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 (A∩Bc)∪(B∩Ac). Thus we haveP((A∩Bc)∪(B∩Ac))=P(A∩Bc) +P(B∩Ac)(disjoint union)=[P(A)-P(A∩B)] + [P(B)-P(A∩B)](Theorem1.2.9a)=P(A) +P(B)-2P(A∩B).c. “At least one ofAorB” isA∪B. So we get the same answer as in a).d. “At most one ofAorB” is (A∩B)c, andP((A∩B)c) = 1-P(A∩B).1.5 a.A∩B∩C={a U.S. birth results in identical twins that are female}b.P(A∩B∩C) =190×13×121.6p= (1-u)(1-w),p1=u(1-w) +w(1-u),p2=uw,p=p2⇒u+w= 1p1=p2⇒uw= 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-πr2Aifi= 0πr2Ah(6-i)2-(5-i)252iifi= 1,...,5.b.P(scoringipoints|board is hit)=P(scoringipoints∩board is hit)P(board is hit)P(board is hit)=πr2AP(scoringipoints∩board is hit)=πr2A(6-i)2-(5-i)252i= 1,...,5....
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