1-17
Pr[people get off on unique floors]
1-18
35 P 12 12
35
365 P r r
365
7 P r r
7
.5 by trial and error.
.4
11 P 7 7
11
PROBABILITY AND STATISTICS WITH APPLICATIONS: A PROBLEM SOLVING TEXT
10 P n n
1-19
n 5. Solve 1
1-20
(a) 7 C 6
7! 7 1! 6!
(b)
13 C 12
(c)
13 C 1
10
.5 by trial and error for the minimum n.
13! 13 12! 1!
13! 13 1! 12!
(d) 5 C 2
5! 10 3! 2!
(e) 4 C 4
4! 1 4! 0!
Remember that 0! 1.
(f) Undefined. (g) 5 C 3
5! 10 2! 3!
(h) 4 C 0
4! 1 0! 4!
(i) 5 C 5
5! 1 5! 0!
(j)
5867 C 1
(k)
7 P 4
4!
(l) 0 C 0
5867! 5867 5866! 1! 7! 7 C 3 35 3! 4!
0! 1 0! 0!
1-21
Both expressions equal 220.
1-22
n Cr
n! n! n C nr . r ! ( n r )! ( n r )! r !
Placing r toppings on a pizza is equivalent to keeping n r off the pizza.
5
6
SOLUTIONS TO EXERCISES – CHAPTER 1
1-23
1-24
6 C 4
The multiplication principle says that the total number of possible pizzas is Number of ways to select toppings ways to select size ways to select crust 13 C 2 3 C 1 2 C 1
468 7 C 1 6 C 1 3 C 1 2 C 1 252 7 C 4 3 C 1 2 C 1 210
(a) (b) (c)
1-25
(a) (b)
15 C 5 4 C 1 5 C 2 6 C 2
1-26
11 C 2 3 C 1 5 C 1
1-27
14 C 3 3 C 1 7 C 2 5 C 1
1-28
(a)
10 C 2 20 C 2 8550
(b) To match exactly one white ball, your lottery ticket must match one of the two “good” white balls. There are 2 C 1 2 ways to do this. The second white ball could be any of the remaining eight “bad” balls. 16 2 C 1 8 C 1 2 C 2 8550 8550
match 1 match 1 match both good bad white blue balls white
1-29
(a)
(b)
1-30
10 C 1
6
60 C 6
.01997
choose 1 piece from 4 choose 2 choose one of the chosen pieces for choose 5 variety w/2 variety w/2 varieties varieties pieces pieces 4 10 C2 6 C5 5 C1 10 C1
4 C3
60 C 6
5 C3
6 C4 4 10 15
centers forwards guards
4 C1
.2697
5 C2
6 C2
centers forwards guards
starters
PROBABILITY AND STATISTICS WITH APPLICATIONS: A PROBLEM SOLVING TEXT
1-31
1-32
15 15! 5! 3! 7! 5,3,7
15 C5
start 5 of the 15
10 C3
7 C 7
3 of remaining 7 of the final 10 are on the bench 7 do not dress
4 4! 12. The sample space of possible words is 2!1!1! 2,1,1
The number of words is
{hoop, hopo, hpoo, pooh, poho, phoo, ohop, ohpo, opho, opoh, ooph, oohp} Pr(chosing HOOP)
1 12
1-33
5 5 3 1 5! 30 2, 2,1 2! 2! 1! 2 2 1
1-34
12 12! 1 , 1 , 3 , 1 , 1 , 1 , 1 , 2 , 1 1! 1! 3! 1! 1! 1! 1! 2!1! P’s e’s n’s s’s y’s l ’s v’s a’s i’s
1-35
(a)
Deal 5 cards to the first poker player. There are 52 C 5 ways to do this. Next, deal 5 of the remaining 47 cards to the second player and continue in this fashion. The total number of ways to distribute six 5-card poker hands is 52! 52 C5 47 C5 42 C5 37 C5 32 C5 27 C5 5! 5! 5! 5! 5! 5! 22!
(b)
The 52 playing cards need to partitioned into 6 piles of 5 and 1 pile of 22. There are
52 52! ways to do this. 5! 5! 5! 5! 5! 5! 22! 5,5,5,5,5,5,22
this equals 1
1-36
(a)
25 C5
20 C 5
frat party soup kitchen
15 C15 stay home
25! 20! 15! 25! 20! 5! 15! 5! 15! 5! 5! 15!
25 . 5,5,15
In terms of a multinomial coefficient,
7
8
SOLUTIONS TO EXERCISES – CHAPTER 1
(b) This is similar to part (a), but we need to take into account that the 5 person teams are essentially the same. That is, suppose the hoop game was team blue versus team green. If the teams are {Amy, Bree, Candie, Diane, Edna} versus {Zoe, Yolanda, Xena, Willaminia, Venus}, then it does NOT matter which is team green and which group is team blue. 25 C5
20 C 5 2
23 C 7 2
1-37
30 C7
1-38
There are
9 9! 2! 4! 3! 2,4,3
9 C2
7 C4 3 C 3 total ways to assign the jobs.
assign 2 women both bad jobs
2 C2
Pr(women get both bad jobs)
assign 4 of the 7 assign the remaining men the average jobs 3 men the good jobs
7 C4 9 C2
7 C4 3 C 3
3 C 3
1 . 36
2 7 2 7 1 Note: This can also be solved as since the problem doesn’t involve 36 9 2 the distinction between average and good jobs. Section 1.3 Sampling and Distribution
1-39
There are 16 different side dishes. The customer must select different sides in a specific order: 16 P 3 16 15 14 3360
first side second side third side
1-40
Order does not matter. So, for example, the following 3-side choices are equivalent: {(1) corn & dressing & mac-N-cheese, (2) corn & mac-N-cheese & dressing, (3) mac-N-cheese & corn & dressing, (4) mac-N-cheese & dressing & corn, (5) dressing & mac-N-cheese & corn, (6) dressing & corn & mac-N-cheese} 3360 There are 16 C 3 560 such orders of 3 side dishes 6
1-41
163 4096
PROBABILITY AND STATISTICS WITH APPLICATIONS: A PROBLEM SOLVING TEXT
1-42
This is equivalent to distributing 3 indistinguishable balls (selections) into 16 fixed urns (side-dishes) without exclusion. 16 31 C3 18 C 3 816
1-43
Ordered samples without replacement. See Exercise 1-39 and solution.
1-44
Unordered samples with replacement. See Exercise 1-42 and solution.
