Prob Solution Manual (Probability Statistics)

Instructor Solution ManualProbability and Statistics for Engineers and Scientists(3rd Edition)AnthonyHayter1InstructorSolutionManualThis instructor solution manual to accompany the third edition ofProbability and Statistics for Engineers and Scientists by Anthony Hayterprovides worked solutions and answers to all of the problems given in the textbook. The studentsolutionmanual providesworkedsolutionsandanswerstoonlytheodd-numberedproblemsgiven at the end of the chapter sections. In addition to the material contained in the studentsolutionmanual, thisinstructormanual thereforeprovidesworkedsolutionsandanswerstothe even-numbered problems given at the end of the chapter sections together with all of thesupplementary problems at the end of each chapter.2Contents1 ProbabilityTheory 71.1 Probabilities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 71.2 Events . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 91.3 Combinations of Events . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 131.4 Conditional Probability . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 161.5 Probabilities of Event Intersections . . . . . . . . . . . . . . . . . . . . . . . . . . 221.6 Posterior Probabilities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 281.7 Counting Techniques . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 321.9 Supplementary Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 372 RandomVariables 492.1 Discrete Random Variables . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 492.2 Continuous Random Variables . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 542.3 The Expectation of a Random Variable . . . . . . . . . . . . . . . . . . . . . . . 582.4 The Variance of a Random Variable . . . . . . . . . . . . . . . . . . . . . . . . . 622.5 Jointly Distributed Random Variables . . . . . . . . . . . . . . . . . . . . . . . . 682.6 Combinations and Functions of Random variables . . . . . . . . . . . . . . . . . . 772.8 Supplementary Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 863 DiscreteProbabilityDistributions 953.1 The Binomial Distribution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 953.2 The Geometric and Negative Binomial Distributions . . . . . . . . . . . . . . . . 993.3 The Hypergeometric Distribution. . . . . . . . . . . . . . . . . . . . . . . . . . . 1023.4 The Poisson Distribution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1053.5 The Multinomial Distribution. . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1073.7 Supplementary Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1094 ContinuousProbabilityDistributions 1134.1 The Uniform Distribution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1134.2 The Exponential Distribution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1164.3 The Gamma Distribution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1194.4 The Weibull Distribution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1214.5 The Beta Distribution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1234.7 Supplementary Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12534 CONTENTS5 TheNormalDistribution 1295.1 Probability Calculations using the Normal Distribution . . . . . . . . . . . . . . 1295.2 Linear Combinations of Normal Random Variables . . . . . . . . . . . . . . . . . 1355.3 Approximating Distributions with the Normal Distribution . . . . . . . . . . . . 1405.4 Distributions Related to the Normal Distribution. . . . . . . . . . . . . . . . . . 1445.6 Supplementary Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1486 DescriptiveStatistics 1576.1 Experimentation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1576.2 Data Presentation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1596.3 Sample Statistics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1616.6 Supplementary Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1647 StatisticalEstimationandSamplingDistributions 1677.2 Properties of Point Estimates . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1677.3 Sampling Distributions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1707.4 Constructing Parameter Estimates . . . . . . . . . . . . . . . . . . . . . . . . . . 1767.6 Supplementary Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1778 InferencesonaPopulationMean 1838.1 Condence Intervals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1838.2 Hypothesis Testing . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1898.5 Supplementary Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1969 ComparingTwoPopulationMeans 2059.2 Analysis of Paired Samples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2059.3 Analysis of Independent Samples . . . . . . . . . . . . . . . . . . . . . . . . . . . 2099.6 Supplementary Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21810 DiscreteDataAnalysis 22510.1Inferences on a Population Proportion . . . . . . . . . . . . . . . . . . . . . . . . 22510.2Comparing Two Population Proportions . . . . . . . . . . . . . . . . . . . . . . . 23210.3Goodness of Fit Tests for One-way Contingency Tables . . . . . . . . . . . . . . . 24010.4Testing for Independence in Two-way Contingency Tables . . . . . . . . . . . . . 24610.6Supplementary Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25111 TheAnalysisofVariance 26311.1One Factor Analysis of Variance . . . . . . . . . . . . . . . . . . . . . . . . . . . 26311.2Randomized Block Designs . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27311.4Supplementary Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 28112 SimpleLinearRegressionandCorrelation 28712.1The Simple Linear Regression Model . . . . . . . . . . . . . . . . . . . . . . . . . 28712.2Fitting the Regression Line . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 28912.3Inferences on the Slope Parameter1 . . . . . . . . . . . . . . . . . . . . . . . . . 29212.4Inferences on the Regression Line. . . . . . . . . . . . . . . . . . . . . . . . . . . 29612.5Prediction Intervals for Future Response Values. . . . . . . . . . . . . . . . . . . 29812.6The Analysis of Variance Table . . . . . . . . . . . . . . . . . . . . . . . . . . . . 30012.7Residual Analysis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 302CONTENTS 512.8Variable Transformations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 30312.9Correlation Analysis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 30512.11Supplementary Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 30613 MultipleLinearRegressionandNonlinearRegression 31713.1Introduction to Multiple Linear Regression . . . . . . . . . . . . . . . . . . . . . 31713.2Examples of Multiple Linear Regression . . . . . . . . . . . . . . . . . . . . . . . 32013.3Matrix Algebra Formulation of Multiple Linear Regression . . . . . . . . . . . . . 32213.4Evaluating Model Accuracy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 32713.6Supplementary Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 32814 MultifactorExperimentalDesignandAnalysis 33314.1Experiments with Two Factors . . . . . . . . . . . . . . . . . . . . . . . . . . . . 33314.2Experiments with Three or More Factors . . . . . . . . . . . . . . . . . . . . . . 33614.3Supplementary Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 34015 NonparametricStatisticalAnalysis 34315.1The Analysis of a Single Population . . . . . . . . . . . . . . . . . . . . . . . . . 34315.2Comparing Two Populations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 34715.3Comparing Three or More Populations. . . . . . . . . . . . . . . . . . . . . . . . 35015.4Supplementary Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 35416 QualityControlMethods 35916.2Statistical Process Control . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 35916.3Variable Control Charts . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 36116.4Attribute Control Charts . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 36316.5Acceptance Sampling . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 36416.6Supplementary Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 36517 ReliabilityAnalysisandLifeTesting 36717.1System Reliability . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 36717.2Modeling Failure Rates . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 36917.3Life Testing . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 37217.4Supplementary Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3746 CONTENTSChapter1ProbabilityTheory1.1 Probabilities1.1.1 S= {(head, head, head), (head, head, tail), (head, tail, head), (head, tail, tail),(tail, head, head), (tail, head, tail), (tail, tail, head), (tail, tail, tail)}1.1.2 S = {0 females, 1 female, 2 females, 3 females, . . . , n females}1.1.3 S= {0,1,2,3,4}1.1.4 S= {January 1, January 2, .... , February 29, .... , December 31}1.1.5 S= {(on time, satisfactory), (on time, unsatisfactory),(late, satisfactory), (late, unsatisfactory)}1.1.6 S= {(red, shiny), (red, dull), (blue, shiny), (blue, dull)}1.1.7 (a)p1p= 1p = 0.5(b)p1p= 2p =23(c) p = 0.25p1p=131.1.8 0.13 + 0.24 + 0.07 + 0.38 +P(V ) = 1P(V ) = 0.1878 CHAPTER1. PROBABILITYTHEORY1.1.9 0.08 + 0.20 + 0.33 +P(IV ) +P(V ) = 1P(IV ) +P(V ) = 1 0.61 = 0.39Therefore, 0 P(V ) 0.39.IfP(IV ) = P(V ) thenP(V ) = 0.195.1.1.10 P(I) = 2 P(II) andP(II) = 3 P(III) P(I) = 6 P(III)Therefore,P(I) +P(II) +P(III) = 1so that(6 P(III)) + (3 P(III)) +P(III) = 1.Consequently,P(III) =110,P(II) = 3 P(III) =310andP(I) = 6 P(III) =610.1.2. EVENTS 91.2 Events1.2.1 (a) 0.13 +P(b) + 0.48 + 0.02 + 0.22 = 1P(b) = 0.15(b) A = {c, d} so thatP(A) = P(c) +P(d) = 0.48 + 0.02 = 0.50(c) P(A

) = 1 P(A) = 1 0.5 = 0.501.2.2 (a) P(A) = P(b) +P(c) +P(e) = 0.27 so P(b) + 0.11 + 0.06 = 0.27and hence P(b) = 0.10(b) P(A

) = 1 P(A) = 1 0.27 = 0.73(c) P(A

) = P(a) +P(d) +P(f) = 0.73 so 0.09 +P(d) + 0.29 = 0.73and hence P(d) = 0.351.2.3 Over a four year period including one leap year, the number of days is(3 365) + 366 = 1461.The number of January days is 4 31 = 124and the number of February days is (3 28) + 29 = 113.The answers are therefore1241461and1131461.1.2.4 S= {1, 2, 3, 4, 5, 6}Prime = {1, 2, 3, 5}All the events in Sare equally likely to occur and each has a probability of16so thatP(Prime) =P(1) +P(2) +P(3) +P(5) =46=23.1.2.5 See Figure 1.10.The event that the score on at least one of the two dice is a prime number consistsof the following 32 outcomes:{(1,1), (1,2), (1,3), (1,4), (1,5), (1,6), (2,1), (2,2), (2,3), (2,4) (2,5), (2,6), (3,1), (3,2),(3,3), (3,4), (3,5), (3,6), (4,1), (4,2), (4,3), (4,5), (5,1), (5,2), (5,3), (5,4), (5,5), (5,6),(6,1), (6,2), (6,3), (6,5)}Each outcome in Sis equally likely to occur with a probability of136so thatP(at least one score is a prime number) = 32 136=3236=89.The complement of this event is the event that neither score is a prime number whichincludes the following four outcomes:10 CHAPTER1. PROBABILITYTHEORY{(4,4), (4,6), (6,4), (6,6)}Therefore,P(neither score prime) =136 +136 +136 +136=19.1.2.6 In Figure 1.10 let (x, y) represent the outcome that the score on the red die is x andthescoreonthebluedieisy. Theeventthatthescoreonthereddieisstrictlygreater than the score on the blue die consists of the following 15 outcomes:{(2,1), (3,1), (3,2), (4,1), (4,2), (4,3), (5,1), (5,2), (5,3), (5,4), (6,1), (6,2), (6,3),(6,4), (6,5)}The probability of each outcome is136so the required probability is 15 136=512.This probability is less than 0.5 because of the possibility that both scores are equal.The complement of this event is the event that the red die has a score lessthanorequal to the score on the blue die which has a probability of 1 512=712.1.2.7 P(or )=P(A) + P(K) + . . . + P(2)+P(A) + P(K) + . . . + P(2)=152 +. . . +152=2652=121.2.8 P(draw an ace) = P(A) +P(A) +P(A) +P(A)=152 +152 +152 +152=452=1131.2.9 (a) Let the four players be named A, B, C, and T for Terica, and let the notation(X, Y ) indicate that playerXis the winner and playerYis the runner up.The sample space consists of the 12 outcomes:S= {(A,B), (A,C), (A,T), (B,A), (B,C), (B,T), (C,A), (C,B), (C,T), (T,A),(T,B), (T,C)}The event Terica is winner consists of the 3 outcomes {(T,A), (T,B), (T,C)}.Sinceeachoutcomein Sisequallylikelytooccurwithaprobabilityof112itfollows thatP(Terica is winner) =312=14.(b) The event Terica is winner or runner up consists of 6 out of the 12 outcomesso thatP(Terica is winner or runner up) =612=12.1.2. EVENTS 111.2.10 (a) See Figure 1.24.P(Type I battery lasts longest)= P((II, III, I)) +P((III, II, I))= 0.39 + 0.03 = 0.42(b) P(Type I battery lasts shortest)= P((I, II, III)) +P((I, III, II))= 0.11 + 0.07 = 0.18(c) P(Type I battery does not last longest)= 1 P(Type I battery lasts longest)= 1 0.42 = 0.58(d) P(Type I battery last longer than Type II)= P((II, I, III)) +P((II, III, I)) +P((III, II, I))= 0.24 + 0.39 + 0.03 = 0.661.2.11 (a) See Figure 1.25.Theeventbothassemblylinesareshut down consistsof thesingleoutcome{(S,S)}.Therefore,P(both assembly lines are shut down) = 0.02.(b) The event neither assembly line is shut down consists of the outcomes{(P,P), (P,F), (F,P), (F,F)}.Therefore,P(neither assembly line is shut down)= P((P, P)) +P((P, F)) +P((F, P)) +P((F, F))= 0.14 + 0.2 + 0.21 + 0.19 = 0.74.(c) The event at least one assembly line is at full capacity consists of the outcomes{(S,F), (P,F), (F,F), (F,S), (F,P)}.Therefore,P(at least one assembly line is at full capacity)= P((S, F)) +P((P, F)) +P((F, F)) +P((F, S)) +P((F, P))= 0.05 + 0.2 + 0.19 + 0.06 + 0.21 = 0.71.(d) The event exactly one assembly line at full capacity consists of the outcomes{(S,F), (P,F), (F,S), (F,P)}.Therefore,P(exactly one assembly line at full capacity)= P((S, F)) +P((P, F)) +P((F, S)) +P((F, P))12 CHAPTER1. PROBABILITYTHEORY= 0.05 + 0.20 + 0.06 + 0.21 = 0.52.Thecomplementof neitherassemblylineisshut down istheeventat least oneassembly line is shut down which consists of the outcomes{(S,S), (S,P), (S,F), (P,S), (F,S)}.The complement of at least one assembly line is at full capacity is the event neitherassembly line is at full capacity which consists of the outcomes{(S,S), (S,P), (P,S), (P,P)}.1.2.12 The sample space isS= {(H,H,H), (H,T,H), (H,T,T), (H,H,T), (T,H,H), (T,H,T), (T,T,H), (T,T,T)}with each outcome being equally likely with a probability of18.The event two heads obtained in succession consists of the three outcomes{(H,H,H), (H,H,T), (T,H,H)}so thatP(two heads in succession) =38.1.3. COMBINATIONSOFEVENTS 131.3 CombinationsofEvents1.3.1 Theevent Acontains theoutcome0whiletheemptyset does not containanyoutcomes.1.3.2 (a) See Figure 1.55.P(B) = 0.01 + 0.02 + 0.05 + 0.11 + 0.08 + 0.06 + 0.13 = 0.46(b) P(B C) = 0.02 + 0.05 + 0.11 = 0.18(c) P(AC) = 0.07+0.05+0.01+0.02+0.05+0.08+0.04+0.11+0.07+0.11 = 0.61(d) P(A B C) = 0.02 + 0.05 = 0.07(e) P(A B C) = 1 0.03 0.04 0.05 = 0.88(f) P(A

