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# HW1_Solutions - IE:3610 Stochastic Modeling(Fall 2015...

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IE:3610 Stochastic Modeling (Fall 2015) Homework 1 Solutions 1. Problem 24-1 (a, b) from Chapter 24 (on ICON) A cube has its six sides colored red, white, blue, green, yellow, and violet. It is assumed that these six sides are equally likely to show when the cube is tossed. The cube is tossed once. (a) Describe the sample space. (b) Consider the random variable that assigns the number 0 to red and white, the number 1 to green and blue, and the number 2 to yellow and violet. What is the distribution of this random variable? Solution: (a) S = { Red, White, Blue, Green, Yellow, Violet } (b) X = 0 if Read or White 1 if Green or Blue 2 if Yellow or Violet Probability of either of the faces turning up is 1/6. That is, P(Red) = P(White) = P(Blue) = P(Green) = P(Yellow) = P(Violet) = 1 / 6. Then probability mass function for X is: P( X = 0) = P(Read S White) = P(Read) + P(White) = 1/6 + 1/6 = 1/3 P( X = 1) = P(Green S Blue) = P(Green) + P(Blue) = 1/6 + 1/6 = 1/3 P( X = 2) = P(Yellow S Violet) = P(Yellow) + P(Violet) = 1/6 + 1/6 = 1/3 Therefore, the probabilities are equally distributed. 2. Problem 24-2 (c) from Chapter 24 (on ICON) Suppose the sample space Ω consists of the four points { ω 1 , ω 2 , ω 3 , ω 4 } , and the associated probabilities over the events are given by P( ω 1 ) = 1 3 , P( ω 2 ) = 1 5 , P( ω 3 ) = 3 10 , P( ω 4 ) = 1 6 . Define the random variable X 1 by X 1 ( ω 1 ) = 1, X 1 ( ω 2 ) = 1, X 1 ( ω 3 ) = 4, X 1 ( ω 4 ) = 5, 1

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and the random variable X 2 by X 2 ( ω 1 ) = 1, X 2 ( ω 2 ) = 1, X 2 ( ω 3 ) = 1, X 2 ( ω 4 ) = 5.
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