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Unformatted text preview: CS 70 Fall 2006 — Solutions to Homework #11 November 30, 2006 1. Machine Failures. We have that X 1 ∼ Geometric ( p 1 ) , X 2 ∼ Geometric ( p 2 ) are the independent r.v.’s representing the number of runs until the first failure for machines M 1 and M 2 respectively. If X = min { X 1 , X 2 } then for x ∈ { 1 , 2 , . . . } Pr ( X ≥ x ) = Pr (min { X 1 , X 2 } ≥ x ) = Pr (( X 1 ≥ x ) ∧ ( X 2 ≥ x )) = Pr ( X 1 ≥ x ) · Pr ( X 2 ≥ x ) , by the independence of X 1 and X 2 = (1 p 1 ) x 1 (1 p 2 ) x 1 , by the tail probability of a Geometric [see Thm 21.2 part (2)] = ((1 p 1 ) (1 p 2 )) x 1 = (1 p 1 p 2 + p 1 p 2 ) x 1 This is the tail probability of a Geometric with parameter p 1 + p 2 p 1 p 2 and since X ∈ { 1 , 2 , . . . } is the support of a Geometric r.v. it must be that X ∼ Geometric ( p 1 + p 2 p 1 p 2 ) as claimed. 2. Geometric Distribution. Let T be the total number of attempts Mr. Bond must make before he escapes from his imprisonment. I.e. the first T 1 attempts are failures—where he either enters the AC duct or the sewer pipe—and the final attempt at trial T seems him enter the door. Thus T corresponds to the number of coin tosses landing tails until a head turns up, where a tail (Bond fails) comes up with probability 2 / 3 while a head (007 escapes) has probability 1 / 3. Hence T ∼ Geometric (1 / 3). Define X t for t ∈ { 1 , 2 , . . . , T 1 } to be the time Bond spends on failed trial t . Then X the total time Bond spends trying to escape is a r.v. satisfying X = T 1 X t =1 X t Clearly the failure trial times X 1 , . . . , X T 1 are i.i.d. with distribution X t = ( 2 with probability 0 . 5 5 with probability 0 . 5 and so E [ X 1 ] = . . . = E [ X T 1 ] = 2 · Pr ( X t = 2) + 0...
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 Fall '08
 HAULLGREN
 Normal Distribution, Probability, Xt, X Geometric, distribution Xt

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