1-45
Pr(order 3 servings of the same side)
1-46
(a)
351 C5
9
16 816
7 C 5
(b) The parent gives each child one dollar bill. She is left with two dollar bills which she can distribute to her 3 children: 4 C 2 . 1-47
(a) The fundamental theorem of counting implies the solution equals: Number of ways to distribute the Snickers Number of ways to distribute the gum 13 C10 10 C 7 (b) After giving each trick-or-treater a Snickers and a pack of gum, she has 6 Snickers bars and 3 packs of gum to distribute in any way she likes. 6 4 1 C6 3 4 1 C3 9 C6 6 C3
14 C 10 ways to
1-48
Each child could receive multiple pickled beets. There are give beets.
1-49
A bridge hand consists of 13 cards dealt from a standard deck of 52 different cards. 52 C 13
1-50
52 C 5
1-51
It depends upon the number of chicken sandwiches. If you do not order a chicken sandwich, then you have 5 dollars to distribute to 5 different $1 items. There are 9 C 5 ways to do this. If you order exactly one chicken sandwich, then you are left with 3 dollars with which to buy $1 items. You either get zero, one, or two chicken sandwiches. 9 C5 7 C3 5 C 1
1-52
36 C 21 45 C 21
2.101 1022
10 51 C10
10
SOLUTIONS TO EXERCISES – CHAPTER 1
Section 1.4 More Applications
1-53
(a)
( x 3) 4 1x 4 4 x 3 31 6 x 2 32 4 x 33 1 34
x 4 12 x 3 54 x 2 108 x 81 (b)
(2 x y) 5 1 (2 x) 5 5 (2 x) 4 y 10 (2 x) 3 y 2
10 (2 ) 2 y 3 5 (2 x) y 4 1 y 5 32 x 5 80 x 4 y 80 x3 y 2 40 x 2 y3 10 xy4 y5 (c)
(4 x 5 y )3 ((4 x) ( 5 y ))3
1 (4 x)3 3 (4 x) 2 ( 5 y)1 3 (4 x)1 ( 5 y) 2 1 ( 5 y)3 64 x 3 240 x 2 y 300 xy 2 125 y3
1-54
n 1 and r 1
The key here is to write out
n 1 , and then combine using a common r
denominator.
n 1 n 1 (n 1)! (n 1)! (n r )! (r 1)! ( n r 1)! r ! r 1 r (n 1)! r (n 1)! ( n r ) (n r )! r ( r 1)! ( n r ) ( n r 1)! r ! ( n 1)! r (n 1)! ( n r ) (n r )! r ! ( n r )! r ! n (n 1)! r ( n 1)! (n r ) n! . ( n r )! r ! ( n r )! r ! r
1 and y 1. (1 1) n 2 n n C0 n C1 n C2 n Cn .
1-55
Let
1-56
Selecting r items from n possible toppings to put on the pie is equivalent to selecting n r to keep off of the pie.
1-57
n n n 1 Since 4 5 5 , the coefficient of n 1 3,876 11, 628 15, 504 . 5
x 5 y n 4 x 5 y ( n 1) 5 is
PROBABILITY AND STATISTICS WITH APPLICATIONS: A PROBLEM SOLVING TEXT
11
n 5 n 4 11,628 3 . Thus n 19 and the coefficient of Alternatively, 5 3,876 n 4 191 20 20 x 5 y n 4 x 5 y15 in the expansion of y x y is 15,504 . 5 1-58
(a) ( x 2 y 5 z ) 3 1 x 3 3 x 2 ( 2 y ) 3 x 2 (5 z )
3 x (2 y ) 2 3 x (5 y ) 2 6 x (2 y ) (5 z ) 1 ( 2 y )3 3 ( 2 y )2 (5 z ) 3 (2 y ) (5 z ) 2 1 (5 z )3 x 3 6 x 2 y 12 xy 2 8 y 3 15x 2 z 60xyz 60 y 2 z 75 xz 2 150 yz 2 125 z 3 (b) ( w x y 2 z ) 2 1 w 2 1 x 2 1 ( y ) 2 1 (2 z ) 2
2 w x 2 w ( y) 2 w (2 z ) 2 x ( y ) 2 x (2 z ) 2 ( y ) (2 z )
1-59
Detailed calculations follow the exercise in the text.
1-60
Each of the five dice has 6 possible outcomes. So the total number of ways to roll five dice is 65. (a) Pr(Five dice-of-a-kind)
6 C1 5 C 5 5
6
.00077.
4 of what kind?
(b) Pr(Four dice-of-a-kind)
6 C1
ways to select 4 of the 5 die to match
5 C4 5
the remaining die must be different
5 C 1
6
52 4 .01929. 6 (c) There are 2 possible straights, 1-2-3-4-5 and 2-3-4-5-6. It remains to count how many ways these straights can arise in a roll of 5 dice. 5 4 3 2 1 2 1 1 1 1 1 2 5! .03086 Pr(straight) = 65 65
12
SOLUTIONS TO EXERCISES – CHAPTER 1
(d) A full house would have three dice be of one value (1, 2, 3, 4, 5, or 6) and two dice of a second value. 6 5 5 2
1 3 1 2 6 10 5 1 Pr(Full house) = 5 .03858 5 6
6
(e) The only way to get nothing is to have dice of the form: 1-2-3-4-6, 1-2-3-5-6, 1-2-4-5-6, or 1-3-4-5-6. It remains to count the ways these can sequence can be arranged on 5 dice. 4 5! 480 Pr(nothing) = .06173 65 65
6 5 5 2 1 1 3 2 1 1 6 10 10 2 (f) Pr(Three dice-of-a-kind) = 5 .15432 5 6
6
6 5 3 4 1 2 2 2 1 1 15 10 3 4 (g) Pr(Two pair) = 5 .23148 5 6
6
6 5 5 3 2 1 1 2 3 1 1 1 6 10 10 3 2 (h) Pr(one pair) = 5 .46296 5 6
6
It is neat that you are more likely to get 3 dice-of a kind than you are to roll nothing. Did you check that the sum of these probabilities is one?
1-61 Solutions tabulated in the text.
1-62
Expected Winnings 49,999,999
1 120,526,770
99,999
49,638,178 120,526,770
117,184,724 41 (1) 120,526,770 120,526,770
.4118.
That is, the expected winning is negative 41 cents per play.
PROBABILITY AND STATISTICS WITH APPLICATIONS: A PROBLEM SOLVING TEXT
Section 1.5 Chapter 1 Sample Examination
1.
Order does not matter. Both solutions are
2.
799 10 4
3.
Order matters here.
4.
15 C 2
(a)
1365
13 C1
13
6.
(a)
15 C1
14 C 2
give bop-on-nose away first
give slinky's away to 2 of the remaining 14 students
51 C 17 2 pair and 13 other cards without the old maid
(b)
65,780.
