chapter 1

Chapter 2: Probabilityd. P(passes inspection) = P({not flawed} ∪ {flawed and passes})= P(not flawed) + P(flawed and passes)e. P(flawed | passed) == .9 + P(passes | flawed)• P(flawed) = .9+(1 – p) 3 (.1)3P(flawed ∩ passed).1(1 − p)=P(passed).9 + .1(1 − p)3.1(.5).9 + .1(.5)For p = .5, P(flawed | passed) = = . 01373384.2000 = , P(B) = P(A ∩ B) + P(A′ ∩ B)10,000a. P(A) = . 0219992000= P(B|A) P(A) + P(B|A′) P(A′) = • (.2) + • (.8) = . 299999999P(A ∩ B) = .039984; since P(A ∩ B) ≠ P(A)P(B), the events are not independent.b. P(A ∩ B) = .04. Very little difference. Yes.1 2c. P(A) = P(B) = .2, P(A)P(B) = .04, but P(A ∩ B) = P(B|A)P(A) = ⋅ = . 0222 , so the9 10two numbers are quite different.In a, the sample size is small relative to the “population” size, while here it is not.85. P(system works) = P( 1 – 2 works ∩ 3 – 4 – 5 – 6 works ∩ 7 works)= P( 1 – 2 works) • P( 3 – 4 – 5 – 6 works) •P( 7 works)= (.99) (.9639) (.9) = .8588With the subsystem in figure 2.14 connected in parallel to this subsystem,P(system works) = .8588+.927 – (.8588)(.927) = .989785