100% satisfaction guarantee Immediately available after payment Both online and in PDF No strings attached
logo-home
Complete Solution Manual Probability and Statistics for Computer Scientists 2nd Edition Baron Questions & Answers with rationales $16.99   Add to cart

Exam (elaborations)

Complete Solution Manual Probability and Statistics for Computer Scientists 2nd Edition Baron Questions & Answers with rationales

 1904 views  20 purchases
  • Course
  • Probability
  • Institution
  • Probability

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+

Preview 4 out of 143  pages

  • June 21, 2023
  • 143
  • 2022/2023
  • Exam (elaborations)
  • Questions & answers
  • probability
book image

Book Title:

Author(s):

  • Edition:
  • ISBN:
  • Edition:
  • Probability
  • Probability
avatar-seller
gradexam
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

The benefits of buying summaries with Stuvia:

Guaranteed quality through customer reviews

Guaranteed quality through customer reviews

Stuvia customers have reviewed more than 700,000 summaries. This how you know that you are buying the best documents.

Quick and easy check-out

Quick and easy check-out

You can quickly pay through credit card or Stuvia-credit for the summaries. There is no membership needed.

Focus on what matters

Focus on what matters

Your fellow students write the study notes themselves, which is why the documents are always reliable and up-to-date. This ensures you quickly get to the core!

Frequently asked questions

What do I get when I buy this document?

You get a PDF, available immediately after your purchase. The purchased document is accessible anytime, anywhere and indefinitely through your profile.

Satisfaction guarantee: how does it work?

Our satisfaction guarantee ensures that you always find a study document that suits you well. You fill out a form, and our customer service team takes care of the rest.

Who am I buying these notes from?

Stuvia is a marketplace, so you are not buying this document from us, but from seller gradexam. Stuvia facilitates payment to the seller.

Will I be stuck with a subscription?

No, you only buy these notes for $16.99. You're not tied to anything after your purchase.

Can Stuvia be trusted?

4.6 stars on Google & Trustpilot (+1000 reviews)

79373 documents were sold in the last 30 days

Founded in 2010, the go-to place to buy study notes for 14 years now

Start selling
$16.99  20x  sold
  • (0)
  Add to cart