15 P 3
give slinky's give bop-on-nose away to 2 of to 1 of the the 15 students remaining 13
5.
26 C 5
25 C2 2 C 2 2 C2 23 C13 2 C1
two pair, the old maid, and 12 other cards
12
25 C2 2 C2 2 C2 1 C1 2 C2 23 C12 2 C1 51 C 17
52 C 5
(b)
4 C1
13 C4
39 C 1
which suit? ways to select 4 the remaining cards from the card selected suit
7.
(a) 7 C3 6 C 2 (b) Pr(5 kids selected included 3 soccer and 2 football)
8.
Using the multiplication rule gives 3 2 3 18 .
9.
49 C5
7 C3
6 C 2
13 C 5
.408.
42 C 1 80,089,128. The Jackpots were not getting large enough, so there
were more balls added to decrease the chances of winning and therefore increasing cumulative Grand Prizes. 10.
(2 x y)5 32 x 5 80 x 4 y 80 x 3 y 2 40 x 2 y 3 10 xy 4 y 5
13
14
11.
SOLUTIONS TO EXERCISES – CHAPTER 1
There are 15 terms in the multinomial expansion.
4
2 y 5 z 1 x 4 4 x3 (2 y ) 4 x3 (5 z) 6 x 2 (2 y) 2 1 (5 z ) 4
12 cards have faces
12.
12 C 3
Pr(all 3 cards have faces)
ways to match w hite socks
13.
Pr(matching socks)
.00995.
52 C 3
8 C2
ways to match black socks
4 C 2
12 C 2
34 66
1 Pr(one white and one black sock) 1
14.
There are a total of 100 birds, therefore
100 C 6 ways
8 4 . 66
to take six birds.
Joe could take 2 Canada geese (and one from each of the other prey), or 2 ducks, or 2 eagles, or 2 cranes, or 2 flamingos. This can be done in
25 C1 40 C1 10 C1 5 C1 20 C1 25 C2 40 C1 10 C1 5 C1 20 C1 25 C1 40 C2 10 C1 5 C1 20 C1 25 C1 40 C1 10 C2 5 C1 20 C1 25 C1 40 C1 10 C1 5 C2 47,500,000 ways.
20 C 2
Pr(at least one bird of each variety)
47,500,000 .03985. 1,192,052,400
5 . 13
15.
Pr(passing on first attempt)
16.
Use the multiplication rule with the following steps: Step 1. Step 2. Step 3. Step 4.
Choose any 3 rows (order immaterial) - # ways 6 C 3 20 . List the rows in ascending order for definiteness. Choose a column for the first listed row - # ways 5 . Choose a column for the 2nd listed row - # ways 4 . Choose a column for the 3rd listed row - # ways 3 .
Answer 20 5 4 3 1200 (C)
CHAPTER 2: GENERAL R ULES OF PROBABILITY Section 2.2 Set Theory Survival Kit
2-1
2-2
2-3
(a)
O {1,3,5, 7,9} and P {2,3,5,7}.
(b)
The odd primes less than 10 are O P {3,5, 7}.
(c)
The set of odd numbers or prime numbers less than 10 is O P {1,2,3,5,7,9}.
(d)
The odd numbers less than 10 that are not prime is the set O P {1,9}.
(a)
A B {1, 2}.
(b)
B { ,water}, so N ( B) 2.
(c)
A B {1, 2, , Jamaal,gum}, so ( A B ) {water}.
(a)
The Luisi family pets can be summarized as: Food
Named
snake steer chicken rabbit
dog cat frog
rock
(b)
The number of pets of each type is summarized as: Food
Named
2 1
3
2
15
16
2-4
SOLUTIONS TO EXERCISES – CHAPTER 2
The shaded section denotes the indicated set.
A
A
B
B
A B
A B
A
B
B
B
2-5
From Subsection 1.4.3, we already have the probability of the winning prizes. Pr(losing) 1 Pr(winning something) Pr(losing) 1
1, 712,304 1 120,576,770 120,576,770
probability of winning the grand prize
probabilty of matching only the powerball
117,184,724 . 120,576, 770
PROBABILITY AND STATISTICS WITH APPLICATIONS: A PROBLEM SOLVING TEXT
2-6
17
Doubles (1,1),(2,2),(3,3),(4,4),(5,5),(6,6) . Sum is divisible by 3 {(1, 2),(1,5),(2,1),(2, 4),(3,3),(3,6), (4,2),(4,5),(5,1),(5,4),(6,3),(6,6)}. Doubles Sum is divisible by 3 {(3,3),(6, 6)}
Doubles Sum is divisible by 3 {(1,1),(1, 2),(1,5),(2,1),(2, 2),(2, 4),(3,3),(3,6), (4,2),(4,4),(4,5),(5,1),(5,4),(5,5),(6,3),(6,6)}. Pr(Doubles Sum is divisible by 3)
2-7
16 6 12 2 36 36 36 36
Let K equal the percent of king size mattresses sold, Q equal the percent of queen size mattresses sold, and T equal the percent of twin size mattresses sold. 1 ( K T ), and K 3T . 4 Solving for the variables, K 60%, Q 20%, and T 20%. We are given: K Q T 100%, Q
The probability that the next mattress sold is king or queen size is K Q 80% (C).
2-8
A {2,4,6,8,10}
B {1,4,6,8,9,10} A B {1,2,3,5,7,9} A B {3,5,7) A B {1,2,4,6,8,9,10} A B {4,6,8,10}
A B {4,6,8,10} 2-9
A B A B is represented by the shaded area. A
B
18
2-10
SOLUTIONS TO EXERCISES – CHAPTER 2
( A B )
B
A
B
A
A
2-11
A
B A B
A
B
B C A B A C
2-12 B
A
N
N ( A
A \ B
N
A B
N B \ A
B ) N ( A \ B ) N ( A B ) N ( B \ A) N ( A \ B ) N ( A B ) N ( B \ A) N ( A B ) N ( A B) N ( A) N ( B ) N ( A B )
PROBABILITY AND STATISTICS WITH APPLICATIONS: A PROBLEM SOLVING TEXT
2-13
A B C is the shaded area
A
B
C
A
2-14
B C '.
Method 1: Using the Axioms N ( A) N ( B )
N (U ) N ( A). 10 N (U ) N ( B ). 6
From the inclusion-exclusion relationship, N (A B ) N (A B ) N ( A B ) 6 10 12 4.
Method 2: Venn Diagram
A
B
2
4
8
Method 3: Venn Box diagram (Subsection 2.2.4)
B B
A
A
4
6
2
8 14
6
10 10 20
6
N (A )
N (B ).
