hw3+fall+2011+solutions

# hw3+fall+2011+solutions - ECI114 HW3 Due 5p.m Wed Oct 26th...

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ECI114 HW3 Due 5p.m Wed, Oct. 26 th Q1. a) Here cylinders are selected without replacement. And n/N = 5/15 > 0.1. Thus, X is a hypergeometric random variable, i.e. it uses the “multiple combinations” rule. - - = - - = 5 15 5 6 15 6 ) ( x x n N x n r N x r x p where x = 0, 1, 2, 3, 4, 5. b) Pr[X=2] = p(2) = - 5 15 2 5 9 2 6 = 0.41958. c) Pr[X = 0] = p(0) = - 5 15 0 5 9 0 6 = 0.04196, Pr[X = 1] = p(1) = - 5 15 1 5 9 1 6 = 0.25175. Pr[X = 2] = p(2) = 6 9 2 5 2 15 5    ÷ ÷ -   ÷ = 0.41958, Pr[X = 3] = p(3) = - 5 15 3 5 9 3 6 = 0.23976, Pr[X = 4] = p(4) = - 5 15 4 5 9 4 6 = 0.04496, Pr[X = 5] = p(5) = - 5 15 5 5 9 5 6 = 0.00200, Pr[X 2] = Pr[X=0] + Pr[X=1] + Pr[X=2] = 0.04196 + 0.25175 + 0.41958 = 0.7133. Or, Pr[X 2] = 1 - Pr[X=3] - Pr[X=4] - Pr[X=5] = 1 - 0.23976 - 0.04496 - 0.00200 = 0.7133. Q2. a) The probability of obtaining the correct answer on problem one is: p = 1/3 = .333. 1

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The appropriate RV can be defined as Bernoulli, geometric, or binomial: Bernoulli: Let X = 1 if the answer to problem 1 is correct, and 0 else (then you are finding Pr[X=1], i.e. that the answer to problem 1 is correct). geometric: Let X = # questions answered till the first correct answer is given (then you are finding Pr[X=1], i.e. that the first correct answer is given on the first question). binomial: Let X = # correct answers out of n=1 Bernoulli trial, which is essentially a Bernoulli RV (then you are finding Pr[X=1], i.e. that the answer on the single question is correct) b) Let X denote the number of problems completed until the first answer is correct; then X is a geometric random variable with p = 1/3. p(4) = (1 – 1/3)
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## This note was uploaded on 01/19/2012 for the course ECI 114 taught by Professor Mocktarian during the Fall '08 term at UC Davis.

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hw3+fall+2011+solutions - ECI114 HW3 Due 5p.m Wed Oct 26th...

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