chapter 1

Chapter 2: Probability101.a. The law of total probability gives3P ( late | E) ⋅ P(P(late) = ∑ i E ii=1= (.02)(.40) + (.01)(.50) + (.05)(.10) = .018)b. P(E 1 ′ | on time) = 1 – P(E 1 | on time)P(E1∩ on.time)(.98)(.4)= 1−= 1 − = . 601P(on.time).982102. Let B denote the event that a component needs rework. Then3P(B) = ∑i=1P ( B|A ) ⋅ P() = (.05)(.50) + (.08)(.30) + (.10)(.20) = .069i A i(.05)(.50)Thus P(A 1 | B) = = . 362.069(.08)(.30)P(A 2 | B) = = . 348.069(.10)(.20)P(A 3 | B) = = . 290.069103.(365)(364)...(356)a. P(all different) = . 883(365)10=P(at least two the same) = 1 - .883 = .117b. P(at least two the same) = .476 for k=22, and = .507 for k=23c. P(at least two have the same SS number) = 1 – P(all different)=(1000)(999)...(991)1−10(1000)= 1 - .956 = .044Thus P(at least one “coincidence”) = P(BD coincidence ∪ SS coincidence)= .117 + .044 – (.117)(.044) = .15691