SolutionHW2

# SolutionHW2 - STAT500 HW#2_solutions Solutions to Homework...

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STAT500 HW#2_solutions Solutions to Homework 2 1) (fifth) a) A={6} n(A) / n(S) = 1/6 b) B={2,4,6} n(B) / n(S) = 3/6=1/2 c) C={3,4,5,6} n(C) / n(S) = 4/6=2/3 d) D={4,6} n(D) / (S)=2/6=1/3 (sixth) a) A={6} n(A) / n(S) = 1/6 b) B={1,3,5} n(B) / n(S) = 3/6=1/2 c) C={4,5,6} n(C) / n(S) = 3/6=1/2 d) D={4,6} n(D) / (S)=2/6=1/3 2) X∩Y Mutually exclusive? X∩Y =Ø? P(X∩Y) P(X)P(Y) Independent? P(X∩Y)=P(X)P(Y)? A∩B {6} X 1/6 (1/6)(3/6) X A∩C {6} X 1/6 (1/6)(4/6) X A∩D {6} X 1/6 (1/6)(2/6) X B∩C {4,6} X 2/6=1/3 (3/6)(4/6)=1/3 O B∩D {4.6} X 2/6=1/3 (3/6)(1/3)=1/6 X C∩D {4,6} X 2/6=1/3 (4/6)(2/6)=2/9 X Event B and event C are independent. (OR, Event A, B, and C are not independent, because P(A∩B∩C) = P({6})=1/ 6 1/18 = P(A)*P(B)*P(C). ) None of the pairs have empty intersections, therefore none of the events are mutually exclusive. 3) a) P(A)=0.20+0.15+0.30=0.65, P(B)=0.10+0.30=0.40. b) P(both A and B occur) = P(A∩B) =P({5})=0.30 c) P(either A or B occurs) = P(A U B)=P({1,3,4,5})=0.20+0.15+0.10+0.30=0.75 d) P(A∩B)=0.30 ≠ 0.26=0.65*0.40=P(A)P(B), A and B are not independent. e) A∩B={5}≠Ø, A and B are not mutually exclusive. 4) a) P (professional would accept promotion) = 0.7435. P (professional rejects promotion) = 1 - 0.7435 = 0.2565. b) P (professional is part of two-career marriage) = 0.46. P (professional is part of one-career marriage) = 0.37.

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STAT500 HW#2_solutions 5) First let's list the events and probabilities. Let event A = pay in full any month (or first month) and let B = pay in full the next month (or second month). Then we have the following probabilities:
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SolutionHW2 - STAT500 HW#2_solutions Solutions to Homework...

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