Hw05soln - ECE 310 Spring 2005 HW 5 solutions Problem E4.1 p3 = 1 P{1 2 = 0.35 p1 = 1 P{2 3 = 0.15 p2 = 1 p1 p3 = 0.5 P{1 3 = p1 p3 = 0.5 Problem

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ECE 310, Spring 2005, HW 5 solutions Problem E4.1 p 3 = 1 - P ( { ω 1 2 } ) = 0 . 35 p 1 = 1 - P ( { ω 2 3 } ) = 0 . 15 p 2 = 1 - p 1 - p 3 = 0 . 5 P ( { ω 1 3 } ) = p 1 + p 3 = 0 . 5 Problem E4.4 Let A be the event that less than two errors occurred: P ( A ) = 1 X k =0 ± 5 k (0 . 1) k (0 . 9) 5 - k = (0 . 9) 5 + 5(0 . 1)(0 . 9) 4 0 . 92 Problem E4.10 p c = 2 k =0 ( 100 k ) (0 . 01) k (0 . 99) 100 - k 0 . 92 The most probable number of errors is b (101)(0 . 01) c = 1 Problem E4.16 Lifetime L should be modelled with geometric distribution; we know β = m m +1 = 10 11 , thus: P ( L = n ) = (1 - 10 11 )( 10 11 ) n = 10 n 11 n +1 , n N Problem E4.19 We know that T has geometric distribution: P ( T = k ) = (1 - β ) β k - 1 , k = 1 , 2 ,... For k = 1, we know P ( T = 1) = P ( { 5 , 6 } ) = 1 3 , so: (1 - β ) = 1 3 β = 2 3 P ( T = k ) = 1 3 ( 2 3 ) k - 1 = 2 k - 1 3 k , k = 1 , 2 ,... Problem E4.20 We must use Poisson (100). Hence: P ( s ) = 120 k =80 e - 100 100 k k ! 0 . 96 Problem E4.21 a) Use the Poisson model. Let A be the desired event, then: P ( A ) = e - γT ( γT ) 0 0! = e - γT b) Using the result of section 4.4.4: k * = b γT c P k * = e - γT
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This homework help was uploaded on 09/26/2007 for the course ECE 3100 taught by Professor Haas during the Spring '05 term at Cornell University (Engineering School).

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Hw05soln - ECE 310 Spring 2005 HW 5 solutions Problem E4.1 p3 = 1 P{1 2 = 0.35 p1 = 1 P{2 3 = 0.15 p2 = 1 p1 p3 = 0.5 P{1 3 = p1 p3 = 0.5 Problem

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