B) = 0.08 + 0.06 + 0.11 + 0.13 = 0.38(g) P(B

C) = 0.04 + 0.03 + 0.05 + 0.11 + 0.05 + 0.02 + 0.08 + 0.04 + 0.11 + 0.07+ 0.07 + 0.05 = 0.72(h) P(A (B C)) = 0.07 + 0.05 + 0.01 + 0.02 + 0.05 + 0.08 + 0.04 + 0.11 = 0.43(i) P((A B) C) = 0.11 + 0.05 + 0.02 + 0.08 + 0.04 = 0.30(j) P(A

C) = 0.04 + 0.03 + 0.05 + 0.08 + 0.06 + 0.13 + 0.11 + 0.11 + 0.07 + 0.02+ 0.05 + 0.08 + 0.04 = 0.87P(A

C)

= 1 P(A

C) = 1 0.87 = 0.131.3.4 (a) A B = {females with black hair}(b) A C

= {all females and any man who does not have brown eyes}(c) A

B C = {males with black hair and brown eyes}(d) A (B C) = {females with either black hair or brown eyes or both}1.3.5 Yes, because a card must be drawn from either a red suit or a black suit but it cannotbe from both at the same time.No, because the ace of hearts could be drawn.14 CHAPTER1. PROBABILITYTHEORY1.3.6 P(A B) = P(A) +P(B) P(A B) 1so thatP(B) 1 0.4 + 0.3 = 0.9.Also,P(B) P(A B) = 0.3so that0.3 P(B) 0.9.1.3.7 SinceP(A B) = P(A) +P(B) P(A B)it follows thatP(B) = P(A B) P(A) +P(A B)= 0.8 0.5 + 0.1 = 0.4.1.3.8 S= {1, 2, 3, 4, 5, 6} where each outcome is equally likely with a probability of16.The eventsA,B, andB

areA = {2, 4, 6},B= {1, 2, 3, 5} andB

= {4, 6}.(a) A B = {2} so thatP(A B) =16(b) A B = {1, 2, 3, 4, 5, 6} so thatP(A B) = 1(c) A B

= {4, 6} so thatP(A B

) =26=131.3.9 Yes,thethreeeventsaremutuallyexclusivebecausetheselectedcardcanonlybefrom one suit.Therefore,P(A B C) = P(A) +P(B) +P(C) =14 +14 +14=34.A

istheeventaheart isnot obtained (orsimilarlytheeventaclub, spade, ordiamond is obtained ) so thatBis a subset ofA

.1.3.10 (a) A B = {A, A}(b) A C = {A, A, A, A, K, K, K, K, Q, Q, Q, Q,J, J, J, J}(c) B C

= {A, 2, . . . , 10, A, 2, . . . , 10}(d) B

C = {K, K, Q, Q, J, J}A (B

C) = {A, A, A, A, K, K, Q, Q, J, J}1.3. COMBINATIONSOFEVENTS 151.3.11 Let the eventO be an on time repair and let the eventSbe a satisfactory repair.It is known thatP(O S) = 0.26,P(O) = 0.74 andP(S) = 0.41.We want to ndP(O

S

).Since the eventO

S

can be written (O S)

it follows thatP(O

S

) = 1 P(O S)= 1 (P(O) +P(S) P(O S))= 1 (0.74 + 0.41 0.26) = 0.11.1.3.12 LetR be the event that a red ball is chosen and letSbe the event that a shiny ballis chosen.It is known thatP(R S) =55200,P(S) =91200andP(R) =79200.Therefore, the probability that the chosen ball is either shiny or red isP(R S) = P(R) +P(S) P(R S)=79200 +91200 55200=115200= 0.575.The probability of a dull blue ball isP(R

S

) = 1 P(R S)= 1 0.575 = 0.425.1.3.13 LetA be the event that the patient is male,letBbe the event that the patient isyounger than thirty years of age, and let C be the event that the patient is admittedto the hospital.It is given thatP(A) = 0.45,P(B) = 0.30,P(A

B

C) = 0.15,andP(A

B) = 0.21.The question asks forP(A

B

C

).Notice thatP(A

B

) = P(A

) P(A

B) = (1 0.45) 0.21 = 0.34so thatP(A

B

C

) = P(A

B

) P(A

B

C) = 0.34 0.15 = 0.19.16 CHAPTER1. PROBABILITYTHEORY1.4 ConditionalProbability1.4.1 See Figure 1.55.(a) P(A | B) =P(AB)P(B)=0.02+0.05+0.010.02+0.05+0.01+0.11+0.08+0.06+0.13= 0.1739(b) P(C | A) =P(AC)P(A)=0.02+0.05+0.08+0.040.02+0.05+0.08+0.04+0.018+0.07+0.05= 0.59375(c) P(B | A B) =P(B(AB))P(AB)=P(AB)P(AB)= 1(d) P(B | A B) =P(B(AB))P(AB)=P(B)P(AB)=0.460.46+0.320.08= 0.657(e) P(A | A B C) =P(A(ABC))P(ABC)=P(A)P(ABC)=0.3210.040.050.03= 0.3636(f) P(A B | A B) =P((AB)(AB))P(AB)=P(AB)P(AB)=0.080.7= 0.11431.4.2 A = {1, 2, 3, 5} andP(A) =46=23P(5 | A) =P(5A)P(A)=P(5)P(A)=(16)(23)=14P(6 | A) =P(6A)P(A)=P()P(A)= 0P(A | 5) =P(A5)P(5)=P(5)P(5)= 11.4.3 (a) P(A | red suit) =P(Ared suit)P(red suit)=P(A)P(red suit)=(152)(2652)=126(b) P(heart | red suit) =P(heartred suit)P(red suit)=P(heart)P(red suit)=(1352)(2652)=1326=12(c) P(red suit | heart) =P(red suitheart)P(heart)=P(heart)P(heart)= 1(d) P(heart | black suit) =P(heartblack suit)P(black suit)=P()P(black suit)= 0(e) P(King | red suit) =P(Kingred suit)P(red suit)=P(K, K)P(red suit)=(252)(2652)=226=113(f) P(King | red picture card) =P(Kingred picture card)P(red picture card)1.4. CONDITIONALPROBABILITY 17=P(K, K)P(red picture card)=(252)(652)=26=131.4.4 P(A) is smaller thanP(A | B).Event B is a necessary condition for event A and so conditioning on event B increasesthe probability of eventA.1.4.5 There are 54 blue balls and so there are 150 54 = 96 red balls.Also, there are 36 shiny, red balls and so there are 96 36 = 60 dull, red balls.P(shiny | red) =P(shiny red)P(red)=(36150)(96150)=3696=38P(dull | red) =P(dull red)P(red)=(60150)(96150)=6096=581.4.6 Let the eventO be an on time repair and let the eventSbe a satisfactory repair.It is known thatP(S | O) = 0.85 andP(O) = 0.77.The question asks forP(O S) which isP(O S) = P(S | O) P(O) = 0.85 0.77 = 0.6545.1.4.7 (a) It depends ontheweather patterns intheparticular locationthat is beingconsidered.(b) It increases since there are proportionallymore blackhairedpeople amongbrown eyed people than there are in the general population.(c) It remains unchanged.(d) It increases.1.4.8 Over a four year period including one leap year the total number of days is(3 365) + 366 = 1461.Of these, 4 12 = 48 days occur on the rst day of a month and so the probabilitythat a birthday falls on the rst day of a month is481461= 0.0329.Also, 4 31 = 124 days occur in March of which 4 days are March 1st.Consequently, the probability that a birthday falls on March 1st. conditional that itis in March is4124=131= 0.0323.18 CHAPTER1. PROBABILITYTHEORYFinally, (328) +29 = 113 days occur in February of which 4 days are February 1st.Consequently, the probability that a birthday falls on February 1st. conditional thatit is in February is4113= 0.0354.1.4.9 (a) Let A be the event that Type I battery lasts longest consisting of the outcomes{(III, II, I), (II, III, I)}.Let BbetheeventthatTypeIbatterydoesnot fail rst consistingof theoutcomes {(III,II,I), (II,III,I), (II,I,III), (III,I,II)}.The eventA B= {(III,II,I), (II,III,I)} is the same as eventA.Therefore,P(A | B) =P(AB)P(B)=0.39+0.030.39+0.03+0.24+0.16= 0.512.(b) LetCbe the event that TypeIIbatteryfailsrstconsisting of the outcomes{(II,I,III), (II,III,I)}.Thus,A C = {(II, III, I)} and thereforeP(A | C) =P(AC)P(C)=0.390.39+0.24= 0.619.(c) Let D be the event that Type II battery lasts longest consisting of the outcomes{(I,III,II), (III,I,II)}.Thus,A D = and thereforeP(A | D) =P(AD)P(D)= 0.(d) LetEbetheeventthatTypeIIbatterydoesnotfail rst consistingoftheoutcomes {(I,III,II), (I,II,III), (III,II,I), (III,I,II)}.Thus,A E = {(III, II, I)} and thereforeP(A | E) =P(AE)P(E)=0.030.07+0.11+0.03+0.16= 0.081.1.4.10 See Figure 1.25.(a) Let A be the event both lines at full capacity consisting of the outcome {(F,F)}.LetBbe the event neither line is shut down consisting of the outcomes{(P,P), (P,F), (F,P), (F,F)}.Thus,A B = {(F, F)} and thereforeP(A | B) =P(AB)P(B)=0.19(0.14+0.2+0.21+0.19)= 0.257.(b) Let C be the event at least one line at full capacity consisting of the outcomes{(F,P), (F,S), (F,F), (S,F), (P,F)}.Thus,C B = {(F, P), (F, F), (P, F)} and thereforeP(C | B) =P(CB)P(B)=0.21+0.19+0.20.74= 0.811.1.4. CONDITIONALPROBABILITY 19(c) Let D be the event that one line is at full capacity consisting of the outcomes{(F,P), (F,S), (P,F), (S,F)}.LetEbe the event one line is shut down consisting of the outcomes{(S,P), (S,F), (P,S), (F,S)}.Thus,D E = {(F, S), (S, F)} and thereforeP(D | E) =P(DE)P(E)=0.06+0.050.06+0.05+0.07+0.06= 0.458.(d) LetG be the event that neither line is at full capacity consisting of theoutcomes {(S,S), (S,P), (P,S), (P,P)}.LetHbetheeventthatatleastonelineisatpartial capacityconsistingofthe outcomes {(S,P), (P,S), (P,P), (P,F), (F,P)}.Thus,F G = {(S, P), (P, S), (P, P)} and thereforeP(F | G) =P(FG)P(G)=0.06+0.07+0.140.06+0.07+0.14+0.2+0.21= 0.397.1.4.11 LetL, WandHbetheeventsthatthelength, widthandheightrespectivelyarewithin the specied tolerance limits.It is giventhat P(W) =0.86, P(L W H) =0.80, P(L W H

) =0.02,P(L

W H) = 0.03 andP(W H) = 0.92.(a) P(W H) = P(L W H) +P(L

W H) = 0.80 + 0.03 = 0.83P(H) = P(W H) P(W) +P(W H) = 0.92 0.86 + 0.83 = 0.89P(W H | H) =P(WH)P(H)=0.830.89= 0.9326(b) P(L W) = P(L W H) +P(L W H

) = 0.80 + 0.02 = 0.82P(L W H | L W) =P(L WH)P(L W)=0.800.82= 0.97561.4.12 Let A be the event that the gene is of type A, and let D be the event that the geneis dominant.P(D | A

) = 0.31P(A

D) = 0.22Therefore,P(A) = 1 P(A

)= 1 P(A

D)P(D|A

)= 1 0.220.31= 0.2901.4.13 (a) LetEbe the event that the componentpassesonperformance,letA be theevent that the componentpassesonappearance, and letCbe the event thatthe component passes on cost.P(A C) = 0.4020 CHAPTER1. PROBABILITYTHEORYP(E A C) = 0.31P(E) = 0.64P(E

A

C

) = 0.19P(E

A C

) = 0.06Therefore,P(E

A

C) = P(E

A

) P(E

A

C

)= P(E

) P(E

A) 0.19= 1 P(E) P(E

A C) P(E

A C

) 0.19= 1 0.64 P(A C) +P(E A C) 0.06 0.19= 1 0.64 0.40 + 0.31 0.06 0.19 = 0.02(b) P(E A C | A C) =P(EAC)P(AC)=0.310.40= 0.7751.4.14 (a) LetTbe the event good taste, letSbe the event good size, and letA be theevent good appearance.P(T) = 0.78P(T S) = 0.69P(T S

A) = 0.05P(S A) = 0.84Therefore,P(S | T) =P(TS)P(T)=0.690.78= 0.885.(b) Notice thatP(S

A

) = 1 P(S A) = 1 0.84 = 0.16.Also,P(T S

) = P(T) P(T S) = 0.78 0.69 = 0.09so thatP(T S

A

) = P(T S

) P(T S

A) = 0.09 0.05 = 0.04.Therefore,P(T | S

A

) =P(TS

A

)P(S

A

)=0.040.16= 0.25.1.4.15 P(delay) = (P(delay | technical problems) P(technical problems))+ (P(delay | no technical problems) P(no technical problems))= (1 0.04) + (0.33 0.96) = 0.35681.4.16 LetSbetheeventthatachipsurvives500temperaturecyclesandletAbetheevent that the chip was made by company A.P(S) = 0.42P(A | S