19
20
2-16
SOLUTIONS TO EXERCISES – CHAPTER 2
N ( A
B C D)
There are There are There are
N ( B ) N (C ) N ( D ) N ( A B ) N ( A C ) N ( A D ) N ( B C ) N ( B D ) N (C D ) N ( A B C ) N ( A B D ) N ( A C D ) N ( B C D) N ( A B C D )
N ( A)
4 ways to select one of the four regions. 4 C 2 6 intersections of two of the four regions. 4 C 3 4 intersections of three of the four regions. 4 C 1
2-17
A four region Venn diagram could be drawn as:
2-18
There are seven male children without purple hair in this family.
Adults
Females
3
5 4
7
6
2 0
1
Purple hair
PROBABILITY AND STATISTICS WITH APPLICATIONS: A PROBLEM SOLVING TEXT
2-19
(B) 50% 40% 10% of the population owns exactly one of auto and home.
Auto owner 40%
Home
20% 10% 30%
2-20
(D)
There are 880 young married females.
Young
Males 720
880
870
600 800
2410 3190
530
Married
2-21
(D) 52% of these viewers watch none of the three sports.
Gymnastics 12%
Baseball 6% 8%
2%
52%
5% Soccer
11% 4%
21
22
2-22
SOLUTIONS TO EXERCISES – CHAPTER 2
(A)
5% of the patients need both lab work and are referred to a specialist. The percentages in bold are given in the problem. Try this just using a Venn diagram.
Referred to Specialist Not Referred to Specialist
2-23
Lab work 5% 35% 40%
No lab work 25% 35% 60%
30% 70%
100%
(D) Pr( A) .60. The key here is to note that Pr( A B ) 0.7 implies that Pr( A B) 0.3. Similarly, Pr A B ' 0.9 implies Pr A ' B 0.1. These are versions of De Morgan’s laws. We do not have enough information to find Pr A B or Pr A B , but fortunately we are just asked to find Pr A 1 .1.3 .6 B
A .1 .3
2-24
(D)
We have two equations with two unknown variables x and y (see box
diagram below). The first comes from the fact that the sum of the probability of four disjoint outcomes must equal 1. The second equation comes from the statement of the problem. 22% x y 12% 100%
implies that
x y 66%.
22% x 14% implies that
y x 14%.
probabilty visit chiropractor
exceeds by 14%
22% y
x 26%
probabilty visit P.T.
and y 40%. Pr(visit P.T.) 22% 26% 48%.
Chiropractor No chiropractor
Visit P.T 22%
No P.T y
x
12% y 12%
22% x
22% y
x 12% 100%
PROBABILITY AND STATISTICS WITH APPLICATIONS: A PROBLEM SOLVING TEXT
2-25
23
Draw the Mickey Mouse diagram, with Auto, Homeowners and Renters labeled as shown below. Since H and R are mutually exclusive the subsets comprising their intersection have cardinality zero. Statement (i) gives N ( A B C ) 17 . Statement (v) gives N ( H \ A) 11 . The remaining unknown subsets are labeled x, y , z , w .
A 17
H y
z
11
0 0 w R
From (ii) x y z 64 . Also, 100 17 11 ( x y z ) w 17 11 64 w w 8 . From (iii), y 11 2( x 8) , and, From (iv), x y 35 . Solving the last two equations simultaneously gives x N ( A R ) 10
(B)
Section 2.3 Conditional Probability
2-26
There are 3 levels of choices in the following tree. The first branch concerns whether the professor put the batteries in an empty pocket or in the pocket with the two new batteries. The second level concerns which pocket the professor withdraws batteries from. In the third level, if the four batteries are in the same pocket, we will use counting techniques from chapter 1 to determine the likelihood of grabbing either two batteries that do not work, two batteries that do work, or one battery that works along with one battery that does not work.
24
SOLUTIONS TO EXERCISES – CHAPTER 2
Select two dead batteries
Professor selects pocket with dead batteries Professor places dead batteries in empty pocket
Professor places dead batteries in pocket with 2 good batteries
Select two good batteries
Professor selects pocket with good batteries
1
1
Select two dead batteries
Professor selects pocket with all batteries
1
Select one dead and one good battery Select two good batteries
3 1 1 5 14 . 1 1 4 2 4 6 24
(a)
Pr(get at least one good battery)
(b)
Pr(two good batteries at least one good battery)
Pr(two good at least one good battery) Pr(at least one good battery)
Pr(two good) Pr(at least one good battery)
3 1 1 1 1 1 4 6 10 . 4 2 14 14 24 (c)
Pr(batteries in same pocket at least one good battery)
The audio device will play
Pr(batteries in same pocket at least one good battery) Pr(at least one good battery)
1 5 1 6 5 . 4 14 14 24
The audio device will play
PROBABILITY AND STATISTICS WITH APPLICATIONS: A PROBLEM SOLVING TEXT
2-27
(a)
1 3
.
If you originally select the door with the new car, then Monte can select from either of two doors to show you a goat. If you select a door concealing a goat, Monte only has one door that he can open. Consider the following diagram. 1
Monte shows you the goat below
You pick a door hiding a goat
Monte shows you the goat above
You pick a door hiding a goat You pick a door hiding a new automobile
If you change doors, you win a new car
If you change doors, you win a new car
1
Monte shows you the first goat
If you change doors, you win a goat
Monte shows you the second goat
If you change doors, you win a goat
Pr(win auto change doors)
2-28
25
This is the scenario where we will keep our original door. You will win the car if and only if you selected the correct door originally (it just doesn’t matter what Monte does). There is a one in 3 chance of this. Pr(win auto keep original door)
(b)
1 3
1 1
1 3
1 1
2 3
.
