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machine_sol

# machine_sol - Pstat160 Solution vending machine problem...

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Pstat160 Winter 2010 Solution vending machine problem Part 1. The can chose the state space to be S = { 0 , 1 , 2 } with 0 = no machine down 1 = one machine down 2 = two machine down We denote by X n = # number of machine down, and claim that this define a Markov chain on S . Indeed, this way the current state encode all the information needed to describe the probability to move to a new state. The memoryless property of the geometric distribution implies that the probability that a machine is fixed on a given day (if it has not yet been fixed) is 1 / 5 = . 2. Below, we list all possible transitions, and the corresponding event is in { } . Check each time that the probability add-up to 1! Fixing the departing state to be: i=0: None of the machines are down. 0 0 := { Both machine don’t break } , prob= . 9 2 0 1 := { One machine only breaks } , prob = 2 × . 9 × . 1 = . 18 0 2 := { Both machine break } , prob = . 1 2 = . 01 i=1: One machine is down 1 0 := { Broken machine fixed, other doesn’t break, prob= . 2 × . 9 = . 18 1 1 := { Broken one fixed and other breaks } S {

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machine_sol - Pstat160 Solution vending machine problem...

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