Dr. Hackney STA Solutions pg 5

Dr. Hackney STA Solutions pg 5 - 1-2Solutions Manual for...

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Unformatted text preview: 1-2Solutions Manual for Statistical Inferenceb. AorBbut not both is (ABc)(BAc). Thus we haveP((ABc)(BAc))=P(ABc) +P(BAc)(disjoint union)=[P(A)-P(AB)] + [P(B)-P(AB)](Theorem1.2.9a)=P(A) +P(B)-2P(AB).c. At least one ofAorB isAB. So we get the same answer as in a).d. At most one ofAorB is (AB)c, andP((AB)c) = 1-P(AB).1.5 a.ABC={a U.S. birth results in identical twins that are female}b.P(ABC) =19013121.6p= (1-u)(1-w),p1=u(1-w) +w(1-u),p2=uw,p=p2u+w= 1p1=p2uw= 1/3.These two equations implyu(1-u) = 1/3, which has no solution in the real numbers. Thus,the probability assignment is not legitimate.1.7 a.P(scoringipoints) =1-r2Aifi= 0r2Ah(6-i)2-(5-i)252iifi= 1,...,5.b.P(scoringipoints|board is hit)=P(scoringipointsboard is hit)P(board is hit)P(board is hit)=r2AP(scoringipointsboard is hit)=r2A(6-i)2-(5-i)252i= 1,...,5....
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