Math

Stat515 HW#1 Solutions Prof H.K Hsieh 09/12/2012 1. A = {FF}, B = {MM}, C = {MF, FM, MM}. Then, A?B = [pic], [pic]={FF,MM}, A?C= [pic], [pic]= S, B?C = {MM}, [pic] = C, [pic]= {MF, FM}, 2.8 a. 36 + 6 = 42b. 36-3=33c. 63-36-6=18 2.10 a. S = {A, B, AB, O} b. P({A}) = 0.41, P({B}) = 0.10, P({AB}) = 0.04, P({O}) = 0.45. c. P({A} or {AB}) = P({A}) + P({AB}) = .41+.04=0.45, since the events A and AB atomic number 18 mutu everyy exclusive. 15. . Since the events argon mutually exclusive and P(S) = P(E1) + + P(E4) = 1. So, P(E2) = 1 .01 .09 .81 = .09. b. P(at least one hit) = P(E1) + P(E2) + P(E3) = .01+.09+.09= .19. 16. a. 1/3b. 1/3 + 1/15 = 6/15c. 1/3 + 1/16 = 19/48 d. 1 (1/3+1/15+1/3+1/16) =1. 191/240= 49/240 28. . Denote the four lavdidates as A1, A2, A3, and M. a. Since coordinate is not important, the outcomes are {{A1,A2}, {A1,A3}, {A1M}, {A2,A3}, {A2,M}, {A3,M}}. b. assuming equally promising outcomes, all have fortune 1/6. c. P(minority hired) = P(A1M) + P(A2M) + P(A3M) = 1/6+1/6+1/6=3/6=.5 30. allow w1 denote the head start wine, w2 the guerilla, and w3 the third. a. Each try localize is an tenacious triple indicating the ranking.

b. 3!=6 triples: (w1,w2,w3), (w1,w3,w2), (w2,w1,w3), (w2,w3,w1), (w3,w1,w2), (w3,w2,w1) c. By symmetry, without handout of generality, we can assume w1 is the best, w2 is the second best, and w3 is the worst. There are 3 legitimate triples where w1 is not worse than w2, namely, (w1,w2,w3), (w1,w3,w2), (w3,w1,w2). So, the probability is 3/6=1/2=0.5. 32. Let 1 represent a client quest style 1, and 2 represent a client quest style 2. a. The sample space consists of the following 16 four-tuples: 1111, 1112, 1121, 1211, 2111, 1122, 1212, 2112, 1221, 2121, 2211, 2221, 2212, 2122, 1222, 2222 b. If the styles are equally in demand, the aubergeing should be...If you want to overtake a full essay, order it on our website: Ordercustompaper.com

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