) = 0.731.4. CONDITIONALPROBABILITY 21Therefore,P(A

S

) = P(S

) P(A

| S

) = (1 0.42) (1 0.73) = 0.1566.22 CHAPTER1. PROBABILITYTHEORY1.5 ProbabilitiesofEventIntersections1.5.1 (a) P(both cards are picture cards) =1252 1151=1322652(b) P(both cards are from red suits) =2652 2551=6502652(c) P(one card is from a red suit and one is from black suit)= (P(rst card is red) P(2nd card is black | 1st card is red))+ (P(rst card is black) P(2nd card is red | 1st card is black))=_2652 2651_+_2652 2651_ =6762652 2 =26511.5.2 (a) P(both cards are picture cards) =1252 1252=9169The probability increases with replacement.(b) P(both cards are from red suits) =2652 2652=14The probability increases with replacement.(c) P(one card is from a red suit and one is from black suit)= (P(rst card is red) P(2nd card is black | 1st card is red))+ (P(rst card is black) P(2nd card is red | 1st card is black))=_2652 2652_+_2652 2652_ =12The probability decreases with replacement.1.5.3 (a) No, they are not independent.Notice thatP((ii)) =313 = P((ii) | (i)) =1151.(b) Yes, they are independent.Notice thatP((i) (ii)) = P((i)) P((ii))sinceP((i)) =14P((ii)) =313andP((i) (ii)) = P(rst card a heart picture (ii))+P(rst card a heart but not a picture (ii))=_352 1151_+_1052 1251_ =1532652=352.(c) No, they are not independent.Notice thatP((ii)) =12 = P((ii) | (i)) =2551.1.5. PROBABILITIESOFEVENTINTERSECTIONS 23(d) Yes, they are independent.Similar to part (b).(e) No, they are not independent.1.5.4 P(all four cards are hearts) = P(lst card is a heart)P(2nd card is a heart | lst card is a heart)P(3rd card is a heart | 1st and 2nd cards are hearts)P(4th card is a heart | 1st, 2nd and 3rd cards are hearts)=1352 1251 1150 1049= 0.00264P(all 4 cards from red suits) = P(1st card from red suit)P(2nd card is from red suit | lst card is from red suit)P(3rd card is from red suit | 1st and 2nd cards are from red suits)P(4th card is from red suit | 1st, 2nd and 3rd cards are from red suits)=2652 2551 2450 2349= 0.055P(all 4 cards from dierent suits) = P(1st card from any suit)P(2nd card not from suit of 1st card)P(3rd card not from suit of 1st or 2nd cards)P(4th card not from suit of 1st, 2nd, or 3rd cards)= 1 3951 2650 1349= 0.1051.5.5 P(all 4 cards are hearts) = (1352)4=1256The probability increases with replacement.P(all 4 cards are from red suits) = (2652)4=116The probability increases with replacement.P(all 4 cards from dierent suits) = 1 3952 2652 1352=332The probability decreases with replacement.1.5.6 The eventsA andBare independent so thatP(A |B) =P(A),P(B |A) =P(B),andP(A B) =P(A)P(B).To show that two events are independent it needs to be shown that one of the abovethree conditions holds.(a) Recall thatP(A B) +P(A B

) = P(A)and24 CHAPTER1. PROBABILITYTHEORYP(B) +P(B

) = 1.Therefore,P(A | B

) =P(AB

)P(B

)=P(A)P(AB)1P(B)=P(A)P(A)P(B)1P(B)=P(A)(1P(B))1P(B)=P(A).(b) Similar to part (a).(c) P(A

B

) +P(A

B) = P(A

)so thatP(A

B

) = P(A) P(A

B) = P(A) P(A

)P(B)since the eventsA

andBare independent.Therefore,P(A

B

) = P(A)(1 P(B)) = P(A

)P(B

).1.5.7 The only way that a message will not get through the network is if both branches areclosedatthesametime. Thebranchesareindependentsincetheswitchesoperateindependently of each other.Therefore,P(message gets through the network)= 1 P(message cannot get through the top branch or the bottom branch)= 1 (P(message cannot get through the top branch)P(message cannot get through the bottom branch)).Also,P(message gets through the top branch) = P(switch 1 is open switch 2 is open)= P(switch 1 is open) P(switch 2 is open)= 0.88 0.92 = 0.8096since the switches operate independently of each other.Therefore,P(message cannot get through the top branch)= 1 P(message gets through the top branch)= 1 0.8096 = 0.1904.Furthermore,P(message cannot get through the bottom branch)1.5. PROBABILITIESOFEVENTINTERSECTIONS 25= P(switch 3 is closed) = 1 0.9 = 0.1.Therefore,P(message gets through the network) = 1 (0.1 0.1904) = 0.98096.1.5.8 Giventhebirthdayoftherstperson, thesecondpersonhasadierentbirthdaywith a probability364365.The third person has a dierent birthday from the rst two people with a probability363365, and so the probability that all three people have dierent birthdays is1 364365 363365.Continuing in this manner the probability thatn people all have dierent birthdaysis therefore364365 363365 362365 . . . 366n365andP(at least 2 people out of n share the same birthday)= 1 P(n people all have dierent birthdays)= 1 _364365 363365 . . .366n365_.This probability is equal to 0.117 forn = 10,is equal to 0.253 forn = 15,is equal to 0.411 forn = 20,is equal to 0.569 forn = 25,is equal to 0.706 forn = 30,and is equal to 0.814 forn = 35.The smallest values ofn for which the probability is greater than 0.5 isn = 23.Note that in these calculations it has been assumed that birthdays are equally likelytooccuronanydayof theyear, althoughinpracticeseasonal variationsmaybeobserved in the number of births.1.5.9 P(no broken bulbs) =83100 8299 8198= 0.5682P(one broken bulb) = P(broken, not broken, not broken)+P(not broken, broken, not broken) +P(not broken, not broken, broken)=_17100 8399 8298_+_83100 1799 8298_+_83100 8299 1798_ = 0.3578P(no more than one broken bulb in the sample)= P(no broken bulbs) +P(one broken bulb)= 0.5682 + 0.3578 = 0.926026 CHAPTER1. PROBABILITYTHEORY1.5.10 P(no broken bulbs) =83100 83100 83100= 0.5718P(one broken bulb) = P(broken, not broken, not broken)+P(not broken, broken, not broken) +P(not broken, not broken, broken)=_17100 83100 83100_+_83100 17100 83100_+_83100 83100 17100_ = 0.3513P(no more than one broken bulb in the sample)= P(no broken bulbs) +P(one broken bulb)= 0.5718 + 0.3513 = 0.9231The probability of nding no broken bulbs increases with replacement, but the prob-ability of nding no more than one broken bulb decreases with replacement.1.5.11 P(drawing 2 green balls)= P(1st ball is green) P(2nd ball is green | 1st ball is green)=72169 71168= 0.180P(two balls same color)= P(two red balls) +P(two blue balls) +P(two green balls)=_43169 42168_+_54169 53168_+_72169 71168_ = 0.344P(two balls dierent colors) = 1 P(two balls same color)= 1 0.344 = 0.6561.5.12 P(drawing 2 green balls) =72169 72169= 0.182P(two balls same color)= P(two red balls) +P(two blue balls) +P(two green balls)=_43169 43169_+_54169 54169_+_72169 72169_ = 0.348P(two balls dierent colors) = 1 P(two balls same color)= 1 0.348 = 0.652Theprobabilitythatthetwoballsaregreenincreaseswithreplacementwhiletheprobability of drawing two balls of dierent colors decreases with replacement.1.5.13 P(same result on both throws) = P(both heads) +P(both tails)= p2+ (1 p)2= 2p22p + 1 = 2(p 0.5)2+ 0.5which is minimized whenp = 0.5 (a fair coin).1.5. PROBABILITIESOFEVENTINTERSECTIONS 271.5.14 P(each score is obtained exactly once)= 1 56 46 36 26 16=5324P(no sixes in seven rolls) =_56_7= 0.2791.5.15 (a)_12_5=132(b) 1 56 46=59(c) P(BBR) +P(BRB) +P(RBB)=_12 12 12_+_12 12 12_+_12 12 12_=38(d) P(BBR) +P(BRB) +P(RBB)=_2652 2551 2650_+_2652 2651 2550_+_2652 2651 2550_=13341.5.16 1 (1 0.90)n 0.995is satised forn 3.1.5.17 Claims from clients in the same geographical area would not be independent of eachother since they would all be aected by the same ooding events.1.5.18 (a) P(system works) = 0.88 0.78 0.92 0.85 = 0.537(b) P(system works) = 1 P(no computers working)= 1 ((1 0.88) (1 0.78) (1 0.92) (1 0.85)) = 0.9997(c) P(system works) = P(all computers working)+P(computers 1,2,3 working, computer 4 not working)+P(computers 1,2,4 working, computer 3 not working)+P(computers 1,3,4 working, computer 2 not working)+P(computers 2,3,4 working, computer 1 not working)= 0.537 + (0.88 0.78 0.92 (1 0.85)) + (0.88 0.78 (1 0.92) 0.85)+ (0.88 (1 0.78) 0.92 0.85) + ((1 0.88) 0.78 0.92 0.85)= 0.90328 CHAPTER1. PROBABILITYTHEORY1.6 PosteriorProbabilities1.6.1 (a) The following information is given:P(disease) = 0.01P(no disease) = 0.99P(positive blood test | disease) = 0.97P(positive blood test | no disease) = 0.06Therefore,P(positive blood test) = (P(positive blood test | disease) P(disease))+(P(positive blood test | no disease) P(no disease))= (0.97 0.01) + (0.06 0.99) = 0.0691.(b) P(disease | positive blood test)=P(positive blood test disease)P(positive blood test)=P(positive blood test | disease)P(disease)P(positive blood test)=0.970.010.0691= 0.1404(c) P(no disease | negative blood test)=P(no disease negative blood test)P(negative blood test)=P(negative blood test | no disease)P(no disease)1P(positive blood test)=(10.06)0.99(10.0691)= 0.99971.6.2 (a) P(red) = (P(red | bag 1) P(bag 1)) + (P(red | bag 2) P(bag 2))+ (P(red | bag 3) P(bag 3))=_13 310_+_13 812_+_13 516_ = 0.426(b) P(blue) = 1 P(red) = 1 0.426 = 0.574(c) P(red ball from bag 2) = P(bag 2) P(red ball | bag 2)=13 812=29P(bag 1 | red ball) =P(bag 1 red ball)P(red ball)=P(bag 1)P(red ball | bag 1)P(red ball)=133100.426= 0.235P(bag 2 | blue ball) =P(bag 2 blue ball)P(blue ball)=P(bag 2)P(blue ball | bag 1)P(blue ball)1.6. POSTERIORPROBABILITIES 29=134120.574= 0.1941.6.3 (a) P(Section I) =55100(b) P(grade is A)= (P(A | Section I) P(Section I)) + (P(A | Section II) P(Section II))=_1055 55100_+_1145 45100_ =21100(c) P(A | Section I) =1055(d) P(Section I | A) =P(A Section I)P(A)=P(Section I)P(A | Section I)P(A)=55100105521100=10211.6.4 The following information is given:P(Species 1) = 0.45P(Species 2) = 0.38P(Species 3) = 0.17P(Tagged | Species 1) = 0.10P(Tagged | Species 2) = 0.15P(Tagged | Species 3) = 0.50Therefore,P(Tagged) = (P(Tagged | Species 1) P(Species 1))+ (P(Tagged | Species 2) P(Species 2)) + (P(Tagged | Species 3) P(Species 3))= (0.10 0.45) + (0.15 0.38) + (0.50 0.17) = 0.187.P(Species 1 | Tagged) =P(Tagged Species 1)P(Tagged)=P(Species 1)P(Tagged | Species 1)P(Tagged)=0.450.100.187= 0.2406P(Species 2 | Tagged) =P(Tagged Species 2)P(Tagged)=P(Species 2)P(Tagged | Species 2)P(Tagged)=0.380.150.187= 0.3048P(Species 3 | Tagged) =P(Tagged Species 3)P(Tagged)30 CHAPTER1. PROBABILITYTHEORY=P(Species 3)P(Tagged | Species 3)P(Tagged)=0.170.500.187= 0.45451.6.5 (a) P(fail) = (0.02 0.77) + (0.10 0.11) + (0.14 0.07) + (0.25 0.05)= 0.0487P(C | fail) =0.140.070.0487= 0.2012P(D | fail) =0.250.050.0487= 0.2567The answer is 0.2012 + 0.2567 = 0.4579.(b) P(A | did not fail) =P(A)P(did not fail | A)P(did not fail)=0.77(10.02)10.0487= 0.79321.6.6 P(C) = 0.15P(W) = 0.25P(H) = 0.60P(R | C) = 0.30P(R | W) = 0.40P(R | H) = 0.50Therefore,P(C | R

) =P(R

|C)P(C)P(R

|C)P(C)+P(R

|W)P(W)+P(R

|H)P(H)=(10.30)0.15((10.30)0.15)+((10.40)0.25)+((10.50)0.60)= 0.1891.6.7 (a) P(C) = 0.12P(M) = 0.55P(W) = 0.20P(H) = 0.13P(L | C) = 0.003P(L | M) = 0.009P(L | W) = 0.014P(L | H) = 0.018Therefore,P(H | L) =P(L|H)P(H)P(L|C)P(C)+P(L|M)P(M)+P(L|W)P(W)+P(L|H)P(H)=0.0180.13(0.0030.12)+(0.0090.55)+(0.0140.20)+(0.0180.13)= 0.2241.6. POSTERIORPROBABILITIES 31(b) P(M | L

) =P(L

|M)P(M)P(L

|C)P(C)+P(L

|M)P(M)+P(L

|W)P(W)+P(L

|H)P(H)=0.9910.55(0.9970.12)+(0.9910.55)+(0.9860.20)+(0.9820.13)= 0.5511.6.8 (a) P(A) = 0.12P(B) = 0.34P(C) = 0.07P(D) = 0.25P(E) = 0.22P(M | A) = 0.19P(M | B) = 0.50P(M | C) = 0.04P(M | D) = 0.32P(M | E) = 0.76Therefore,P(C | M) =P(M|C)P(C)P(M|A)P(A)+P(M|B)P(B)+P(M|C)P(C)+P(M|D)P(D)+P(M|E)P(E)=0.040.07(0.190.12)+(0.500.34)+(0.040.07)+(0.320.25)+(0.760.22)= 0.0063(b) P(D | M

