# Hw3_sol - 20 = 0(d P | Y-1 | 2 = 1 = P Y = 0 or Y = 2 = P Y = 0 P Y = 2 = P Y = 2 = 3 16(e P Y is an odd integer = ∑ ∞ k =0 3 4 1 4 2 k = 4 5

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ISyE 2027 Homework 3: Solutions due on Wednesday, June 9 1. - 2. (a) This is a reasonable assumption. (b) p = 1 100 . P ( X = k ) = ( 10 k ) × (1 / 100) k × (99 / 100) 10 - k , for k = 0 , 1 ,..., 10 (c) 1 - P ( X = 0) - P ( X = 1) = 1 - ( 99 100 ) 10 - 10 × 1 100 × ( 99 100 ) 9 0 . 00427 3. (a) Geometric distribution. (b) Binomial distribution. (c) Bernoulli distribution. 4. Since each participation is independent with former participation, M has a Geometric distribution P ( M = k ) = (1 - p ) k - 1 p, k = 1 , 2 ,... . Also the two lotteries are independent with each other. So the probability to win both prize is p 1 p 2 . The parameter p of its distribution is the probability that he wins at least one prize for one participation, which is p = p 1 (1 - p 2 ) + p 2 (1 - p 1 ) + p 1 p 2 = p 1 + p 2 - p 1 p 2 5. (a) P ( Y = 0) = 0 (b) P ( Y 20) = 2 0 k =1 3 4 ( 1 4 ) k - 1 = 1 - ( 1 4 ) 2 0 (c) P ( Y = 0 | Y 20) = P ( Y =0) P ( Y

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Unformatted text preview: 20) = 0 (d) P ( | Y-1 | 2 = 1) = P ( Y = 0, or, Y = 2) = P ( Y = 0) + P ( Y = 2) = P ( Y = 2) = 3 / 16 (e) P ( Y is an odd integer) = ∑ ∞ k =0 3 4 ( 1 4 ) 2 k = 4 / 5 6. (a) P ( Y = 0) = e-3 (b) P ( Y ≤ 1) = P ( Y = 0) + P ( Y = 1) = e-3 + e-3 3 1 = 4 e-3 (c) P ( Y = 0 | Y ≤ 1) = P ( Y =0) P ( Y ≤ 1) = 1 / 4 1 7. (a) P (failure) = P ( S < . 55) = R . 55-∞ f ( x ) dx = R . 5 4 x dx + R . 55 . 5 (4-4 x ) dx = 0 . 595. (b) Suppose F ( q . 5 = 0 . 5). For b ≤ . 5 we obtain F ( b ) = R b 4 x dx = 2 b 2 . Then 2 q 2 . 5 = 0 . 5 and q . 5 = 0 . 5. 2...
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## This note was uploaded on 03/29/2011 for the course ISYE 2027 taught by Professor X during the Spring '11 term at Central GA Tech.

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Hw3_sol - 20 = 0(d P | Y-1 | 2 = 1 = P Y = 0 or Y = 2 = P Y = 0 P Y = 2 = P Y = 2 = 3 16(e P Y is an odd integer = ∑ ∞ k =0 3 4 1 4 2 k = 4 5

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