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Unformatted text preview: IEOR 3106: Introduction to Operations Research: Stochastic Models SOLUTIONS to Part I of the First Midterm Exam, October 5, 2010 You need to show your work. Briefly explain your reasoning. 1. A Game of Chance (15 points) Consider a game of chance played by making independent flips of a single fair coin. On each flip the coin comes up heads with probability 1 / 2. The game is played in stages. There are up to three stages. In the first stage, the player flips the coin two times and counts the number of heads observed in these two flips. Let N 1 be the random number of heads observed in stage 1. The game is over if N 1 = 0. The game continues to a second stage if N 1 ≥ 1. In the second stage, the player flips the coin N 1 times and counts the number of heads observed in these N 1 flips. Let N 2 be the random number of heads observed in stage 2. The game is over if N 2 = 0. If N 2 ≥ 1, then the game continues to a third stage. In the third stage, the player flips the coin N 2 times and counts the number of heads observed in these N 2 flips. Let N 3 be the random number of heads observed in stage 3. Let N be the number of heads observed in the last stage played. For example, N = N 1 = 0 if N 1 = 0, while N = N 3 = 2 if N 1 ≥ 1, N 2 ≥ 1 and N 3 = 2. ———————————————————————— It is convenient to draw a probability tree, as in the lecture notes for the first class: probability tree for coin flipping start 2 stage 1 1 1/4 1/2 1/4 1 2 1 1 2 1/64 1/4 1/4 1/4 1/2 1/32 = 2/64 1/64 1/16 = 4/64 1/16 = 4/64 stage 2 stage 3 1/4 1/2 1/2 1/2 1/2 1/2 1 1 1/2 1/4 = 16/64 1/4 = 16/64 1/8 = 8/64 1/8 = 8/64 1/2 1/16 = 4/64 (a) (5 points) What is the probability distribution of N ? ———————————————————————— From the probability tree, adding the appropriate values on the end of the branches, we get P ( N = 0) = 49 / 64 , P ( N = 1) = 14 / 64 , P ( N = 2) = 1 / 64 . ———————————————————————— (b) (1 point) What is E [ N ]? ———————————————————————— Then E [ N ] = 0 × P ( N = 0) + 1 × P ( N = 1) + 2 × P ( N = 2) = 16 / 64 = 1 / 4 . ———————————————————————— (c) (2 points) What is the variance of N ? ———————————————————————— It is easy to use V ar ( N ) = E [ N 2 ] ( E [ N ]) 2 , where E [ N 2 ] = 0 2 × P ( N = 0) + 1 2 × P ( N = 1) + 2 2 × P ( N = 2) = 18 / 64 = 9 / 32 ....
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 Spring '11
 WARDWHITT
 Operations Research, Probability theory, probability density function, xmax, Foolsgold, Inc, Foolsgold, Inc.

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