) =P(M

|D)P(D)P(M

|A)P(A)+P(M

|B)P(B)+P(M

|C)P(C)+P(M

|D)P(D)+P(M

|E)P(E)=0.680.25(0.810.12)+(0.500.34)+(0.960.07)+(0.680.25)+(0.240.22)= 0.30532 CHAPTER1. PROBABILITYTHEORY1.7 CountingTechniques1.7.1 (a) 7! = 7 6 5 4 3 2 1 = 5040(b) 8! = 8 7! = 40320(c) 4! = 4 3 2 1 = 24(d) 13! = 13 12 11 . . . 1 = 6,227,020,8001.7.2 (a) P72=7!(72)!= 7 6 = 42(b) P95=9!(95)!= 9 8 7 6 5 = 15120(c) P52=5!(52)!= 5 4 = 20(d) P174=17!(174)!= 17 16 15 14 = 571201.7.3 (a) C62=6!(62)!2!=652= 15(b) C84=8!(84)!4!=876524= 70(c) C52=5!(52)!2!=542= 10(d) C146=14!(146)!6!= 30031.7.4 The number of full meals is 5 3 7 6 8 = 5040.The number of meals with just soup or appetizer is (5 + 3) 7 6 8 = 2688.1.7.5 The number of experimental congurations is 3 4 2 = 24.1.7.6 (a) Let the notation (2,3,1,4) represent the result that the player who nished 1stintournament1nished2ndintournament2, theplayerwhonished2ndintournament1nished3rdintournament2, theplayerwhonished3rdintournament 1 nished 1st in tournament 2, and the player who nished 4th intournament 1 nished 4th in tournament 2.Then the result (1,2,3,4) indicates that each competitor received the same rank-ing in both tournaments.1.7. COUNTINGTECHNIQUES 33Altogetherthereare4! =24dierentresults,eachequallylikely,andsothissingle result has a probability of124.(b) The results where no player receives the same ranking in the two tournamentsare:(2,1,4,3), (2,3,4,1), (2,4,1,3), (3,1,4,2), (3,4,1,2) (3,4,2,1), (4,1,2,3), (4,3,1,2),(4,3,2,1)There are nine of these results and so the required probability is924=38.1.7.7 The number of rankings that can be assigned to the top 5 competitors isP205=20!15!= 20 19 18 17 16 = 1,860,480.The number of ways in which the best 5 competitors can be chosen isC205=20!15!5!= 15504.1.7.8 (a) C1003=100!97!3!=10099986= 161700(b) C833=83!80!3!=8382816= 91881(c) P(no broken lightbulbs) =91881161700= 0.568(d) 17 C832= 17 83822= 57851(e) The number of samples with 0 or 1 broken bulbs is91881 + 57851 = 149732.P(sample contains no more than 1 broken bulb) =149732161700= 0.9261.7.9 Cn1k+Cn1k1=(n1)!k!(n1k)! +(n1)!(k1)!(nk)!=n!k!(nk)!_nkn+kn_ =n!k!(nk)!= CnkThis relationship can be interpreted in the following manner.Cnkisthenumberofwaysthatkballscanbeselectedfromnballs. Supposethatone ball is red while the remainingn 1 balls are blue. Either all kballs selectedare blue or one of the selected balls is red. Cn1kis the number of waysk blue ballscan be selected whileCn1k1is the number of ways of selecting the one red ball andk 1 blue balls.1.7.10 (a) The number of possible 5 card hands isC525=52!47!5!=2,598,960.(b) The number of ways to get a hand of 5 hearts isC135=13!8!5!= 1287.(c) The number of ways to get a ush is 4 C135= 4 1, 287 = 5148.34 CHAPTER1. PROBABILITYTHEORY(d) P(ush) =51482,598,960= 0.00198.(e) There are 48 choices for the fth card in the hand and so the number of handscontaining all four aces is 48.(f) 13 48 = 624(g) P(hand has four cards of the same number or picture) =6242,598,960= 0.00024.1.7.11 There aren! ways in whichn objects can be arranged in a line. If the line is madeinto a circle and rotations of the circle are considered to be indistinguishable, thenthere are n arrangements of the line corresponding to each arrangement of the circle.Consequently, there aren!n= (n 1)! ways to order the objects in a circle.1.7.12 The number of ways that six people can sit in a line at a cinema is 6! = 720.See the previous problem.The number of ways that six people can sit around a dinner table is 5! = 120.1.7.13 Consider 5 blocks, one block being Andrea and Scott and the other four blocks beingthe other four people. At the cinema these 5 blocks can be arranged in 5! ways, andthen Andrea and Scott can be arranged in two dierent ways within their block, sothat the total number of seating arrangements is 2 5! = 240.Similarly, the total number of seating arrangements at the dinner table is 24! = 48.If Andrea refuses to sit next to Scott then the number of seating arrangements canbe obtained by subtraction. The total number of seating arrangements at the cinemais 720 240 = 480 and the total number of seating arrangements at the dinner tableis 120 48 = 72.1.7.14 The total number of arrangements ofn balls isn! which needs to be divided byn1!because the rearrangements of then1 balls in box 1 are indistinguishable, and simi-larly it needs to be divided by n2! . . . nk! due to the indistinguishable rearrangementspossible in boxes 2 tok.Whenk= 2theproblemisequivalenttothenumberofwaysofselectingn1balls(orn2 balls) fromn = n1 +n2 balls.1.7.15 (a) Using the result provided in the previous problem the answer is12!3!4!5!= 27720.(b) Suppose that the balls in part (a) are labelled from 1 to 12. Then the positionsof the three red balls in the line (where the places in the line are labelled 1 to1.7. COUNTINGTECHNIQUES 3512) can denote which balls in part (a) are placed in the rst box, the positionsof the four blue balls in the line can denote which balls in part (a) are placed inthe second box, and the positions of the ve green balls in the line can denotewhich balls in part (a) are placed in the third box. Thus, there is a one-to-onecorrespondence between the positioning of the colored balls in part (b) and thearrangements of the balls in part (a) so that the problems are identical.1.7.1614!3!4!7!= 1201201.7.1715!3!3!3!3!3!= 168,168,0001.7.18 The total number of possible samples isC6012.(a) Thenumberof samplescontainingonlyitemswhichhaveeitherexcellentorgood quality isC4312.Therefore, the answer isC4312C6012=4360 4259 . . . 3249= 0.0110.(b) The number of samples that contain three items of excellent quality, three itemsof good quality, three items of poor quality and three defective items isC183C253C123C53=4,128,960,000.Therefore, the answer is4,128,960,000C6012= 0.00295.1.7.19 The ordering of the visits can be made in 10! = 3,628,800 dierent ways.The number of dierent ways the ten cities be split into two groups of ve cities isC105= 252.1.7.20_262__263_ = 8450001.7.21 (a)_398__528_=3952 3851 3750 3649 3548 3447 3346 3245= 0.082(b)_132__132__132__132__528_= 0.04936 CHAPTER1. PROBABILITYTHEORY1.7.22_52__304__52__408_= 0.03561.9. SUPPLEMENTARYPROBLEMS 371.9 SupplementaryProblems1.9.1 S = { 1, 1.5, 2, 2.5, 3, 3.5, 4, 4.5, 5, 5.5, 6}1.9.2 If thefourcontestantsarelabelledA, B, C, Dandthenotation(X, Y )isusedtoindicate that contestant X is the winner and contestant Yis the runner up, then thesample space is:S = {(A, B), (A, C), (A, D), (B, A), (B, C), (B, D),(C, A), (C, B), (C, D), (D, A), (D, B), (D, C)}1.9.3 One way is to have the two team captains each toss the coin once. If one obtains ahead and the other a tail,then the one with the head wins (this could just as wellbedonetheotherwayaroundsothattheonewiththetail wins, aslongasitisdecided beforehand). If both captains obtain the same result, that is if there are twoheadsortwotails,thentheprocedurecouldberepeateduntildierentresultsareobtained.1.9.4 See Figure 1.10.There are 36 equally likely outcomes,16 of which have scores diering by no morethan one.Therefore,P(the scores on two dice dier by no more than one) =1636=49.1.9.5 The number of ways to pick a card is 52.The number of ways to pick a diamond picture card is 3.Therefore,P(picking a diamond picture card) =352.1.9.6 With replacement:P(drawing two hearts) =1352 1352=116= 0.0625Without replacement:P(drawing two hearts) =1352 1251=351= 0.0588The probability decreases without replacement.1.9.7 A = {(1, 1), (1, 2), (1, 3), (2, 1), (2, 2), (3, 1)}B = {(1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6)}38 CHAPTER1. PROBABILITYTHEORY(a) A B = {(1, 1), (2, 2)}P(A B) =236=118(b) A B = {(1, 1), (1, 2), (1, 3), (2, 1), (2, 2), (3, 1), (3, 3), (4, 4), (5, 5), (6, 6)}P(A B) =1036=518(c) A

B = {(1, 1), (1, 4), (1, 5), (1, 6), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6),(3, 2), (3, 3), (3, 4), (3, 5), (3, 6), (4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6),(5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)}P(A

B) =3236=891.9.8 See Figure 1.10.Let the notation (x, y) indicate that the score on the red die isx and that the scoreon the blue die isy.(a) The event the sum of the scores on the two dice is eightconsists of the outcomes:{(2, 6), (3, 5), (4, 4), (5, 3), (6, 2)}Therefore,P(red die is 5 | sum of scores is 8)=P(red die is 5 sum of scores is 8)P(sum of scores is 8)=(136)(536)=15.(b) P(either score is 5 | sum of scores is 8) = 2 15=25(c) The event the score on either die is 5consists of the 11 outcomes:{(1, 5), (2, 5), (3, 5), (4, 5), (5, 5), (6, 5), (5, 6), (5, 4), (5, 3), (5, 2), (5, 1)}Therefore,P(sum of scores is 8 | either score is 5)=P(sum of scores is 8 either score is 5)P(either score is 5)=(236)(1136)=211.1.9.9 P(A) = P(either switch 1 or 4 is open or both)= 1 P(both switches 1 and 4 are closed)= 1 0.152= 0.9775P(B) = P(either switch 2 or 5 is open or both)1.9. SUPPLEMENTARYPROBLEMS 39= 1 P(both switches 2 and 5 are closed)= 1 0.152= 0.9775P(C) = P(switches 1 and 2 are both open) = 0.852= 0.7225P(D) = P(switches 4 and 5 are both open) = 0.852= 0.7225IfE = C D thenP(E) = 1 (P(C

) P(D

))= 1 (1 0.852)2= 0.923.Therefore,P(message gets through the network)= (P(switch 3 is open) P(A) P(B)) + (P(switch 3 closed) P(E))= (0.85 (1 0.152)2) + (0.15 (1 (1 0.852)2)) = 0.9506.1.9.10 The sample space for the experiment of two coin tosses consists of the four equallylikely outcomes:{(H, H), (H, T), (T, H), (T, T)}Three out of these four outcomes contain at least one head, so thatP(at least one head in two coin tosses) =34.The sample space for four tosses of a coin consists of 24= 16 equally likely outcomesof which the following 11 outcomes contain at least two heads:{(HHTT), (HTHT), (HTTH), (THHT), (THTH), (TTHH),(HHHT), (HHTH), (HTHH), (THHH), (HHHH)}Therefore,P(at least two heads in four coin tosses) =1116which is smaller than the previous probability.1.9.11 (a) P(blue ball) = (P(bag 1) P(blue ball | bag 1))+ (P(bag 2) P(blue ball | bag 2))+ (P(bag 3) P(blue ball | bag 3))+ (P(bag 4) P(blue ball | bag 4))=_0.15 716_+_0.2 818_+_0.35 919_+_0.3 711_ = 0.5112(b) P(bag 4 | green ball) =P(green ball bag 4)P(green ball)=P(bag 4)P(green ball | bag 4)P(greenball)=0.30P(green ball)= 040 CHAPTER1. PROBABILITYTHEORY(c) P(bag 1 | blue ball) =P(bag 1)P(blue ball | bag 1)P(blue ball)=0.157160.5112=0.06560.5112= 0.1281.9.12 (a) S = {1, 2, 3, 4, 5, 6, 10}(b) P(10) = P(score on die is 5) P(tails)=16 12=112(c) P(3) = P(score on die is 3) P(heads)=16 12=112(d) P(6) = P(score on die is 6) + (P(score on die is 3) P(tails))=16 + (16 12)=14(e) 0(f) P(score on die is odd | 6 is recorded)=P(score on die is odd 6 is recorded)P(6 is recorded)=P(score on die is 3)P(tails)P(6 is recorded)=(112)(14)=131.9.13 54= 62545= 1024In this case 54< 45, and in generalnn12< nn21when 3 n1< n2.1.9.1420!5!5!5!5!= 1.17 101020!4!4!4!44!= 3.06 10111.9.15 P(X = 0) =14P(X = 1) =12P(X = 2) =14P(X = 0 | white) =18P(X = 1 | white) =12P(X = 2 | white) =381.9. SUPPLEMENTARYPROBLEMS 41P(X = 0 | black) =12P(X = 1 | black) =12P(X = 2 | black) = 01.9.16 LetA be the event that the order is from a rst time customerand letBbe the event that the order is dispatched within one day.It is given thatP(A) = 0.28,P(B | A) = 0.75, andP(A

B

) = 0.30.Therefore,P(A

B) = P(A

) P(A

B

)= (1 0.28) 0.30 = 0.42P(A B) = P(A) P(B | A)= 0.28 0.75 = 0.21P(B) = P(A

B) +P(A B)= 0.42 + 0.21 = 0.63andP(A | B) =P(A B)P(B)=0.210.63=13.1.9.17 It is given thatP(Puccini) = 0.26P(Verdi) = 0.22P(other composer) = 0.52P(female | Puccini) = 0.59P(female | Verdi) = 0.45andP(female) = 0.62.(a) SinceP(female) = (P(Puccini) P(female | Puccini))+ (P(Verdi) P(female | Verdi))+ (P(other composer) P(female | other composer))it follows that0.62 = (0.26 0.59) + (0.22 0.45) + (0.52 P(female | other composer))so thatP(female | other composer) = 0.7069.42 CHAPTER1. PROBABILITYTHEORY(b) P(Puccini | male) =P(Puccini)P(male | Puccini)P(male)=0.26(10.59)10.62= 0.2811.9.18 The total number of possible samples isC9210.(a) The number of samples that do not contain any bers of polymer B isC7510.Therefore, the answer isC7510C9210=7592 7491 . . . 6683= 0.115.(b) The number of samples that contain exactly one ber of polymer B is 17C759.Therefore, the answer is17C759C9210= 0.296.(c) The number of samples that contain three bers of polymer A, three bers ofpolymer B, and four bers of polymer C isC433C173C324.Therefore, the answer isC433C173C324C9210= 0.042.1.9.19 Thetotal numberof possiblesequencesof headsandtailsis25=32, witheachsequencebeingequallylikely. Of these, sixteendontincludeasequenceof threeoutcomes of the same kind.Therefore, the required probability is1632= 0.5.1.9.20 (a) Calls answered by an experienced operator that last over ve minutes.(b) Successfullyhandledcallsthatwereansweredeitherwithintensecondsorbyan inexperienced operator (or both).(c) Callsansweredaftertensecondsthatlastedmorethanveminutesandthatwere not handled successfully.(d) Callsthatwereeitheransweredwithintensecondsandlastedlessthanveminutes,or that were answered by an experienced operator and were handledsuccessfully.1.9.21 (a)20!7!7!6!= 133, 024, 3201.9. SUPPLEMENTARYPROBLEMS 43(b) If the rst and the second job are assigned to production line I, the number ofassignments is18!5!7!6!= 14, 702, 688.If the rst and the second job are assigned to production line II, the number ofassignments is18!7!5!6!= 14, 702, 688.If the rst and the second job are assigned to production line III, the numberof assignments is18!7!7!4!= 10, 501, 920.Therefore, the answer is14, 702, 688 + 14, 702, 688 + 10, 501, 920 = 39, 907, 296.(c) The answer is 133, 024, 320 39, 907, 296 = 93, 117, 024.1.9.22 (a)_133__523_=1352 1251 1150= 0.0129(b)_41__41__41__523_=1252 851 450= 0.00291.9.23 (a)_484__524_=4852 4751 4650 4549= 0.719(b)_41__483__524_=4448474652515049= 0.256(c)_152_3=11406081.9.24 (a) True44 CHAPTER1. PROBABILITYTHEORY(b) False(c) False(d) True(e) True(f) False(g) False1.9.25 LetWbe the event that the team wins the gameand letSbe the event that the team has a player sent o.P(W) = 0.55P(S