Consider the following Box Diagram with the information in (i), (ii) and (iii) filled in: High Regular Irregular
a 14
Low
Normal
22
c b
15 100
b 100 14 22 64 From (iv), a Pr[Irregular High] Pr[High|Irregular] Pr[Irregular] Similarly, from (v), c
Regular Irregular Column Total
1 8
1 3
15% 5%
64% 8%. Completing the Box Diagram gives, High 9 5 14
Pr[Regular Low] 20% (E)
Low 20 2 22
Normal 56 8 64
Row Total 85 15 100
1
1
1
1
26
2-29
SOLUTIONS TO EXERCISES – CHAPTER 2
(C) Let x equal the probability that the male is a smoker, given that he does NOT have a circulation problem. Pr(circulation problem smoker) Pr(smoker)
Pr(circulation problem smoker)
.25 2 x .5 2 . .25 2 x .75 x 1.25 5
Smoker 25%
Male has circulation problem
NonSmoker
Male does NOT have circulation problem
Smoker 75% NonSmoker
2-30
(C) Pr(smoker died)
2-31
(B)
Pr(smoker died) .10 .05 .005 36%. Pr(died) .10 .05 .90 .01 .014
N
Pr(1 N 4) Pr( N 1 N 4) Pr( N 4)
1 2
0
1 6
1 12
1 20
1 30
1 6
1 12
1 20
2 . 5
1 1 30
2 3 4 5 …
Pr(
n)
1 1 1 2 2 1 1 23 6 1 1 3 4 12 1 1 4 5 20 1 1 5 6 30 1 1 6 7 42 …
PROBABILITY AND STATISTICS WITH APPLICATIONS: A PROBLEM SOLVING TEXT
2-32
27
(a) Pr(Penguins win series) .55 .35 .55 .65 .55 .45 .35 .55 .47575. (b) Pr(Devils win at least one game) 1 .55 .35 .8075. 35% Pittsburgh wins the series 2-0 Pittsburgh wins New Jersey wins
55% Pittsburgh wins
Game 1 played in Pittsburgh
Pittsburgh wins 45%
Pittsburgh wins
35%
New Jersey wins
New Jersey wins
New Jersey wins the series 2-1 45% 55%
Pittsburgh wins the series 2-1
45%
New Jersey wins the series 2-1
Game 3 played in Pittsburgh, if necessary
Game 2 played in New Jersey
2-33
New Jersey wins
65%
New Jersey wins
Pittsburgh wins the series 2-1
55%
Pittsburgh wins
65% New Jersey wins the series 2-0
(B) This is a standard Bayes’ formula problem, easily solved by drawing the appropriate tree diagram. The data may also be displayed in a Box diagram similar to the one in Exercise 2-28:
Survive Die Column Total d a
Critical
Serious
Stable
a
b
c
10
30
d
100
100 10 30 60 Pr[Die Critical] Pr[Die | Critical] Pr[Critical]
Similarly, b (10%) (30%) the box gives,
Survive Die Column Total Pr serious survive
3% , and
Critical 6 4 10
c
Serious 27 3 30
Pr serious survive Pr survive
(40%) (10%) 4%.
(1%) (60%)
Stable 59.4 .6 60
Row Total
27 92.4
.6%. Completing
Row Total 92.4 7.6 100
29.2% (B)
28
2-34
SOLUTIONS TO EXERCISES – CHAPTER 2
Pr(exactly 1 hit in 3 at-bats) .25 .6 .75 .75 .25 .6 .75 .75 .25 .365625.
.400 Hit
.400
Hit
No hit
Hit
.250
No hit
.250
Hit
No hit
.750
First at-bat
No hit
Hit
.750
Second at-bat
.750 .400
Hit No hit
.250
.600
No hit
Hit
.600
.600 .250
No hit .750 Third at-bat
2-35
Pr(call Packy “dad” Packy is your father)
Pr(call “dad” Packy is your father) Pr(Packy is your father) .01 .90 .1538. .01 .90 .99 .05
2-36 One Car Sports Car
Multiple Cars
Row Total
a
20
64
100
No Sports Car Column Total
Using (iv), a Pr[Sports Car Multiple Cars]
Pr[Sports Car | Multiple Cars]Pr[Multiple Cars] (15%) (64%) 9.6%
The complete Box Diagram is now easily calculated:
PROBABILITY AND STATISTICS WITH APPLICATIONS: A PROBLEM SOLVING TEXT
One Car
Multiple Cars
Row Total
Sports Car
10.4
9.6
20
No Sports Car
25.6
54.4
80
Column Total
36
64
100
Thus, Pr One Car No Sports Car 25.6%
(C)
Note: One may also display the information in a tree diagram: Insure a sports car
.15
Does NOT insure a sports car
.64 Insure more than 1 car
.85
Insure a sports car
Insure exactly 1 car
.36
Does NOT insure a sports car
Section 2.4 Independence
2-37
.511 Male
.511
Female
Male
.489 Female
.511 Male
First person chosen .489
Female Second person chosen
489
29
30
2-38
SOLUTIONS TO EXERCISES – CHAPTER 2
You are encouraged to observe that this is structurally the same as problem 2-37.
Heads
Heads
Tails
Heads
Tails First coin flip
Tails Second coin flip
Number of Heads Probability
0 1 9
1
2
2
1 2 3 3
4 9
2-39 Make .384
.384 .616
Miss
Misses his 3 point attempt
Make .616
John Stockton’s first attempt
Miss
.384 .616
second attempt
2
4 9
.384
Make
Makes his 3 point attempt
2 3
paths through the probability tree that have one head and one tail
.616
Makes exactly 2
Make
.384
Makes exactly 2
Miss
.616
Make
.384
Miss
.616
Make
.384
Miss
.616
Miss
Makes exactly 2
third attempt
Pr(Stockton makes exactly two 3-pointer)
3
.384
.384
.616
ways to make exactly two 3-pointer on three attepts
probability of making a 3-pointer
probability of making a 3-pointer
probabilit y of missing a 3-pointer
.2725.
PROBABILITY AND STATISTICS WITH APPLICATIONS: A PROBLEM SOLVING TEXT
31
12 13 . Pr( B) . The events A and B are dependent. 51 52
2-40
Pr( B | A)
2-41
(a) Since Pr( B ) b, we know Pr( B) 1 b. To show that A and B are independent, we observe Pr( B) Pr( A) Pr( B) a (1 b).
A A
B ab b(1 a )
B a(1 b)
a
(1 a )(1 b)
1 a
b
1 b
1
2-42
Not necessarily.
2-43
(a)
U HHH , HHT , HTH , HTT ,THH , THT , TTH ,TTT .
(b)
A {HHH , HHT , TTH , TTT }, B { HHH , HTH , THT , TTT }, and C {HHH , HTT , THH , TTT }. Thus, Pr( A) Pr( B) Pr( C )
(c) Pr( A B) Pr( HHH , TTT )
1 . 2 2 1 . 8 4
Similarly,
Pr[ A C ] Pr[ B C ] Pr[ HHH , TTT ]
(d)
Yes, because for example, Pr( B)
(e)
Pr( A B C ) Pr[ HHH , TTT ]
(f)
No, since Pr( B C )
1 4
1 4 1 8
1 4
1 4
.
1 2
1 2
Pr( A) Pr( B).
.
1 2
1 2
1 2
Pr[ A] Pr[ B] Pr[ C ].
2-44
Pr( A B ). Using independence, solve (.2 ) (.3 x ) x. There are two solutions, x .2 or x .3. Pr( A B) .7 or .8.
2-45
(a)
Let
Since the coin is fair (all outcomes are equally likely) and we do not need to track probabilities, listing the sample space will be sufficient.
