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MATH 425 Final: MATH 425 Final Exam 2 Fall 2015 Solutions

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turquoisegnu128

31 Jan 2019

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Hence the probability is (cid:0) 39: in (x + 2y + 3z)5, the coe cient of xy2z2 is (cid:0) 5. 13(cid:1) possible subsets in all, but of those with no spade, there are only (cid:0) 39. = 17063919/1334062100 = 0. 01279: (a) a man has two children. Given that at least one of them is a boy, the probability that they are both boys is 1/3. (b) a man has two children. Then p (e) = 3/5, p (f ) = 7/10, but p (ef ) = 2/5 6= 21/50, so they are not independent. (but close!) 10 = 3: cov(x +y, 2x +y ) = cov(x, 2x +y ) +cov(y, 2x +y ) = cov(x, 2x) +cov(x, y ) + Then the number of di erent books bought is x = x1 + + x10. Hence e[xi] = 1 (9/10)5, and so the expected number of di erent books bought is 10(1 (9/10)5) = 4. 095.

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Related Questions

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