) = 0.85P(W | S

) = 0.60SinceP(W) = P(W S) +P(W S

)= P(W S) + (P(W | S

) P(S

))it follows that0.55 = P(W S) + (0.60 0.85).Therefore,P(W S) = 0.04.1.9.26 (a) LetNbe the event that the machine is newand letG be the event that the machine has good quality.P(N G

) =120500P(N

) =230500Therefore,P(N G) = P(N) P(N G

)= 1 230500 120500=150500= 0.3.(b) P(G | N) =P(NG)P(N)=0.31230500=591.9. SUPPLEMENTARYPROBLEMS 451.9.27 (a) LetMbe the event male,letEbe the event mechanical engineer,and letSbe the event senior.P(M) =113250P(E) =167250P(M

E

) =52250P(M

E S) =19250Therefore,P(M | E

) = 1 P(M

| E

)= 1 P(M

E

)P(E

)= 1 52250167= 0.373.(b) P(S | M

E) =P(M

E S)P(M

E)=P(M

E S)P(M

)P(M

E

)=1925011352= 0.2241.9.28 (a) LetTbe the event that the tax form is led on time,letSbe the event that the tax form is from a small business,and letA be the event that the tax form is accurate.P(T S A) = 0.11P(T

S A) = 0.13P(T S) = 0.15P(T

S A

) = 0.21Therefore,P(T | S A) =P(TS A)P(S A)=P(TS A)P(TS A)+P(T

S A)=0.110.11+0.13=1124.(b) P(S

) = 1 P(S)= 1 P(T S) P(T

S)= 1 P(T S) P(T

S A) P(T

S A

)= 1 0.15 0.13 0.21 = 0.511.9.29 (a) P(having exactly two heart cards) =C132C392C524= 0.213(b) P(having exactly two heart cards and exactly two club cards)=C132C132C524= 0.02246 CHAPTER1. PROBABILITYTHEORY(c) P(having 3 heart cards | no club cards)= P(having 3 heart cards from a reduced pack of 39 cards)=C133C261C394= 0.091.9.30 (a) P(passing the rst time) = 0.26P(passing the second time) = 0.43P(failing the rst time and passing the second time)= P(failing the rst time) P(passing the second time)= (1 0.26) 0.43 = 0.3182(b) 1 P(failing both times) = 1 (1 0.26) (1 0.43) = 0.5782(c) P(passing the rst time | moving to the next stage)=P(passing the rst time and moving to the next stage)P(moving to the next stage)=0.260.5782= 0.451.9.31 The possible outcomes are (6, 5, 4, 3, 2), (6, 5, 4, 3, 1), (6, 5, 4, 2, 1), (6, 5, 3, 2, 1),(6, 4, 3, 2, 1), and (5, 4, 3, 2, 1).Each outcome has a probability of165so that the required probability is665=164=11296.1.9.32 P(at least one uncorrupted le) = 1 P(both les corrupted)= 1 (0.005 0.01) = 0.999951.9.33 LetCbe the event that the pump is operating correctlyand letL be the event that the light is on.P(L | C

) = 0.992P(L | C) = 0.003P(C) = 0.996Therefore, using Bayes theoremP(C

| L) =P(L | C

)P(C

)P(L | C

)P(C

)+P(L | C)P(C)=0.9920.004(0.9920.004)+(0.0030.996)= 0.57.1.9. SUPPLEMENTARYPROBLEMS 471.9.34_42__42__43__43__5210_=127,465,3201.9.35 (a)_73__113_=711 610 59=733(b)_71__42__113_=14551.9.36 (a) The probability of an infected person having strain A isP(A) = 0.32.The probability of an infected person having strain B isP(B) = 0.59.The probability of an infected person having strain C isP(C) = 0.09.P(S | A) = 0.21P(S | B) = 0.16P(S | C) = 0.63Therefore, the probability of an infected person exhibiting symptoms isP(S) = (P(S | A) P(A)) + (P(S | B) P(B)) + (P(S | C) P(C))= 0.2183andP(C | S) =P(S | C)P(C)P(S)=0.630.090.2183= 0.26.(b) P(S

) = 1 P(S) = 1 0.2183 = 0.7817P(S

| A) = 1 P(S | A) = 1 0.21 = 0.79Therefore,P(A | S

) =P(S

| A)P(A)P(S

)=0.790.320.7817= 0.323.(c) P(S

) = 1 P(S) = 1 0.2183 = 0.781748 CHAPTER1. PROBABILITYTHEORYChapter2RandomVariables2.1 DiscreteRandomVariables2.1.1 (a) Since0.08 + 0.11 + 0.27 + 0.33 +P(X = 4) = 1it follows thatP(X = 4) = 0.21.(c) F(0) = 0.08F(1) = 0.19F(2) = 0.46F(3) = 0.79F(4) = 1.002.1.2xi-4 -1 0 2 3 7pi0.21 0.11 0.07 0.29 0.13 0.192.1.3xi1 2 3 4 5 6 8 9 10pi136236236336236436236136236F(xi)136336536836103614361636173619364950 CHAPTER2. RANDOMVARIABLESxi12 15 16 18 20 24 25 30 36pi436236136236236236136236136F(xi)2336253626362836303632363336353612.1.4 (a)xi0 1 2pi0.5625 0.3750 0.0625(b)xi0 1 2F(xi) 0.5625 0.9375 1.000(c) The valuex = 0 is the most likely.Without replacement:xi0 1 2pi0.5588 0.3824 0.0588F(xi) 0.5588 0.9412 1.000Again,x = 0 is the most likely value.2.1. DISCRETERANDOMVARIABLES 512.1.5xi-5 -4 -3 -2 -1 0 1 2 3 4 6 8 10 12pi136136236236336336236536136436336336336336F(xi)1362364366369361236143619362036243627363036333612.1.6 (a)xi-6 -4 -2 0 2 4 6pi18181828181818(b)xi-6 -4 -2 0 2 4 6F(xi)1828385868781(c) The most likely value isx = 0.2.1.7 (a)xi0 1 2 3 4 6 8 12pi0.061 0.013 0.195 0.067 0.298 0.124 0.102 0.140(b)xi0 1 2 3 4 6 8 12F(xi) 0.061 0.074 0.269 0.336 0.634 0.758 0.860 1.000(c) The most likely value is 4.52 CHAPTER2. RANDOMVARIABLESP(not shipped) = P(X 1) = 0.0742.1.8xi-1 0 1 3 4 5pi161616161616F(xi)162636465612.1.9xi1 2 3 4pi2531015110F(xi)2571091012.1.10 Since

i=11i2=26it follows thatP(X = i) =62i2is a possible set of probability values.However, since

