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hw8-sol - AMS 311(Fall 2010 Joe Mitchell PROBABILITY THEORY...

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AMS 311 (Fall, 2010) Joe Mitchell PROBABILITY THEORY Homework Set # 8 – Solution Notes (1). (21 points) Consider the maze shown below. There are three cells (Cell 1, Cell 2, and Cell 3) and two deadly (quite permanent) outcomes (Death By Poison, and the dreaded Death By Guillotine). A rat is initially placed in cell 1. When the rat enters Cell i , he wanders around within the cell for X i minutes, where X i is uniformly distributed between i and i 2 , and then he exits the cell by picking one of the doors at random (e.g., if there are 3 doors, he picks each with probability 1/3). (a). (7 points) Find the probability that the rat dies by poison. (Recall that he starts in Cell 1.) Let p i = P ( dies by poison | starts in cell i ). Condition on the first move: p 1 = (2 / 5)1 + (2 / 5) p 3 + (1 / 5)0 p 2 = (2 / 2) p 3 p 3 = (1 / 5)1 + (2 / 5) p 1 + (2 / 5) p 2 Solve to get p 1 , p 2 , and p 3 ( p 1 = 8 / 11, p 2 = p 3 = 9 / 11). The final answer is p 1 = 8 / 11 (since we know the rat starts in Cell 1). (b). (7 points) What is the expected number of minutes that the rat lives? The expected time the rat spends in Cell i before leaving it is E ( X i ) = i + i 2 2 minutes, since X i is Uniform between i and i 2 . Thus, E ( X 1 ) = 1, E ( X 2 ) = 3, and E ( X 3 ) = 6. Let t i = E ( number of minutes he lives | he starts in cell i ). Condition on the first move: t 1 = (2 / 5)( E ( X 1 ) + 0) + (1 / 5)( E ( X 1 ) + 0) + (2 / 5)( E ( X 1 ) + t 3 ), t 2 = (2 / 2)( E ( X 2 ) + t 3 ), t 3 = (1 / 5)( E ( X 3 ) + 0) + (2 / 5)( E ( X 3 ) + t 1 ) + (2 / 5)( E ( X 3 ) + t 2 ). Rewriting, and using the expectations E ( X 1 ) = 1, E ( X 2 ) = 3, and E ( X 3 ) = 6, we get t 1 = 1 + (2 / 5) t 3 , t 2 = 3 + t 3 , t 3 = 6 + (2 / 5) t 1 + (2 / 5) t 2 . Solve for the t i ’s (Wolfram alpha gives t 1 = 87 / 11, t 2 = 223 / 11, t 3 = 190 / 11); the final answer is t 1 = 87 / 11. (c). (7 points) What is the probability the rat visits Cell 2 before he dies? Let q i = P ( rat visits cell 2 before he dies | he starts in cell i ). Condition on the first move: q 1 = (1 / 5)0 + (2 / 5)0 + (2 / 5) q 3 q 2 = 1 q 3 = (1 / 5)0 + (2 / 5) q 1 + (2 / 5) q 2 Solve for the q i
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