U HHH , HHT , HTH , HTT ,THH , THT , TTH ,TTT . A first coin is tails {THH ,THT ,TTH ,TTT }. B the third flip is heads {HHH , HTH ,THH ,TTH }. (b)
Pr( B | A)
(c)
Yes, the events are independent.
2 4
1 2
.
32
2-46
SOLUTIONS TO EXERCISES – CHAPTER 2
(a)
Now is when tracking probabilities through a tree diagram is critical.
(b)
Pr( B | A)
(c)
2-47
Pr( A B) .4 .6 .6 .4 .4 .6 60%. Pr( A) .4 Yes, the events are independent. You can see this by computing Pr( B) Pr(third flip is heads) .60.
Let O be the event of using orthodontic work, F fillings and E extractions. Using (i) and (ii) we have: O
O
Row Total
a
c
b
1
F F Column Total 1 / 2
a Pr F O 1 Pr F O 1 2 / 3 1 / 3 b 1 1 / 2 1 / 2 . By independence, a bc c a / b 2 / 3 . Completing the O and F box gives:
O
O
Row Total
F
1/ 6 1/ 6
1/ 3
F
1/ 3 1/ 3
2/3
Column Total 1 / 2 1 / 2
1
Similarly, using (i) and (iii) and the independence of O and E gives: O
O
Row Total
E
1/ 4 1/ 4
1/ 2
E
1/ 4 1/ 4
1/ 2
Column Total 1 / 2 1 / 2
1
We now use Pr F 1 / 3 and Pr E 1 / 2 , plus (iv) to fill out the F and E box: F
F
Row Total
E
1/ 8
3/8
1/ 2
E
5 / 24
7 / 24
1/ 2
Column Total
1/ 3
2/3
1
Note: F and E are not independent.
Pr E F 1 Pr E F 1 7 / 24 17 / 24 (D)
PROBABILITY AND STATISTICS WITH APPLICATIONS: A PROBLEM SOLVING TEXT
2-48
33
We take the statement that a “different brand of pregnancy test was purchased” to imply that these test results are independent. Pr(both tests are incorrect) .01 .02 0.0002.
2-49
First we fill in the missing information. Major
Business Science Totals (a)
Grades
A 2 7 9
Pr(student assigned ‘B’)
B C 3 3 7 1 10 4
D 5 0 5
E 1 2 3
Totals 14 17 31
10 . 31 students - business majors
(b) (c) (d)
(e)
2-50
31 14 number of science majors 17 number of students 31 31 7 Pr(science major assigned ‘B’) . 31 This is most efficiently calculated by looking at the table. We are told that the student is a business major. Therefore the student lives in the shaded row of the table. 3 Pr(assigned ‘C’ business major) . 14 2 Pr(business major assigned 'A') . 9 Pr(science major)
(f)
No, because Pr(assigned ‘C’ business major) Pr(assigned ‘C’).
(a)
We will use counting techniques (combinations) that we studied in chapter 1. The number of ways to select 2 batteries from a flashlight containing 5 batteries is 5 C 2 10.
Pr(both batteries lost their charge)
2 C 2 5 C 2
1 . 10
3 C1 2 C 1
(b)
Pr(exactly one battery lost its charge)
(c)
Draw a tree diagram with appropriate probabilities.
5 C 2
6 . 10
2
2 4 Pr(both batteries tested lost their charge) 16%. 5 25 Pr(exactly one tested battery lost its charge) 2
2 3 12 48%. 5 5 25
34
2-51
SOLUTIONS TO EXERCISES – CHAPTER 2
Let Pr(purchase disability coverage) P[ D ] x , then Pr(purchase collision coverage) Pr[C ] 2x. 0.15 2 x 2 implies that x .274 and 2 x .548 . The completed Box Diagram is: C
C
Row Total
D
.15
.124
.274
D
.398 .328
.726
Column Total .548 .452
1
Pr(select neither insurance) Pr C D .328. (B) 2-52
Pr(select two consecutive face cards)
12 52
, or using combinations 11 51
Pr(select two consecutive face cards)
2-53
2-54
number of ways to deal 2 face cards number of ways to deal 2 cards
(a)
Pr(next roulette spin is black)
(b)
Pr(next eight roulette spins are black)
18 38
12 C 2 52 C 2
.
.
18 38
8
.00253.
(E) Pr(no severe and at most one moderate) (0.5)2 0.5 0.4 0.4 0.5 .65. minor moderate 0.5
0.5
0.4
severe minor
0.1
minor moderate
0.4
moderate
0.5
0.4
severe 0.1 severe minor 0.1
moderate
First Accident
0.5
0.4
severe 0.1 Second Accident
PROBABILITY AND STATISTICS WITH APPLICATIONS: A PROBLEM SOLVING TEXT
2-55
(A) Let B denote the number of blue balls in urn 2. 16 B 6.4 .6 B (0.6) . 16 B 16 B 16 B Solve for B to find there are four blue balls in urn 2. 16 16 B Pr(both ball same color) .44 (0.4)
Red ball Red ball
Blue ball
Red ball Blue ball Urn 1
Blue ball Urn 2
Section 2.5 Bayes’ Theorem
2-56
Draw the standard tree diagram to track probabilities.
Ace
Ace
Non-Ace
Non-Ace
Ace
First card
Non-Ace Second card
(a) (b)
number of aces 4 . number of cards 52 4 Pr(second card is an ace first card not ace) . 51 Pr(first card is an ace)
35
36
SOLUTIONS TO EXERCISES – CHAPTER 2
(c) Pr(first card is an ace second card not ace)
Pr(first card is an ace second card not ace) Pr(second card not ace) 4 52
4 48 52 51 48 48 51 52
47 51
4 . 51
2-57 We are here if first two cards have faces
Face
Non-Face Face Face Non-Face Face
Non-Face
Face
Non-Face Face
Non-Face First card
Non-Face
Face
Second card
Non-Face Third card
(a)
Number of Face Cards Probability
0 9,880 22,100 2,860 22,100
1 9,360 22,100
(b)
Pr(at least 2 face cards)
(c)
Pr(3rd card is face first two cards are face cards)
2 2,640 22,100
3 220 22,100
.
Pr(all three face cards
first two cards are face cards)
10 50
.
