Unformatted text preview: Math 151 HW 3 Solutions Problems 2. P ( X = 1) = 1 / 36, P ( X = 2) = 2 / 36, P ( X = 3) = 2 / 36, P ( X = 4) = 3 / 36, P ( X = 5) = 2 / 36, P ( X = 6) = 4 / 36, P ( X = 8) = 2 / 36, P ( X = 9) = 1 / 36, P ( X = 10) = 2 / 36, P ( X = 12) = 4 / 36, P ( X = 15) = 2 / 36, P ( X = 16) = 1 / 36, P ( X = 18) = 2 / 36, P ( X = 20) = 2 / 36, P ( X = 24) = 2 / 36, P ( X = 25) = 1 / 36, P ( X = 30) = 2 / 36, P ( X = 36) = 1 / 36 8. We will do only (a) and (c). The other two are similar. (a) P ( X = 1) = 1 / 36, P ( X = 2) = 3 / 36, P ( X = 3) = 5 / 36, P ( X = 4) = 7 / 36, P ( X = 5) = 9 / 36, P ( X = 6) = 11 / 36 (c) P ( X = 2) = 1 / 36, P ( X = 3) = 2 / 36, P ( X = 4) = 3 / 36, P ( X = 5) = 4 / 36, P ( X = 6) = 5 / 36, P ( X = 7) = 6 / 36, P ( X = 8) = 5 / 36, P ( X = 6) = 4 / 36, P ( X = 10) = 3 / 36, P ( X = 11) = 2 / 36, P ( X = 12) = 1 / 36 20. (a) P ( X > 0) = P ( X = 1) = 18 38 + 20 38 ( 18 38 ) 2 =0.59 (b)and(c) E [ X ] = 1 P ( X = 1) + ( 1) P ( X = 1) + ( 3) P ( X = 3) = . 1 USD Since on the average she loses 0.1 USD it is not ”winning” strategy. Misleading part of the strategySince on the average she loses 0....
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 Winter '08
 Liu
 Math, Probability, Probability theory, Binomial distribution, 1 K, USD, 0 K, 1 j

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