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STAT_333_Test_Solutions

# STAT_333_Test_Solutions - STAT 333 Test SOLUTIONS 1 Harry...

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STAT 333 Test SOLUTIONS 1. Harry Potter is trapped in the Department of Mysteries. Three doors lead out from his room, one returning back to the centre after 6 minutes, one returning back to the centre after 9 minutes, and one leading to his goal (the Hall of Prophecies) after 20 minutes. Unfortunately, every time he enters the centre room, the doors randomly spin around so he cannot remember which doors have been tried already, so he chooses randomly. [3] a) Let X = the number of minutes to reach his goal. Find E[X]. Let D be the door Harry chooses. E[X|D = 1] = E[6 + X 1 ] = 6 + E[X] since X 1 ~ X E[X|D = 2] = E[9 + X 1 ] = 9 + E[X] since X 1 ~ X E[X|D = 3] = 20 So E[X] = E[E[X|D] = E[X|D = 1]P(D = 1) + E[X|D = 2]P(D = 2) + E[X|D = 3]P(D = 3) = (6 + E[X]) (1/3) + (9 + E[X]) (1/3) + 20 (1/3) = 35/3 + (2/3)E[X] So E[X] = 35 minutes [3] b) Find Var(X). E[X 2 |D = 1] = E[(6 + X 1 ) 2 ] = 36 + 12E[X] + E[X 2 ] since X 1 ~ X E[X 2 |D = 2] = E[(9 + X 1 ) 2 ] = 81 + 18E[X] + E[X 2 ] since X 1 ~ X E[X 2 |D = 3] = 20 2 = 400 So E[X 2 ] = E[E[X 2 |D] = E[X 2 |D = 1]P(D = 1) + E[X 2 |D = 2]P(D = 2) + E[X 2 |D = 3]P(D = 3) = (36 + 12E[X] + E[X 2 ])(1/3) + (81 + 18E[X] + E[X 2 ])(1/3) +400(1/3) = 1567/3 + (2/3)E[X 2 ] since E[X] = 35 So E[X 2 ] = 1567 And Var(X) = E[X 2 ] – E[X] 2 = 1567 – 35 2 = 342 [2] c) Harry’s friend Hermione figures out a way to label the doors so they will not try the same door twice. Describe how you would find E[X] and Var(X) in this case. (No calculations are necessary.) There are a number of ways to do this. The easiest way would be to list out all possible paths Harry could take. There are only 5 cases if he can never choose the same door twice: 1, 2, 3 (prob 1/6), 1, 3 (prob 1/6), 2, 1, 3 (prob 1/6), 2, 3 (prob 1/6), and 2 (prob 1/3). Then the expected value and variance can be calculated from first principles.

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STAT_333_Test_Solutions - STAT 333 Test SOLUTIONS 1 Harry...

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