PROBABILITY AND STATISTICS WITH APPLICATIONS: A PROBLEM SOLVING TEXT
(d)
Pr(all three face cards at least two face cards)
(e)
220 Pr(all three face cards) 22,100 220 . 2,860 Pr(at least two face cards) 2,860 22,100 Solution Method 1: Tree diagram and Bayes’ Theorem
37
Pr(at least 2 face cards last card has face)
Pr(at least 2 face cards last card has face) Pr(last card has face) this is the probability that the first and third cards have faces
12 11 10 12 40 11 40 12 11 52 51 50 52 51 50 52 51 50 12 11 10 12 40 11 40 12 11 40 39 12 52 51 50 52 51 50 52 51 50 52 51 50 This equals
12 . Does it make sense to you? 52
11880 .3882. 30600
Solution Method 2: Using Combinations
Imagine that you know that the third card is a face card. That leaves 11 face cards in a deck of 51 total cards. Now imagine dealing the first two cards from this reduced deck. The question asks us to find the probability that at least one of these cards is a face card.
Pr(at least one face) 1 Pr(two non-face cards) 1
40 C 2 51 C 2
.3882.
38
SOLUTIONS TO EXERCISES – CHAPTER 2
Section 2.6 Credibility
2-58
Good driver
Bad driver Type of driver
Accident or not in year 1 Accident or not in year 2
(a) Pr( A1 A2 ) .8 (.1) 2 .2 (.5) 2 .058. Pr( A2 A1 ) .8 (.1) 2 .2 (.5) 2 .058 29 (b) Pr( A2 | A1 ) . Pr( A1 ) .8 (.1) .2 (.5) .18 90
Pr(G A1 A2 ) .8 (.1) 2 .008 4 . (c) Pr(G | A1 A2 ) Pr( A1 A2 ) .058 29 .8 (.1) 2 .2 (.5) 2 .2 (1 .52 ) .15 .497. (d) Pr( B At least one accident) .2 (1 .52 ) .8 (1 .9 2 ) .302 (e) Pr(G | A1 A2 )
Pr(G A1 A2 ) Pr( A1 A2 )
.8 (.9) 2 .648 .9284. .698 .8 (.9) 2 .2 (.5) 2
2-59
(D)
Pr(1997 1997 or 1998 or 1999)
2-60
(D)
Draw a tree diagram.
Pr(ultra-preferred dies)
.16 (.05) .455. .16 (.05) .18 (.02) .20 (.03)
(0.10) (.001) .0141. (0.10) (.001) (0.40) (.005) (0.50) (.01)
PROBABILITY AND STATISTICS WITH APPLICATIONS: A PROBLEM SOLVING TEXT
2-61
(D)
Draw a tree diagram.
Pr(young adult collision) 2-62
(D)
Let
(.16)(.08) .22. (.08)(.15) (.16)(.08) (.45)(.04) (.31)(.05)
denote the probability that a non-smoker dies. Dies
Heavy Smoker
Lives
Dies Light Smoker Lives NonSmoker Dies Type of smoker
Lives
Live or Die?
Pr( heavy smoker died)
2-63
(B)
(.2)(4 x) 8 .42. (.2)(4 x) (.3)(2 x ) (.5) x 19
Pr(16 20 accident)
(.08)(.06) .158 (.08)(.06) (.15)(.03) (.49)(.02) (.28)(.04)
39
40
SOLUTIONS TO EXERCISES – CHAPTER 2
Section 2.7 Chapter 2 Sample Examination
1. B
A B
number of full houses number of 5 card hands C C C C 3744 13 1 4 3 12 1 4 2 . 2598960 52 C 5
2.
Pr(dealt a full house)
3.
You may wish to create a Venn diagram. The key is to note that that you must have some insurance to be a client.
Auto Insurance No Auto Insurance
Dismemberment No Insurance Dismemberment 17 45 20 0 37 45
62 20 82
(a) N (both auto and dismemberment) 17. (b) N (exactly one kind of insurance) 65 20 45. (c) No, because Pr(auto dismember)
17 62 37 Pr(auto) Pr(dismember). 82 82 82
PROBABILITY AND STATISTICS WITH APPLICATIONS: A PROBLEM SOLVING TEXT
4.
(a) Venn Diagram
U College class
11
9
8
4 Female
Engineers Venn Box Diagram
Female Male
NonEngineer Engineer 9 4 13 19 8 11 17 15 32
9 . 32 21 11 (c) Pr(female engineer) 1 . 32 32 4 (d) Pr(female non-engineer) . 15 (e) No, since Pr(male engineer) Pr(male) Pr(engineer). (b) Pr(female engineer)
5.
(a) Pr(dealt 3 Kings)
number of ways to be dealt 3 Kings number of ways to be dealt 3 cards
(b) Pr(NOT dealt 3 Kings) 1
4 C 3 52 C 3
22096 . 22100
(c) Pr(3rd card King 1st and 2 nd card King) (d) Pr(3 Kings at least 2 Kings)
4 C 3
4 C3
2 . 50
4 C2 48 C 1
4 . 292
4 C 3 52 C 3
4 . 22100
41
42
6.
SOLUTIONS TO EXERCISES – CHAPTER 2
(a) There are 16 elements in the set. The bold numbers denote a sum of 6.
U 22, 23, 24, 25, 32, 33, 34, 35, 42, 43, 44, 45, 52, 53, 54, 55. (b) Pr(sum is nine)
2 . 16
(c) Pr(at least one die is 5)
7 16
0 . 9 0 (e) Pr(at least one 5 sum is 6) . 3 (d) Pr(sum is 6 at least one 5)
7.
(a)
Sum is 4 1 1 2
3
2
Sum is 4
3 Sum is 4 1 2
Urn A
3 Urn B
1 2 2 1 1 2 6 . 4 5 4 5 4 5 20 1 1 2 1 1 12 (c) Pr(at least one 2) . 4 5 4 4 5 20 (b) Pr(sum is 4)
(d) Pr(sum is 4 ball from Urn A is 2) Pr(ball from Urn B is 2) 2 1 4 5 1. (e) Pr(ball from Urn B is 2 sum is 4) 6 3 20
1 . 5
PROBABILITY AND STATISTICS WITH APPLICATIONS: A PROBLEM SOLVING TEXT
8.
(a)
Gender Female Male
Coke 2 5 7
Favorite Drink Diet Pepsi Ripple Water 3 8 4 1 2 0 4 10 4
Red Bull 1 4 5
43
18 12 30
5 . 30 N (female milk) 8 (c) Pr(female milk) . N (milk) 10 (b) Pr(Red Bull)
d) Pr(male milk)
9.
20 . 30
(a) Pr( A) 0.4. (b) Pr( A) 0.6. (c) Pr( A B ) .25. (d) Pr( A | B)
Pr( A B) .25 . Pr( B) .65
(e) Pr( A B ) 0.8. (f) Pr( B | A)
Pr( B A) .25 . Pr( A) .40
(g) No, they are dependent. Pr( A B ) Pr( A) Pr( B ).