i=11idoes not converge, it follows thatP(X = i) =ciis not a possible set of probability values.2.1.11 (a) The state space is {3, 4, 5, 6}.(b) P(X = 3) = P(MMM) =36 25 14=120P(X = 4) = P(MMTM) +P(MTMM) +P(TMMM) =320P(X = 5) = P(MMTTM) +P(MTMTM) +P(TMMTM)2.1. DISCRETERANDOMVARIABLES 53+P(MTTMM) +P(TMTMM) +P(TTMMM) =620Finally,P(X = 6) =12since the probabilities sum to one, or since the nal appointment made is equallylikely to be on a Monday or on a Tuesday.P(X 3) =120P(X 4) =420P(X 5) =1020P(X 6) = 154 CHAPTER2. RANDOMVARIABLES2.2 ContinuousRandomVariables2.2.1 (a) Continuous(b) Discrete(c) Continuous(d) Continuous(e) Discrete(f) This depends on what level of accuracy to which it is measured.It could be considered to be either discrete or continuous.2.2.2 (b) _641xln(1.5)dx =1ln(1.5) [ln(x)]64=1ln(1.5) (ln(6) ln(4)) = 1.0(c) P(4.5 X 5.5) = _5.54.51xln(1.5)dx=1ln(1.5) [ln(x)]5.54.5=1ln(1.5) (ln(5.5) ln(4.5)) = 0.495(d) F(x) = _x41y ln(1.5)dy=1ln(1.5) [ln(y)]x4=1ln(1.5) (ln(x) ln(4))for 4 x 62.2.3 (a) Since_02_1564 +x64_dx =716and_30_38 +cx_dx =98 +9c2it follows that716 +98 +9c2= 1which givesc = 18.(b) P(1 X 1) = _01_1564 +x64_dx +_10_38 x8_dx=691282.2. CONTINUOUSRANDOMVARIABLES 55(c) F(x) = _x2_1564 +y64_dy=x2128 +15x64+716for 2 x 0F(x) =716 +_x0_38 y8_dy= x216 +3x8+716for 0 x 32.2.4 (b) P(X 2) = F(2) =14(c) P(1 X 3) = F(3) F(1)=916 116=12(d) f(x) =dF(x)dx=x8for 0 x 42.2.5 (a) SinceF() = 1 it follows thatA = 1.ThenF(0) = 0 gives 1 +B = 0 so thatB = 1 andF(x) = 1 ex.(b) P(2 X 3) = F(3) F(2)= e2e3= 0.0855(c) f(x) =dF(x)dx= exforx 02.2.6 (a) Since_0.50.125A (0.5 (x 0.25)2)dx = 1it follows thatA = 5.5054.(b) F(x) = _x0.125f(y)dy= 5.5054_x2 (x0.25)330.06315_for 0.125 x 0.5(c) F(0.2) = 0.20356 CHAPTER2. RANDOMVARIABLES2.2.7 (a) SinceF(0) = A+Bln(2) = 0andF(10) = A+Bln(32) = 1it follows thatA = 0.25 andB =1ln(16)= 0.361.(b) P(X> 2) = 1 F(2) = 0.5(c) f(x) =dF(x)dx=1.083x+2for 0 x 102.2.8 (a) Since_100A (e101)d = 1it follows thatA = (e1011)1= 4.54 105.(b) F() = _0f(y)dy=e10e10e1011for 0 10(c) 1 F(8) = 0.00022.2.9 (a) SinceF(0) = 0 andF(50) = 1it follows thatA = 1.0007 andB = 125.09.(b) P(X 10) = F(10) = 0.964(c) P(X 30) = 1 F(30) = 1 0.998 = 0.002(d) f(r) =dF(r)dr=375.3(r+5)4for 0 r 502.2.10 (a) F(200) = 0.1(b) F(700) F(400) = 0.652.2. CONTINUOUSRANDOMVARIABLES 572.2.11 (a) Since_1110Ax(130 x2)dx = 1it follows thatA =4819.(b) F(x) = _x104y(130y2)819dy=4819_65x2x444000_for 10 x 11(c) F(10.5) F(10.25) = 0.623 0.340 = 0.28358 CHAPTER2. RANDOMVARIABLES2.3 TheExpectationofaRandomVariable2.3.1 E(X) = (0 0.08) + (1 0.11) + (2 0.27) + (3 0.33) + (4 0.21)= 2.482.3.2 E(X) =_1 136_+_2 236_+_3 236_+_4 336_+_5 236_+_6 436_+_8 236_+_9 136_+_10 236_+_12 436_+_15 236_+_16 136_+_18 236_+_20 236_+_24 236_+_25 136_+_30 236_+_36 136_= 12.252.3.3 With replacement:E(X) = (0 0.5625) + (1 0.3750) + (2 0.0625)= 0.5Without replacement:E(X) = (0 0.5588) + (1 0.3824) + (2 0.0588)= 0.52.3.4 E(X) =_1 25_+_2 310_+_3 15_+_4 110_= 22.3.5xi2 3 4 5 6 7 8 9 10 15pi113113113113113113113113113413E(X) =_2 113_+_3 113_+_4 113_+_5 113_+_6 113_+_7 113_+_8 113_+_9 113_+_10 113_+_15 413_= $8.77If $9 is paid to play the game, the expected loss would be 23 cents.2.3. THEEXPECTATIONOFARANDOMVARIABLE 592.3.6xi1 2 3 4 5 6 7 8 9 10 11 12pi67277287297210721172672572472372272172E(X) =_1 672_+_2 672_+_3 672_+_4 672_+_5 672_+_6 672_+_7 672_+_8 672_+_9 672_+_10 672_+_11 672_+_12 672_= 5.252.3.7 P(three sixes are rolled) =16 16 16=1216so thatE(net winnings) =_$1215216_+_$4991216_= $1.31.If you can play the game a large number of times then you should play the game asoften as you can.2.3.8 The expected net winnings will be negative.2.3.9xi0 1 2 3 4 5pi0.1680 0.2816 0.2304 0.1664 0.1024 0.0512E(payment) = (0 0.1680) + (1 0.2816) + (2 0.2304)+ (3 0.1664) + (4 0.1024) + (5 0.0512)= 1.9072E(winnings) = $2 $1.91 = $0.09The expected winnings increase to 9 cents per game.Increasing the probability of scoring a three reduces the expected value of thedierence in the scores of the two dice.60 CHAPTER2. RANDOMVARIABLES2.3.10 (a) E(X) = _64x1xln(1.5)dx = 4.94(b) SolvingF(x) = 0.5 givesx = 4.90.2.3.11 (a) E(X) = _40xx8dx = 2.67(b) SolvingF(x) = 0.5 givesx =8 = 2.83.2.3.12 E(X) = _0.50.125x 5.5054 (0.5(x 0.25)2)dx = 0.3095SolvingF(x) = 0.5 givesx = 0.3081.2.3.13 E(X) = _100e1011(e101)d = 0.9977SolvingF() = 0.5 gives = 0.6927.2.3.14 E(X) = _500375.3 r(r+5)4dr = 2.44SolvingF(r) = 0.5 givesr = 1.30.2.3.15 Letf(x) be a probability density function that is symmetric about the point,so thatf( +x) = f( x).ThenE(X) = _xf(x)dxwhich under the transformationx = +y givesE(X) = _( +y)f( +y)dy= _f( +y)dy + _0y (f( +y) f( y))dy= ( 1) + 0 = .2.3.16 E(X) = (3 120) + (4 320) + (5 620) + (6 1020)=10520= 5.252.3.17 (a) E(X) = _11104x2(130x2)819dx= 10.4182342.3. THEEXPECTATIONOFARANDOMVARIABLE 61(b) SolvingF(x) = 0.5 gives the median as 10.385.2.3.18 (a) Since_32A(x 1.5)dx = 1it follows thatA_x21.5x32 = 1so thatA = 1.(b) Let the median bem.Then_m2(x 1.5)dx = 0.5so that_x21.5xm2= 0.5which gives0.5m21.5m+ 1 = 0.5.Therefore,m23m+ 1 = 0so thatm =352.Since 2 m 3 it follows thatm =3+52= 2.618.62 CHAPTER2. RANDOMVARIABLES2.4 TheVarianceofaRandomVariable2.4.1 (a) E(X) =_2 13_+_1 16_+_4 13_+_6 16_=116(b) Var(X) =_13 _2 116_2_+_16 _1 116_2_+_13 _4 116_2_+_16 _6 116_2_=34136(c) E(X2) =_13 (2)2_+_16 12_+_13 42_+_16 62_=776Var(X) = E(X2) E(X)2=776 _116_2=341362.4.2 E(X2) = (020.08) + (120.11) + (220.27)+ (320.33) + (420.21) = 7.52ThenE(X) = 2.48 so thatVar(X) = 7.52 (2.48)2= 1.37and = 1.17.2.4.3 E(X2) =_1225_+_22310_+_3215_+_42110_= 5ThenE(X) = 2 so thatVar(X) = 5 22= 1and = 1.2.4.4 See Problem 2.3.9.E(X2) = (020.168) + (120.2816) + (320.1664)+ (420.1024) + (520.0512)= 5.6192ThenE(X) = 1.9072 so thatVar(X) = 5.6192 1.90722= 1.982.4. THEVARIANCEOFARANDOMVARIABLE 63and = 1.41.A small variance is generally preferable if the expected winnings are positive.2.4.5 (a) E(X2) = _64x2 1xln(1.5)dx = 24.66ThenE(X) = 4.94 so thatVar(X) = 24.66 4.942= 0.25.(b) =0.25 = 0.5(c) SolvingF(x) = 0.25 givesx = 4.43.SolvingF(x) = 0.75 givesx = 5.42.(d) The interquartile range is 5.42 4.43 = 0.99.2.4.6 (a) E(X2) = _40x2_x8_dx = 8ThenE(X) =83so thatVar(X) = 8 _83_2=89.(b) =_89= 0.94(c) SolvingF(x) = 0.25 givesx = 2.SolvingF(x) = 0.75 givesx =12 = 3.46.(d) The interquartile range is 3.46 2.00 = 1.46.2.4.7 (a) E(X2) = _0.50.125x25.5054 (0.5 (x 0.25)2)dx = 0.1073ThenE(X) = 0.3095 so thatVar(X) = 0.1073 0.30952= 0.0115.(b) =0.0115 = 0.107(c) SolvingF(x) = 0.25 givesx = 0.217.SolvingF(x) = 0.75 givesx = 0.401.(d) The interquartile range is 0.401 0.217 = 0.184.64 CHAPTER2. RANDOMVARIABLES2.4.8 (a) E(X2) = _1002e1011(e101)d= 1.9803ThenE(X) = 0.9977 so thatVar(X) = 1.9803 0.99772= 0.985.(b) =0.985 = 0.992(c) SolvingF() = 0.25 gives = 0.288.SolvingF() = 0.75 gives = 1.385.(d) The interquartile range is 1.385 0.288 = 1.097.2.4.9 (a) E(X2) = _500375.3 r2(r+5)4dr = 18.80ThenE(X) = 2.44 so thatVar(X) = 18.80 2.442= 12.8.(b) =12.8 = 3.58(c) SolvingF(r) = 0.25 givesr = 0.50.SolvingF(r) = 0.75 givesr = 2.93.(d) The interquartile range is 2.93 0.50 = 2.43.2.4.10 Adding and subtracting two standard deviations from the mean value gives:P(60.4 X 89.6) 0.75Adding and subtracting three standard deviations from the mean value gives:P(53.1 X 96.9) 0.892.4.11 The interval (109.55, 112.05) is ( 2.5c, + 2.5c)so Chebyshevs inequality gives:P(109.55 X 112.05) 1 12.52= 0.842.4.12 E(X2) =_32120_+_42320_+_52620_+_621020_=56720Var(X) = E(X2) (E(X))22.4. THEVARIANCEOFARANDOMVARIABLE 65=56720 _10520_2=6380The standard deviation is _63/80 = 0.887.2.4.13 (a) E(X2) = _11104x3(130x2)819dx= 108.61538Therefore,Var(X) = E(X2) (E(X))2= 108.61538 10.4182342= 0.0758and the standard deviation is 0.0758 = 0.275.(b) SolvingF(x) = 0.8 gives the 80th percentile of the resistance as 10.69,and solvingF(x) = 0.1 gives the 10th percentile of the resistance as 10.07.2.4.14 (a) Since1 = _32Ax2.5dx =A3.5 (33.523.5)it follows thatA = 0.0987.(b) E(X) = _320.0987x3.5dx=0.09874.5(34.524.5) = 2.58(c) E(X2) = _320.0987x4.5dx=0.09875.5(35.525.5) = 6.741Therefore,Var(X) = 6.741 2.582= 0.085and the standard deviation is 0.085 = 0.29.(d) Solving0.5 = _x20.0987y2.5dy=0.09873.5(x3.523.5)givesx = 2.62.2.4.15 E(X) = (1 0.25) + (1 0.4) + (4 0.35)= $1.55E(X2) = ((1)20.25) + (120.4) + (420.35)= 6.2566 CHAPTER2. RANDOMVARIABLESTherefore, the variance isE(X2) (E(X))2= 6.25 1.552= 3.8475and the standard deviation is 3.8475 = $1.96.2.4.16 (a) Since1 = _43Axdx = 2A(2 3)it follows thatA = 1.866.(b) F(x) = _x31.866ydy= 3.732 (x 3)(c) E(X) = _43x1.866xdx=23 1.866 (41.531.5) = 3.488(d) E(X2) = _43x2 1.866xdx=25 1.866 (42.532.5) = 12.250Therefore,Var(X) = 12.250 3.4882= 0.0834and the standard deviation is 0.0834 = 0.289.(e) SolvingF(x) = 3.732 (x 3) = 0.5givesx = 3.48.(f) SolvingF(x) = 3.732 (x 3) = 0.75givesx = 3.74.2.4.17 (a) E(X) = (2 0.11) + (3 0.19) + (4 0.55) + (5 0.15)= 3.74(b) E(X2) = (220.11) + (320.19) + (420.55) + (520.15)= 14.70Therefore,Var(X) = 14.70 3.742= 0.7124and the standard deviation is 0.7124 = 0.844.2.4. THEVARIANCEOFARANDOMVARIABLE 672.4.18 (a) E(X) = _11x(1x)2dx = 13(b) E(X2) = _11x2(1x)2dx =13Therefore,Var(X) = E(X2) (E(X))2=13 19=29and the standard deviation is23= 0.471.(c) Solving_y1(1x)2dx = 0.75givesy = 0.68 CHAPTER2. RANDOMVARIABLES2.5 JointlyDistributedRandomVariables2.5.1 (a) P(0.8 X 1, 25 Y 30)= _1x=0.8_30y=25_39400 17(x1)250(y25)210000_dxdy= 0.092(b) E(Y ) = _3520y_831200 (y25)210000_dy = 27.36E(Y2) = _3520y2_831200 (y25)210000_dy = 766.84Var(Y ) = E(Y2) E(Y )2= 766.84 (27.36)2= 18.27Y=18.274 = 4.27(c) E(Y |X = 0.55) = _3520y_0.073 (y25)23922.5_dy = 27.14E(Y2|X = 0.55) = _3520y2_0.073 (y25)23922.5_dy = 753.74Var(Y |X = 0.55) = E(Y2|X = 0.55) E(Y |X = 0.55)2= 753.74 (27.14)2= 17.16Y |X=0.55 =17.16 = 4.142.5.2 (a) p1|Y =1 = P(X = 1|Y= 1) =p11p+1=0.120.32= 0.37500p2|Y =1 = P(X = 2|Y= 1) =p21p+1=0.080.32= 0.25000p3|Y =1 = P(X = 3|Y= 1) =p31p+1=0.070.32= 0.21875p4|Y =1 = P(X = 4|Y= 1) =p41p+1=0.050.32= 0.15625E(X|Y= 1) = (1 0.375) + (2 0.25) + (3 0.21875) + (4 0.15625)= 2.15625E(X2|Y= 1) = (120.375) + (220.25) + (320.21875) + (420.15625)= 5.84375Var(X|Y= 1) = E(X2|Y= 1) E(X|Y= 1)2= 5.84375 2.156252= 1.1943X|Y =1 =1.1943 = 1.0932.5. JOINTLYDISTRIBUTEDRANDOMVARIABLES 69(b) p1|X=2 = P(Y= 1|X = 2) =p21p2+=0.080.24=824p2|X=2 = P(Y= 2|X = 2) =p22p2+=0.150.24=1524p3|X=2 = P(Y= 3|X = 2) =p23p2+=0.010.24=124E(Y |X = 2) =_1 824_+_2 1524_+_3 124_=4124= 1.7083E(Y2|X = 2) =_12824_+_221524_+_32124_=7724= 3.2083Var(Y |X = 2) = E(Y2|X = 2) E(Y |X = 2)2= 3.2083 1.70832= 0.290Y |X=2 =0.290 = 0.5382.5.3 (a) Since_3x=2_6y=4A(x 3)ydxdy = 1it follows thatA = 1125.(b) P(0 X 1, 4 Y 5)= _1x=0_5y=4(3x)y125dxdy=9100(c) fX(x) = _64(3x)y125dy =2(3x)25for 2 x 3fY (y) = _32(3x)y125dx =y10for 4 x 6(d) The random variablesXandYare independent sincefX(x) fY (y) = f(x, y)and the ranges of the random variables are not related.(e) Since the random variables are independent it follows thatfX|Y =5(x) is equal tofX(x).70 CHAPTER2. RANDOMVARIABLES2.5.4 (a)X\Y 0 1 2 3 pi+01161160 0216111631621606162 02163161166163 0 0116116216p+j2166166162161(b) See the table above.(c) The random variablesXandYare not independent.For example, notice thatp0+p+0 =216 216=14 = p00 =116.(d) E(X) =_0 216_+_1 616_+_2 616_+_3 216_ =32E(X2) =_02216_+_12616_+_22616_+_32216_ = 3Var(X) = E(X2) E(X)2= 3 _32_2=34The random variableYhas the same mean and variance asX.(e) E(XY ) =_1 1 316_+_1 2 216_+_2 1 216_+_2 2 316_+_2 3 116_+_3 2 116_+_3 3 116_=4416Cov(X, Y ) = E(XY ) (E(X) E(Y ))=4416 _32 32_ =12(f) P(X = 0|Y= 1) =p01p+1=(116)(616)=16P(X = 1|Y= 1) =p11p+1=(316)(616)=12P(X = 2|Y= 1) =p21p+1=(216)(616)=132.5. JOINTLYDISTRIBUTEDRANDOMVARIABLES 71P(X = 3|Y= 1) =p31p+1=0(616)= 0E(X|Y= 1) =_0 16_+_1 12_+_2 13_+ (3 0) =76E(X2|Y= 1) =_0216_+_1212_+_2213_+_320_ =116Var(X|Y= 1) = E(X2|Y= 1) E(X|Y= 1)2=116 _76_2=17362.5.5 (a) Since_2x=1_3y=0A(ex+y+e2xy)dxdy = 1it follows thatA = 0.00896.(b) P(1.5 X 2, 1 Y 2)= _2x=1.5_2y=10.00896 (ex+y+e2xy)dxdy= 0.158(c) fX(x) = _300.00896 (ex+y+e2xy)dy= 0.00896 (ex+3e2x3ex+e2x)for 1 x 2fY (y) = _210.00896 (ex+y+e2xy)dx= 0.00896 (e2+y+ 0.5e4ye1+y0.5e2y)for 0 y 3(d) No, sincefX(x) fY (y) = f(x, y).(e) fX|Y =0(x) =f(x,0)fY (0)=ex+e2x28.2872 CHAPTER2. RANDOMVARIABLES2.5.6 (a)X\Y 0 1 2 pi+0251022610261025710212610213102039102261020 06102p+j571023910261021(b) See the table above.(c) No, the random variablesXandYare not independent.For example,p22 = p2+p+2.(d) E(X) =_0 57102_+_1 39102_+_2 6102_ =12E(X2) =_0257102_+_1239102_+_226102_ =2134Var(X) = E(X2) E(X)2=2134 _12_2=2568The random variableYhas the same mean and variance asX.(e) E(XY ) = 1 1 p11 =13102Cov(X, Y ) = E(XY ) (E(X) E(Y ))=13102 _12 12_ = 25204(f) Corr(X, Y ) =Cov(X,Y )Var(X)Var(Y )= 13(g) P(Y= 0|X = 0) =p00p0+=2557P(Y= 1|X = 0) =p01p0+=2657P(Y= 2|X = 0) =p02p0+=657P(Y= 0|X = 1) =p10p1+=23P(Y= 1|X = 1) =p11p1+=132.5. JOINTLYDISTRIBUTEDRANDOMVARIABLES 73P(Y= 2|X = 1) =p12p1+= 02.5.7 (a)X\Y 0 1 2 pi+04164161169161416216061621160 0116p+j9166161161(b) See the table above.(c) No, the random variablesXandYare not independent.For example,p22 = p2+p+2.(d) E(X) =_0 916_+_1 616_+_2 116_ =12E(X2) =_02916_+_12616_+_22116_ =58Var(X) = E(X2) E(X)2=58 _12_2=38= 0.3676The random variableYhas the same mean and variance asX.(e) E(XY ) = 1 1 p11 =18Cov(X, Y ) = E(XY ) (E(X) E(Y ))=18 _12 12_ = 18(f) Corr(X, Y ) =Cov(X,Y )Var(X)Var(Y )= 13(g) P(Y= 0|X = 0) =p00p0+=49P(Y= 1|X = 0) =p01p0+=49P(Y= 2|X = 0) =p02p0+=19P(Y= 0|X = 1) =p10p1+=2374 CHAPTER2. RANDOMVARIABLESP(Y= 1|X = 1) =p11p1+=13P(Y= 2|X = 1) =p12p1+= 02.5.8 (a) Since_5x=0_5y=0A (20 x 2y)dxdy = 1it follows thatA = 0.0032(b) P(1 X 2, 2 Y 3)= _2x=1_3y=20.0032 (20 x 2y)dxdy= 0.0432(c) fX(x) = _5y=00.0032 (20 x 2y)dy = 0.016 (15 x)for 0 x 5fY (y) = _5x=00.0032 (20 x 2y)dx = 0.008 (35 4y)for 0 y 5(d) No, the random variablesXandYare not independent sincef(x, y) =fX(x)fY (y).(e) E(X) = _50x 0.016 (15 x)dx =73E(X2) = _50x20.016 (15 x)dx =152Var(X) = E(X2) E(X)2=152 _73_2=3718(f) E(Y ) = _50y 0.008 (35 4y)dy =136E(Y2) = _50y20.008 (35 4y)dy =203Var(Y ) = E(Y2) E(Y )2=203 _136_2=7136(g) fY |X=3(y) =f(3,y)fX(3)=172y60for 0 y 5(h) E(XY ) = _5x=0_5y=00.0032 xy (20 x 2y)dxdy = 5Cov(X, Y ) = E(XY ) (E(X) (EY ))= 5 _73 136_ = 1182.5. JOINTLYDISTRIBUTEDRANDOMVARIABLES 75(i) Corr(X, Y ) =Cov(X,Y )Var(X)Var(Y )= 0.02762.5.9 (a) P(same score) = P(X = 1, Y= 1) +P(X = 2, Y= 2)+P(X = 3, Y= 3) +P(X = 4, Y= 4)= 0.80(b) P(X< Y ) = P(X = 1, Y= 2) +P(X = 1, Y= 3) +P(X = 1, Y= 4)+P(X = 2, Y= 3) +P(X = 2, Y= 4) +P(X = 3, Y= 4)= 0.07(c)xi1 2 3 4pi+0.12 0.20 0.30 0.38E(X) = (1 0.12) + (2 0.20) + (3 0.30) + (4 0.38) = 2.94E(X2) = (120.12) + (220.20) + (320.30) + (420.38) = 9.70Var(X) = E(X2) E(X)2= 9.70 (2.94)2= 1.0564(d)yj1 2 3 4p+j0.14 0.21 0.30 0.35E(Y ) = (1 0.14) + (2 0.21) + (3 0.30) + (4 0.35) = 2.86E(Y2) = (120.14) + (220.21) + (320.30) + (420.35) = 9.28Var(Y ) = E(Y2) E(Y )2= 9.28 (2.86)2= 1.1004(e) The scores are not independent.For example,p11 = p1+p+1.Thescoreswouldnotbeexpectedtobeindependentsincetheyapplytothetwo inspectors assessments of the same building. If they were independent itwouldsuggestthatoneoftheinspectorsisrandomlyassigningasafetyscorewithout paying any attention to the actual state of the building.(f) P(Y= 1|X = 3) =p31p3+=13076 CHAPTER2. RANDOMVARIABLESP(Y= 2|X = 3) =p32p3+=330P(Y= 3|X = 3) =p33p3+=2430P(Y= 4|X = 3) =p34p3+=230(g) E(XY ) = 4i=1

