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Unformatted text preview: Pstat160 Winter 2010 Solution vending machine problem Part 1. The can chose the state space to be S = { , 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 addup to 1! Fixing the departing state to be: i=0: None of the machines are down. 0 := { Both machine dont break } , prob= . 9 2 1 := { One machine only breaks } , prob = 2 . 9 . 1 = . 18 2 := { Both machine break } , prob = . 1 2 = . 01 i=1: One machine is down 1 0 := { Broken machine fixed, other doesnt break, prob= . 2 . 9 = . 18 1 1 :=...
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This note was uploaded on 01/10/2011 for the course STAT 160A taught by Professor Bonnet during the Winter '10 term at UCSB.
 Winter '10
 bonnet

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