chapter 1

Chapter 2: Probability97.a. P(pass inspection) = P(pass initially ∪ passes after recrimping) = P(pass initially) + P(fails initially ∩ goes to recrimping ∩ is corrected after recrimping)= .95 + (.05)(.80)(.60) (following path “bad-good-good” on tree diagram)= .974b. P(needed no recrimping | passed inspection) =P(passed.initially )P(passed.inspection).95= = . 9754.97498.a. P(both + ) = P(carrier ∩ both + ) + P(not a carrier ∩ both + )=P(both + | carrier) x P(carrier)+ P(both + | not a carrier) x P(not a carrier)= (.90) 2 (.01) + (.05) 2 (.99) = .01058P(both – ) = (.10) 2 (.01) + (.95) 2 (.99) = .89358P(tests agree) = .01058 + .89358 = .90416P(carrier ∩ both.positive)P(both.positive)2(.90) (.01).01058b. P(carrier | both + ve) = = = . 765699. Let A = 1 st functions, B = 2 nd functions, so P(B) = .9, P(A ∪ B) = .96, P(A ∩ B)=.75. Thus,P(A ∪ B) = P(A) + P(B) - P(A ∩ B) = P(A) + .9 - .75 = .96, implying P(A) = .81.P(B ∩ A)P(A).75.81This gives P(B | A) = = = . 926100. P(E 1 ∩ late) = P( late | E 1 )P(E 1 ) = (.02)(.40) = .00890