4j=1i jpij = 9.29Cov(X, Y ) = E(XY ) (E(X) E(Y ))= 9.29 (2.94 2.86) = 0.8816(h) Corr(X, Y ) =Cov(X,Y )VarXVarY=0.88161.05641.1004= 0.82A high positive correlation indicates that the inspectors are consistent.The closer the correlation is to one the more consistent the inspectors are.2.5.10 (a) _2x=0_2y=0_2z=03xyz232dxdydz = 1(b) _1x=0_1.5y=0.5_2z=13xyz232dxdydz =764(c) fX(x) = _2x=0_2y=03xyz232dydz =x2for 0 2 x2.6. COMBINATIONSANDFUNCTIONSOFRANDOMVARIABLES 772.6 CombinationsandFunctionsofRandomvariables2.6.1 (a) E(3X + 7) = 3E(X) + 7 = 13Var(3X + 7) = 32Var(X) = 36(b) E(5X 9) = 5E(X) 9 = 1Var(5X 9) = 52Var(X) = 100(c) E(2X + 6Y ) = 2E(X) + 6E(Y ) = 14Var(2X + 6Y ) = 22Var(X) + 62Var(Y ) = 88(d) E(4X 3Y ) = 4E(X) 3E(Y ) = 17Var(4X 3Y ) = 42Var(X) + 32Var(Y ) = 82(e) E(5X 9Z + 8) = 5E(X) 9E(Z) + 8 = 54Var(5X 9Z + 8) = 52Var(X) + 92Var(Z) = 667(f) E(3Y Z 5) = 3E(Y ) E(Z) 5 = 4Var(3Y Z 5) = (3)2Var(Y ) + (1)2Var(Z) = 25(g) E(X + 2Y+ 3Z) = E(X) + 2E(Y ) + 3E(Z) = 20Var(X + 2Y+ 3Z) = Var(X) + 22Var(Y ) + 32Var(Z) = 75(h) E(6X + 2Y Z + 16) = 6E(X) + 2E(Y ) E(Z) + 16 = 14Var(6X + 2Y Z + 16) = 62Var(X) + 22Var(Y ) + (1)2Var(Z) = 1592.6.2 E(aX +b) = _(ax +b)f(x)dx= a_xf(x)dx+b_f(x)dx= aE(X) +bVar(aX +b) = E((aX +b E(aX +b))2)= E((aX aE(X))2)= a2E((X E(X))2)= a2Var(X)78 CHAPTER2. RANDOMVARIABLES2.6.3 E(Y ) = 3E(X1) = 3Var(Y ) = 32Var(X1) = 92E(Z) = E(X1) +E(X2) +E(X3) = 3Var(Z) = Var(X1) + Var(X2) + Var(X3) = 32The random variablesYandZhave the same meanbutZhas a smaller variance thanY .2.6.4 length = A1 +A2 +BE(length) = E(A1) +E(A2) +E(B) = 37 + 37 + 24 = 98Var(length) = Var(A1) + Var(A2) + Var(B) = 0.72+ 0.72+ 0.32= 1.072.6.5 Let the random variableXi be the winnings from theithgame.ThenE(Xi) =_10 18_+_(1) 78_ =38andE(X2i ) =_10218_+_(1)278_ =1078so thatVar(Xi) = E(X2i ) (E(Xi))2=84764 .The total winnings from 50 (independent) games isY= X1 +. . . +X50andE(Y ) = E(X1) +. . . +E(X50) = 50 38=754= $18.75withVar(Y ) = Var(X1) +. . . + Var(X50) = 50 84764= 661.72so thatY=661.72 = $25.72.2.6.6 (a) E(average weight) = 1.12 kgVar(average weight) =0.03225= 3.6 1052.6. COMBINATIONSANDFUNCTIONSOFRANDOMVARIABLES 79The standard deviation is0.0325= 0.0012 kg.(b) It is required that0.03n 0.005 which is satised forn 36.2.6.7 Let the random variable Xi be equal to 1 if an ace is drawn on the ithdrawing (whichhappenswithaprobabilityof113)andequalto0ifanaceisnotdrawnontheithdrawing (which happens with a probability of1213).Then the total number of aces drawn isY= X1 +. . . +X10.Notice that E(Xi) =113so that regardless of whether the drawing is performed withor without replacement it follows thatE(Y ) = E(X1) +. . . +E(X10) =1013.Also, notice thatE(X2i ) =113so thatVar(Xi) =113 _113_2=12169.If the drawings are made withreplacement then the random variablesXiare inde-pendent so thatVar(Y ) = Var(X1) +. . . + Var(X10) =120169.However,ifthedrawingsaremadewithoutreplacementthentherandomvariablesXi are not independent.2.6.8 FX(x) = P(X x) = x2for 0 x 1(a) FY (y) = P(Y y) = P(X3 y) = P(X y1/3) = FX(y1/3) = y2/3and sofY (y) =23y1/3for 0 y 1E(y) = _10yfY (y)dy = 0.4(b) FY (y) = P(Y y) = P(X y) = P(X y2) = FX(y2) = y4and sofY (y) = 4y3for 0 y 1E(y) = _10yfY (y)dy = 0.8(c) FY (y) = P(Y y) = P(11+X y) = P(X 1y 1)= 1 FX(1y 1) =2y 1y280 CHAPTER2. RANDOMVARIABLESand sofY (y) = 2y2 +2y3for12 y 1E(y) = _10.5yfY (y)dy = 0.614(d) FY (y) = P(Y y) = P(2X y) = P_X ln(y)ln(2)_= FX_ln(y)ln(2)_ =_ln(y)ln(2)_2and sofY (y) =2ln(y)y(ln(2))2for 1 y 2E(y) = _21yfY (y)dy = 1.612.6.9 (a) Since_20A(1 (r 1)2)dr = 1it follows thatA =34.This givesFR(r) =3r24r34for 0 r 2.(b) V=43r3SinceFV (v) = P(V v) = P_43r3v_ = FR__3v4_1/3_it follows thatfV (v) =12(34)2/3v1/3316for 0 v 323.(c) E(V ) = _3230vfV (v)dv =32152.6.10 (a) Since_L0Ax(L x)dx = 1it follows thatA =6L3.2.6. COMBINATIONSANDFUNCTIONSOFRANDOMVARIABLES 81Therefore,FX(x) =x2(3L2x)L3for 0 x L.(b) The random variable corresponding to the dierence between the lengths of thetwo pieces of rod isW= |L 2X|.Therefore,FW(w) = P_L2 w2 X L2+w2_ = FX_L2+w2_FX_L2 w2_=w(3L2w2)2L3andfW(w) =3(L2w2)2L3for 0 w L.(c) E(W) = _L0wfW(w)dw =38L2.6.11 (a) The return has an expectation of $100, a standard deviation of $20,and a variance of 400.(b) The return has an expectation of $100, a standard deviation of $30,and a variance of 900.(c) The return from fund A has an expectation of $50, a standard deviation of $10,and a variance of 100.The return from fund B has an expectation of $50, a standard deviation of $15,and a variance of 225.Therefore, the total return has an expectation of $100 and a variance of 325,so that the standard deviation is $18.03.(d) The return from fund A has an expectation of $0.1x,a standard deviation of $0.02x,and a variance of 0.0004x2.The return from fund B has an expectation of $0.1(1000 x),a standard deviation of $0.03(1000 x),and a variance of 0.0009(1000 x)2.Therefore, the total return has an expectation of $100and a variance of 0.0004x2+ 0.0009(1000 x)2.82 CHAPTER2. RANDOMVARIABLESThis variance is minimized by takingx = $692,and the minimum value of the variance is 276.9which corresponds to a standard deviation of $16.64.This problem illustrates that the variability of the return on an investment can bereduced by diversifying the investment, so that it is spread over several funds.2.6.12 The expected value of the total resistance is5 E(X) = 5 10.418234 = 52.09.The variance of the total resistance is5 Var(X) = 5 0.0758 = 0.379so that the standard deviation is 0.379 = 0.616.2.6.13 (a) The mean isE(X) =_13 E(X1)_+_13 E(X2)_+_13 E(X3)_=_13 59_+_13 67_+_13 72_ = 66The variance isVar(X) =__13_2Var(X1)_+__13_2Var(X2)_+__13_2Var(X3)_=__13_2102_+__13_2132_+__13_242_ =953so that the standard deviation is _95/3 = 5.63.(b) The mean isE(X) = (0.4 E(X1)) + (0.4 E(X2)) + (0.2 E(X3))= (0.4 59) + (0.4 67) + (0.2 72) = 64.8.The variance isVar(X) = _0.42Var(X1)_+_0.42Var(X2)_+_0.22Var(X3)_= _0.42102_+_0.42132_+_0.2242_ = 43.68so that the standard deviation is 43.68 = 6.61.2.6.14 1000 = E(Y ) = a +bE(X) = a + (b 77)102= Var(Y ) = b2Var(X) = b292Solving these equations givesa = 914.44 andb = 1.11,2.6. COMBINATIONSANDFUNCTIONSOFRANDOMVARIABLES 83ora = 1085.56 andb = 1.11.2.6.15 (a) The mean is = 65.90.The standard deviation is5=0.325= 0.143.(b) The mean is 8 = 8 65.90 = 527.2.The standard deviation is 8 =8 0.32 = 0.905.2.6.16 (a) E(A) =E(X1)+E(X2)2=W+W2= WVar(A) =Var(X1)+Var(X2)4=32+424=254The standard deviation is52= 2.5.(b) Var(B) = 2Var(X1) + (1 )2Var(X2) = 92+ 16(1 )2This is minimized when =1625and the minimum value is14425so that the minimum standard deviation is125= 2.4.2.6.17 When a die is rolled once the expectation is 3.5 and the standard deviation is 1.71(see Games of Chance in section 2.4).Therefore, the sum of eighty die rolls has an expectation of 80 3.5 = 280and a standard deviation of 80 1.71 = 15.3.2.6.18 (a) The expectation is 4 33.2 = 132.8 seconds.The standard deviation is 4 1.4 = 2.8 seconds.(b) E(A1 +A2 +A3 +A4B1B2B3B4)= E(A1) +E(A2) +E(A3) +E(A4) E(B1) E(B2) E(B3) E(B4)= (4 33.2) (4 33.0) = 0.8Var(A1 +A2 +A3 +A4B1B2B3B4)= Var(A1) + Var(A2) + Var(A3) + Var(A4)+ Var(B1) + Var(B2) + Var(B3) + Var(B4)= (4 1.42) + (4 1.32) = 14.6The standard deviation is 14.6 = 3.82.(c) E_A1A2+A3+A43_= E(A1) E(A2)3E(A3)3E(A4)3= 0Var_A1A2+A3+A43_84 CHAPTER2. RANDOMVARIABLES= Var(A1) + Var(A2)9+ Var(A3)9+ Var(A4)9=43 1.42= 2.613The standard deviation is 2.613 = 1.62.2.6.19 Let X be the temperature in Fahrenheit and let Ybe the temperature in Centigrade.E(Y ) = E_5(X32)9_ =_5(E(X)32)9_ =_5(11032)9_ = 43.33Var(Y ) = Var_5(X32)9_ =_52Var(X)92_ =_522.2292_ = 1.49The standard deviation is 1.49 = 1.22.2.6.20 Var(0.5X + 0.3X + 0.2X)= 0.52Var(X) + 0.32Var(X) + 0.22Var(X)= (0.521.22) + (0.322.42) + (0.223.12) = 1.26The standard deviation is 1.26 = 1.12.2.6.21 The inequality56n 10 is satised forn 32.2.6.22 (a) E(X1 +X2) = E(X1) +E(X2) = 7.74Var(X1 +X2) = Var(X1) + Var(X2) = 0.0648The standard deviation is 0.0648 = 0.255.(b) E(X1 +X2 +X3) = E(X1) +E(X2) +E(X3) = 11.61Var(X1 +X2 +X3) = Var(X1) + Var(X2) + Var(X3) = 0.0972The standard deviation is 0.0972 = 0.312.(c) E_X1+X2+X3+X44_=E(X1)+E(X2)+E(X3)+E(X4)4= 3.87Var_X1+X2+X3+X44_=Var(X1)+Var(X2)+Var(X3)+Var(X4)16= 0.0081The standard deviation is 0.0081 = 0.09.2.6. COMBINATIONSANDFUNCTIONSOFRANDOMVARIABLES 85(d) E_X3X1+X22_ = E(X3) E(X1)+E(X2)2= 0Var_X3X1+X22_ = Var(X3) + Var(X1)+Var(X2)4= 0.0486The standard deviation is 0.0486 = 0.220.86 CHAPTER2. RANDOMVARIABLES2.8 SupplementaryProblems2.8.1 (a)xi2 3 4 5 6pi115215315415515(b) E(X) =_2 115_+_3 215_+_4 315_+_5 415_+_6 515_ =1432.8.2 (a)xi0 1 2 3 4 5 6F(xi) 0.21 0.60 0.78 0.94 0.97 0.99 1.00(b) E(X) = (0 0.21) + (1 0.39) + (2 0.18) + (3 0.16)+ (4 0.03) + (5 0.02) + (6 0.01)= 1.51(c) E(X2) = (020.21) + (120.39) + (220.18) + (320.16)+ (420.03) + (520.02) + (620.01)= 3.89Var(X) = 3.89 (1.51)2= 1.61(d) The expectation is 1.51 60 = 90.6and the variance is 1.61 60 = 96.6.2.8.3 (a)xi2 3 4 5pi23013301330230(b) E(X) =_2 230_+_3 1330_+_4 1330_+_5 230_ =72E(X2) =_22230_+_321330_+_421330_+_52230_ =383302.8. SUPPLEMENTARYPROBLEMS 87Var(X) = E(X2) E(X)2=38330 _72_2=3160(c)xi2 3 4 5pi210310310210E(X) =_2 210_+_3 310_+_4 310_+_5 210_ =72E(X2) =_22210_+_32310_+_42310_+_52210_ =13310Var(X) = E(X2) E(X)2=13310 _72_2=21202.8.4 LetXi be the value of theithcard dealt.ThenE(Xi) =_2 113_+_3 113_+_4 113_+_5 113_+_6 113_+_7 113_+_8 113_+_9 113_+_10 113_+_15 413_ =11413The total score of the hand isY= X1 +. . . +X13which has an expectationE(Y ) = E(X1) +. . . +E(X13) = 13 11413= 114.2.8.5 (a) Since_111A_32_xdx = 1it follows thatA =ln(1.5)1.5111.5=1209.6.