10. Assume for definiteness that card with 5 credits is in the left (L) pocket and the 4 credit card is in the right ( R) pocket. She could either spend all 5 credits in L, or spend one from L and 4 from R, with the last one coming from R. These actions can be displayed as the following 5 letter words using L and R: LLLLL, LRRRR, RLRRR, RRLRR, RRRLR. 5
5
1 1 5 Each word has probability , and so the answer is 5 . 32 2 2
44
11.
SOLUTIONS TO EXERCISES – CHAPTER 2
Assume first that the left pocket is depleted leaving 3 credits in the right pocket. This means a total of 6 credits are spent, 1 from the right pocket, 5 from left, and the last one used is from the left pocket. This can be portrayed using 6 letter words with 1 R and 5 L’s, ending in L. That means the first 5 positions in the word consist of 1 R and 4 L’s. There are 5 C 1 5 such words. Next, assume that the right pocket is depleted leaving 3 credits in the left pocket. This means we have 6 letter words ending in R with 2 L’s and 3 R’s in the first 5 positions. There are 5 C 2 10 such words. Thus, the total probability is 6
1 15 15 . 2 64
12.
Consider what can happen to the team that is leading 3 games to 1.
Series won 4-1
Win
50% Series won 4-2 Win Lose Series won 4-3
Win Game 5
Lose
Game 6, if necessary
Lose Series lost 3-4 Game 7, if necessary
13.
Pr(win) 1 .43 .936.
PROBABILITY AND STATISTICS WITH APPLICATIONS: A PROBLEM SOLVING TEXT
14.
A
A B
Row Total
a
b
B
1/ 3
Column Total 2 / 3
1
b 1 1 / 3 2 / 3 and a Pr A B Pr A | B Pr B
7 2 7 . Complete 10 3 15
the Box Diagram to give:
A
A
Row Total
B
7 / 15
1/ 5
2/3
B
1/ 5
2 / 15
1/ 3
Column Total
2/3
1/ 3
1
Then Pr( A B ) 1 2 / 15 13 / 15 . 15.
Begin by writing out the sample space of possible dance couples. Ann’s Dance Partner
Betty’s Dance Partner
Ann
Andy
Betty
Boris
Danielle
Dan
Ann
Andy
Betty
Dan
Danielle
Boris
Ann
Boris
Betty
Andy
Danielle
Dan
Ann
Boris
Betty
Dan
Danielle
Andy
Ann
Dan
Betty
Andy
Danielle
Boris
Ann
Dan
Betty
Boris
Danielle
Andy
Pr(3 wives dance with non-spouse)
16.
Danielle’s Dance Partner
2 . 6
U {abcd , abdc , acbd , acdb, adbc, adcb, bacd , badc, bcad, bcda, bdac, bdca cabd , cadb, cbad , cbda, cdab, cdba , dabc , dacb, dbac, dbca, dcab, dcba}. Pr(all wives dances with non-spouse)
9 24
45
46
17.
SOLUTIONS TO EXERCISES – CHAPTER 2
There is one good young teacher. math faculty at School University
1 2
3
6
Old
18.
Bad
Pr(first person wins)
7 6 5 7 6 5 4 3 7 6 5 4 3 2 1 7 . 13 13 12 11 13 12 11 10 9 13 12 11 10 9 8 7
Red ball Red ball White ball
Red ball
First ball selected
White ball
Red ball
Second ball selected, White ball if necessary.
… et cetera
Third ball selected, if necessary.
3
19.
(a)
(b)
5
White ball
4 2 4 2 4 2 Pr(1 person wins) ... 40% 6 6 6 6 6 6 Use geometric series. st
Pr(1st person wins)
4 3 4 3 2 5 42.59%. 6 6 6 6 6 6
PROBABILITY AND STATISTICS WITH APPLICATIONS: A PROBLEM SOLVING TEXT
20.
.12
.10
.10 .12
.12 .10
.28
To solve for , we know: Pr( A B C ) 1 x Pr( A B C A B) x .06. 3 Pr( A B ) x .12 Pr(no risk factor A)
21.
.28 .46. (C) .60
(C) We have four equations with four unknown variables: 1 1 5 , and w x y z x z , x y , y z 1. 4 3 12 Adding the first three equations together, we see 2 x 2 y 2 z 1 ( x y z ) Pr(no coverage) w 1 ( x y z ) 1
1 1 . 2 2
1 . 2
47
48
22.
SOLUTIONS TO EXERCISES – CHAPTER 2
Let A represent the event of heart disease. Let B represent the event of at least one parent with heart disease. Then: B B Column Total
Pr( A B)
A 102 108 210
Row Total 312 625 937
108 .1728. (B) 625
23.
One Car Sports Car No Sports Car Column Total
Multiple Cars a
Row Total 20
70
100
a Pr Sports Car Multiple Cars
Pr Sports Car | Multiple Cars Pr Multiple Cars (15%)(70%) 10.5%. The completed Box Diagram is: Sports Car No Sports Car Column Total
One Car 9.5 20.5 30
Multiple Cars 10.5 59.5 70
Row Total 20 80 100
Pr(one car no sports car) 20.5% (B)
24.
The Auto/Homeowner Box Diagram is: A H 15 35 H Column Total 50
50 0 50
Row Total 65 35 100
Thus, 35% have auto insurance only, 50% have homeowners only and 15% have both. Then
Pr(renew) (.50)
(.40)
auto only renew given insure only auto
(.35) (.60) (.15) (.80) 53%. (D)
PROBABILITY AND STATISTICS WITH APPLICATIONS: A PROBLEM SOLVING TEXT
25.
49
Begin with the Emergency/Operating room Box Diagram.
Pr E O 1 Pr E O 1 85% 15% . E
Row Total
O
15
a
Column Total
25
100
E O
Also, by independence, 15% 25% a a 60% . The complete Box Diagram is: E
E
Row Total
O
30 10
40
O
45 15
60
Column Total
75
25
100
Pr O 40% (D)
26.
(D)
A tree diagram would be nice, but would have infinitely many branches. The key is to figure out how many ways there can be a total of seven accidents in two weeks. There could be 0 accidents the first week and 7 accidents the second week. Or there could be 1 accident the first week and 6 accidents the second week, and so forth. Let ( a , b ) be the pair representing a accidents the first week and b accidents the second week. Pr(exactly 7 accidents)
Pr (0, 7) Pr (1,6) Pr (7, 0)
1 1 1 1 1 1 8 1 . 1 8 2 7 8 1 9 64 2 2 2 2 2 2 2 prob of 1 accidents in first week
prob of 6 accidents in second week
