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Complete Solution Manual Probability and Statistics for Computer Scientists 2nd Edition Baron Questions & Answers with rationales CA$24.68   Add to cart
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Complete Solution Manual Probability and Statistics for Computer Scientists 2nd Edition Baron Questions & Answers with rationales

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Probability and Statistics for Computer Scientists 2nd Edition Baron Solutions Manual Complete Solution Manual Probability and Statistics for Computer Scientists 2nd Edition Baron Questions & Answers with rationales PDF File All Pages All Chapters Grade A+

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  • June 21, 2023
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  • 2022/2023
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  • probability
  • probability and statistics for computer scientist
  • probability and statistics for computer
  • probability and statistics
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Probability and Statistics for Computer Scientists 2nd Edition Baron Solutions Manual Table of Contents Chapter 2 solutions 3 Chapter 3 solutions 14 Chapter 4 solutions 27 Chapter 5 solutions 40 Chapter 6 solutions 46 Chapter 7 solutions 54 Chapter 8 solutions 66 Chapter 9 solutions 71 Chapter 10 solutions 84 Chapter 11 solutions 110 Appendix: Matlab codes for exercises -projects 131 CHAPTER 2 3 Chapter 2 2.1 An outcome is the chosen pair of chips. The sample space in this problem consists of 15 pairs: AB, AC, AD, AE, AF, BC, BD, BE, BF, CD, CE, CF, DE, DF, EF (or 30 pairs if the order of chips in each pair matters, i.e., AB and BA are different pairs). All the outcomes are equally likely because two chips are chosen at random. One outcome is ‘favorable’, when both chips in a pair are defective (two such pairs if the order matters). Thus, P (both chips are defective) = number of favorable outcomes = total number of outcomes 1/15 } } 2.2 Denote the events: We have: M = { problems with a motherboard } H = { problems with a hard drive } Hence, P {M } = 0.4, P {H} = 0.3, and P {M ∩ H} = 0.15. and P {M ∪ H} = P {M } + P {H} − P {M ∩ H} = 0.4 + 0.3 − 0.15 = 0.55, P {fully functioning MB and HD} = 1 − P {M ∪ H} = 2.3 Denote the events, Then I = {the virus enters through the internet } E = {the virus enters through the e-mail} P {E¯ ∩ I¯} = 1 − P {E ∪ I} = 1 − (P {E} + P {I} − P {E ∩ I}) = 1 − (.3 + .4 − .15) = It may help to draw a Venn diagram. 2.4 Denote the events, C = { knows C/C++ } , F = { knows Fortran } . Then (a) P F¯ = 1 − P {F } = 1 − 0.6 = (b) P F¯ ∩ C¯ = 1 − P {F ∪ C} = 1 − (P {F } + P {C} − P {F ∩ C}) = 1 − (0.7 + 0.6 − 0.5) = 1 − 0.8 = 0.2 (c) P {C\F } = P {C} − P {F ∩ C} = 0.7 − 0.5 = 0.2 0.4 0.45 0.45 } } } } ∩ ∩ } { } { } { } } } } { ∩ } 4 INSTRUCTOR ’S SOLUTION MANUAL (d) P {F \C} = P {F } − P {F ∩ C} = 0.6 − 0.5 = P C F 0.5 (e) P {C | F } = = = P {F } 0.6 (f) P {F | C} = P {C ∩ F } = 0.5 = P {C} 0.7 2.5 Denote the events: Then D1 = {first test discovers the error } D2 = {second test discovers the error} D3 = {third test discovers the error} P { at least one discovers } = P {D1 ∪ D2 ∪ D3} = 1 − P D¯1 ∩ D¯2 ∩ D¯3 = 1 − (1 − 0.2)(1 − 0.3)(1 − 0.5) = 1 − 0.28 = We used the complement rule and independence. 2.6 Let A = {arrive on time}, W = {good weather }. We have P {A | W } = 0.8, P A | W¯ = 0.3, P {W } = 0.6 By the Law of Total Probability, P {A} = P {A | W } P {W } + P A | W¯ P W¯ = (0.8)(0 .6) + (0.3)(0 .4) = 0.60 2.7 Organize the data. Let D = detected , I = via internet , E = via e-mail = I. Notice that the question about detection already assumes that the spyware has entered the system. This is the sample space, and this is why P {I} + P {E} = 1. We have P {I} = 0.7, P {E} = 0.3, P {D | I} = 0.6, P {D | E} = 0.8. By the Law of Total Probability, P {D} = (0.6)(0 .7) + (0.8)(0 .3) = 2.8 Let A1 = {1st device fails}, A2 = {2nd device fails}, A3 = {3rd device fails}. P { on time } = P { all function } = P A1 A2 A3 = P A1 P A2 P A3 (independence) = (1 − 0.01)(1 − 0.02)(1 − 0.02) (complement rule) = 0.9508 0.66 0.72 0.7143 0.8333 0.1 0.1792 } } } } } } } } } } } } CHAPTER 2 5 2.9 P {at least one fails} = 1 − P {all work} = 1 − (.96)( .95)( .90) = . 2.10 P {A ∪ B ∪ C} = 1 − P A¯ ∩ B¯ ∩ C¯ = 1 − P A¯ P B¯ P C¯ = 1 − (1 − 0.4)(1 − 0.5)(1 − 0.2) = 0.76 2.11 (a) P {at least one test finds the error} = 1 − P {all tests fail to find the error} = 1 − (1 − 0.1)(1 − 0.2)(1 − 0.3)(1 − 0.4)(1 − 0.5) = 1 − (0.9)(0 .8)(0 .7)(0 .6)(0 .5) = (b) The difference between events in (a) and (b) is the probability that exactly one test finds an error. This probability equals P {exactly one test finds the error} = P {test 1 find the error, the others don’t find} +P {test 2 find the error, the others don’t find} + . . . = (0.1)(1 − 0.2)(1 − 0.3)(1 − 0.4)(1 − 0.5) +(1 − 0.1)(0.2)(1 − 0.3)(1 − 0.4)(1 − 0.5) + . . . = (0.1)(0 .8)(0 .7)(0 .6)(0 .5) + (0.9)(0 .2)(0 .7)(0 .6)(0 .5) +(0.9)(0 .8)(0 .3)(0 .6)(0 .5) + (0.9)(0 .8)(0 .7)(0 .4)(0 .5) +(0.9)(0 .8)(0 .7)(0 .6)(0 .5) = 0.3714 . Then P {at least two tests find the error} = P {at least one test finds the error} −P {exactly one test finds the error} = 0.8488 − 0.3714 = (c) P {all tests find the error} = (0.1)(0 .2)(0 .3)(0 .4)(0 .5) = 2.12 Let Aj = { dog j detects the explosives }. P {at least one dog detects } = 1 − P {all four dogs don’t detect } = 1 − P A¯1 P A¯2 P A¯3 P A¯4 = 1 − (1 − 0.6)4 = 0.9744 2.13 Let Aj be the event {Team j detects a problem }. Then P {at least one team detects } = 1 − P {no team detects } = 1 − P A¯1 ∩ A¯2 ∩ A¯3 = 1 − P A¯1 P A¯2 P A¯3 = 1 − (1 − 0.8)(1 − 0.8)(1 − 0.8) = 0.992 . 2.14 (a) The total number of possible passwords is P (26, 6) = (26)(25)(24)(23)(22)(21) = 165, 765, 600 because there are 26 letters in the alphabet, they should be all different in the 0.0012 0.4774 0.8488

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