(b) F(x) = _x11209.6_32_ydy= 0.01177_32_x0.01765for 1 x 11(c) SolvingF(x) = 0.5 givesx = 9.332.(d) SolvingF(x) = 0.25 givesx = 7.706.SolvingF(x) = 0.75 givesx = 10.305.88 CHAPTER2. RANDOMVARIABLESThe interquartile range is 10.305 7.706 = 2.599.2.8.6 (a) fX(x) = _214x(2 y)dy = 2x for 0 x 1(b) fY (y) = _104x(2 y)dx = 2(2 y) for 1 y 2Sincef(x, y) = fX(x) fY (y) the random variables are independent.(c) Cov(X, Y ) = 0 because the random variables are independent.(d) fX|Y =1.5(x) = fX(x) because the random variables are independent.2.8.7 (a) Since_105A_x +2x_dx = 1it follows thatA = 0.02572.(b) F(x) = _x50.02572_y +2y_dy= 0.0129x2+ 0.0514 ln(x) 0.404for 5 x 10(c) E(X) = _1050.02572 x_x +2x_dx = 7.759(d) E(X2) = _1050.02572 x2_x +2x_dx = 62.21Var(X) = E(X2) E(X)2= 62.21 7.7592= 2.01(e) SolvingF(x) = 0.5 givesx = 7.88.(f) SolvingF(x) = 0.25 givesx = 6.58.SolvingF(x) = 0.75 givesx = 9.00.The interquartile range is 9.00 6.58 = 2.42.(g) The expectation isE(X) = 7.759.The variance isVar(X)10= 0.0201.2.8.8 Var(a1X1 +a2X2 +. . . +anXn +b)= Var(a1X1) +. . . + Var(anXn) + Var(b)= a21Var(X1) +. . . +a2nVar(Xn) + 02.8. SUPPLEMENTARYPROBLEMS 892.8.9 Y=53X 252.8.10 Notice thatE(Y ) = aE(X) +b and Var(Y ) = a2Var(X).Also,Cov(X, Y ) = E( (X E(X)) (Y E(Y )) )= E( (X E(X)) a(X E(X)) )= aVar(X).Therefore,Corr(X, Y ) =Cov(X,Y )Var(X)Var(Y )=aVar(X)Var(X)a2Var(X)=a|a|which is 1 ifa > 0 and is 1 ifa < 0.2.8.11 The expected amount of a claim isE(X) = _18000xx(1800x)972,000,000dx = $900.Consequently, the expected prot from each customer is$100 $5 (0.1 $900) = $5.The expected prot from 10,000 customers is therefore 10, 000 $5 = $50, 000.The prots may or may not be independent depending on the type of insurance andthe pool of customers.If largenatural disastersaectthepool of customersall atoncethentheclaimswould not be independent.2.8.12 (a) The expectation is 5 320 = 1600 seconds.The variance is 5 632= 19845and the standard deviation is 19845 = 140.9 seconds.(b) The expectation is 320 seconds.The variance is63210= 396.9and the standard deviation is 396.9 = 19.92 seconds.2.8.13 (a) The state space is the positive integers from 1 ton,with each outcome having a probability value of1n.(b) E(X) =_1n 1_+_1n 2_+. . . +_1n n_ =n+1290 CHAPTER2. RANDOMVARIABLES(c) E(X2) =_1n 12_+_1n 22_+. . . +_1n n2_ =(n+1)(2n+1)6Therefore,Var(X) = E(X2) (E(X))2=(n+1)(2n+1)6_n+12_2=n2112.2.8.14 (a) LetXTbe the amount of time that Tom spends on the busand letXNbe the amount of time that Nancy spends on the bus.Therefore, the sum of the times isX = XT +XNandE(X) = E(XT) +E(XN) = 87 + 87 = 174 minutes.IfTomandNancyrideondierentbusesthentherandomvariablesXTandXNare independent so thatVar(X) = Var(XT) + Var(XN) = 32+ 32= 18and the standard deviation is 18 = 4.24 minutes.(b) If TomandNancyridetogether onthesamebus thenXT=XNsothatX = 2 XT, twice the time of the ride.In this caseE(X) = 2 E(XT) = 2 87 = 174 minutesandVar(X) = 22Var(X1) = 2232= 36so that the standard deviation is 36 = 6 minutes.2.8.15 (a) Two heads gives a total score of 20.One head and one tail gives a total score of 30.Two tails gives a total score of 40.Therefore, the state space is {20, 30, 40}.(b) P(X = 20) =14P(X = 30) =12P(X = 40) =14(c) P(X 20) =14P(X 30) =34P(X 40) = 1(d) E(X) =_20 14_+_30 12_+_40 14_ = 30(e) E(X2) =_20214_+_30212_+_40214_ = 950Var(X) = 950 302= 502.8. SUPPLEMENTARYPROBLEMS 91The standard deviation is 50 = 7.07.2.8.16 (a) Since_65Axdx =A2 (6252) = 1it follows thatA =211.(b) F(x) = _x52y11dy =x22511(c) E(X) = _652x211dx =2(6353)33=18233= 5.52(d) E(X2) = _652x311dx =645422=67122= 30.5Var(X) = 30.5 _18233_2= 0.083The standard deviation is 0.083 = 0.29.2.8.17 (a) The expectation is 3 438 = 1314.The standard deviation is 3 4 = 6.93.(b) The expectation is 438.The standard deviation is48= 1.41.2.8.18 (a) If a 1 is obtained from the die the net winnings are (3 $1) $5 = $2If a 2 is obtained from the die the net winnings are $2 $5 = $3If a 3 is obtained from the die the net winnings are (3 $3) $5 = $4If a 4 is obtained from the die the net winnings are $4 $5 = $1If a 5 is obtained from the die the net winnings are (3 $5) $5 = $10If a 6 is obtained from the die the net winnings are $6 $5 = $1Each of these values has a probability of16.(b) E(X) =_3 16_+_2 16_+_1 16_+_1 16_+_4 16_+_10 16_ =32E(X2) =_(3)216_+_(2)216_+_(1)216_+_1216_+_4216_+_10216_ =1316The variance is1316_32_2=2351292 CHAPTER2. RANDOMVARIABLESand the standard deviation is_23512= $4.43.(c) The expectation is 10 32= $15.The standard deviation is 10 4.43 = $13.99.2.8.19 (a) False(b) True(c) True(d) True(e) True(f) False2.8.20 E(total time) = 5 = 5 22 = 110 minutesThe standard deviation of the total time is 5 =5 1.8 = 4.02 minutes.E(average time) = = 22 minutesThe standard deviation of the average time is5=1.85= 0.80 minutes.2.8.21 (a) E(X) = (0 0.12) + (1 0.43) + (2 0.28) + (3 0.17) = 1.50(b) E(X2) = (020.12) + (120.43) + (220.28) + (320.17) = 3.08The variance is 3.08 1.502= 0.83and the standard deviation is 0.83 = 0.911.(c) E(X1 +X2) = E(X1) +E(X2) = 1.50 + 1.50 = 3.00Var(X1 +X2) = Var(X1) + Var(X2) = 0.83 + 0.83 = 1.66The standard deviation is 1.66 = 1.288.2.8.22 (a) E(X) = (22 0.3) + (3 0.2) + (19 0.1) + (23 0.4) = 5.1(b) E(X2) = ((22)20.3) + (320.2) + (1920.1) + (2320.4) = 394.7Var(X) = 394.7 5.12= 368.69The standard deviation is 368.69 = 19.2.2.8.23 (a) Since_42f(x)dx = _42Ax2dx =A4= 1it follows thatA = 4.2.8. SUPPLEMENTARYPROBLEMS 93(b) Since14= _y2f(x)dx = _y24x2dx =_2 4y_it follows thaty =167= 2.29.2.8.24 (a) 100 = E(Y ) = c +dE(X) = c + (d 250)1 = Var(Y ) = d2Var(X) = d216Solving these equations givesd =14andc =752ord = 14andc =3252.(b) The mean is 10 250 = 1000.The standard deviation is 10 4 = 12.65.2.8.25 SinceE(c1X1 +c2X2) = c1E(X1) +c2E(X2) = (c1 +c2) 100 = 100it is necessary thatc1 +c2 = 1.Also,Var(c1X1 +c2X2) = c21Var(X1) +c22Var(X2) = (c21144) + (c22169) = 100.Solving these two equations givesc1 = 0.807 andc2 = 0.193orc1 = 0.273 andc2 = 0.727.2.8.26 (a) The mean is 3A = 3 134.9 = 404.7.The standard deviation is 3A =3 0.7 = 1.21.(b) The mean is 2A + 2B = (2 134.9) + (2 138.2) = 546.2.The standard deviation is 0.72+ 0.72+ 1.12+ 1.12= 1.84.(c) The mean is4A+3B7=(4134.9)+(3138.2)7= 136.3.The standard deviation is0.72+0.72+0.72+0.72+1.12+1.12+1.127= 0.34.94 CHAPTER2. RANDOMVARIABLESChapter3DiscreteProbabilityDistributions3.1 TheBinomialDistribution3.1.1 (a) P(X = 3) =_103_0.1230.887= 0.0847(b) P(X = 6) =_106_0.1260.884= 0.0004(c) P(X 2) = P(X = 0) +P(X = 1) +P(X = 2)= 0.2785 + 0.3798 + 0.2330= 0.8913(d) P(X 7) = P(X = 7) +P(X = 8) +P(X = 9) +P(X = 10)= 3.085 105(e) E(X) = 10 0.12 = 1.2(f) Var(X) = 10 0.12 0.88 = 1.0563.1.2 (a) P(X = 4) =_74_0.840.23= 0.1147(b) P(X = 2) = 1 P(X = 2)= 1 _72_0.820.25= 0.9957(c) P(X 3) = P(X = 0) +P(X = 1) +P(X = 2) +P(X = 3) = 0.03349596 CHAPTER3. DISCRETEPROBABILITYDISTRIBUTIONS(d) P(X 6) = P(X = 6) +P(X = 7) = 0.5767(e) E(X) = 7 0.8 = 5.6(f) Var(X) = 7 0.8 0.2 = 1.123.1.3 X B(6, 0.5)xi0 1 2 3 4 5 6pi0.0156 0.0937 0.2344 0.3125 0.2344 0.0937 0.0156E(X) = 6 0.5 = 3Var(X) = 6 0.5 0.5 = 1.5 =1.5 = 1.22X B(6, 0.7)xi0 1 2 3 4 5 6pi0.0007 0.0102 0.0595 0.1852 0.3241 0.3025 0.1176E(X) = 6 0.7 = 4.2Var(X) = 6 0.7 0.3 = 1.26 =1.5 = 1.123.1.4 X B(9, 0.09)(a) P(X = 2) = 0.1507(b) P(X 2) = 1 P(X = 0) P(X = 1) = 0.1912E(X) = 9 0.09 = 0.813.1.5 (a) P_B_8,12_ = 5_ = 0.21873.1. THEBINOMIALDISTRIBUTION 97(b) P_B_8,16_ = 1_ = 0.3721(c) P_B_8,16_ = 0_ = 0.2326(d) P_B_8,23_ 6_ = 0.46823.1.6 P(B(10, 0.2) 7) = 0.0009P(B(10, 0.5) 7) = 0.17193.1.7 Let the random variableXbe the number of employees taking sick leave.ThenX B(180, 0.35).Therefore, the proportion of the workforce who need to take sick leave isY=X180so thatE(Y ) =E(X)180=1800.35180= 0.35andVar(Y ) =Var(X)1802=1800.350.651802= 0.0013.In general, the variance isVar(Y ) =Var(X)1802=180p(1p)1802=p(1p)180which is maximized whenp = 0.5.3.1.8 The random variableYcan be considered to be the number of successes out ofn1 +n2 trials.3.1.9 X B(18, 0.6)(a) P(X = 8) +P(X = 9) +P(X = 10)=_188_0.680.410+_189_0.690.49+_1810_0.6100.48= 0.0771 + 0.1284 + 0.1734 = 0.378998 CHAPTER3. DISCRETEPROBABILITYDISTRIBUTIONS(b) P(X = 0) +P(X = 1) +P(X = 2) +P(X = 3) +P(X = 4)=_180_0.600.418+_181_0.610.417+_182_0.620.416+_183_0.630.415+_184_0.640.414= 0.00133.2. THEGEOMETRICANDNEGATIVEBINOMIALDISTRIBUTIONS 993.2 TheGeometricandNegativeBinomialDistributions3.2.1 (a) P(X = 4) = (1 0.7)30.7 = 0.0189(b) P(X = 1) = (1 0.7)00.7 = 0.7(c) P(X 5) = 1 (1 0.7)5= 0.9976(d) P(X 8) = 1 P(X 7) = (1 0.7)7= 0.00023.2.2 (a) P(X = 5) =_42_(1 0.6)20.63= 0.2074(b) P(X = 8) =_72_(1 0.6)50.63= 0.0464(c) P(X 7) = P(X = 3) +P(X = 4) +P(X = 5) +P(X = 6) +P(X = 7)= 0.9037(d) P(X 7) = 1 P(X = 3) P(X = 4) P(X = 5)