~HW2_Sol - 7

# ~HW2_Sol - 7 - ECEN-646 Homework 2 Solutions 1 C ¸inlar...

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ECEN-646 Homework 2 Solutions 1 C ¸inlar Exercise 4.9, Chapter 1 (a) P ( X t ) = 1 e 0 . 2 t 0 . 2 te 0 . 2 t t 0 P ( X > t ) = 1 P ( X t ) = e 0 . 2 t + 0 . 2 te 0 . 2 t = (1 + 0 . 2 t ) e 0 . 2 t t 0 The problem asks for P ( X > t + s | X > t ) = P ( { X > t + s } ∩ { X > t } ) P ( X > t ) = P ( X > t + s ) P ( X > t ) = (1 + 0 . 2( t + s )) e 0 . 2( t + s ) (1 + 0 . 2 t ) e 0 . 2 t = (1 + 0 . 2( t + s )) e 0 . 2 s (1 + 0 . 2 t ) Substituting t = 3 and s = 1 . 5 we get the probability that having worked successfully for first 3 hours, the component works successfully for at least 1.5 hrs more as, P ( X > 4 . 5) = (1 + 0 . 2 × 4 . 5) e 0 . 2 × 1 . 5 (1 + 0 . 2 × 3) = 1 . 9 e 0 . 3 1 . 6 0 . 8797 2 C ¸inlar Exercise 4.10, Chapter 1 Let A i be the event that the a th i character is printed. Let I denote the impulse chosen. The printing machine’s behaviour can be characterized as P ( A i | I = i ) = p P ( A j | I = i ) = (1 p ) n 1 j negationslash = i P ( I = i ) = 1 n 1 This study resource was shared via CourseHero.com

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(a) P ( I = i | A i , A i ) = P ( A i , A i | I = i ) P ( I = i ) P ( A i , A i ) (1) P ( A i , A i ) = k = n summationdisplay k =1 P ( A i , A i | I = k ) P ( I = k ) = p 2 n + summationdisplay k negationslash = i ( 1 p n 1 ) 2 1 n = p 2 n + (1 p ) 2 n ( n 1) (2) (1)&(2) P ( I = i | A i , A i ) = p 2 /n p 2 /n + (1 p ) 2 n ( n 1) = p 2 p 2 + (1 p ) 2 ( n 1) (b) P ( I = i | A i , A j ) = P ( A i , A j | I = i ) P ( I = i ) P ( A i , A j ) P ( A i , A j | I = k ) = p (1 p ) n 1 k = i, j ( 1 p n 1 ) 2 k negationslash = i, j P ( A i , A j ) = n summationdisplay k =1 P ( A i , A j | I = k ) P ( I = k ) = p (1 p ) ( n 1) 2 n + summationdisplay k negationslash = i,j parenleftbigg 1 p n 1 parenrightbigg 2 1 n = 2 p (1 p ) n ( n 1) + ( n 2)(1 p ) 2 n ( n 1) 2 P ( I = i | A i , A j ) = p (1 p ) /n ( n 1) 2 p (1 p ) n ( n 1) + ( n 2)(1 p ) 2 n ( n 1) 2 = 1 2 + ( n 2)(1 p ) ( n 1) p (c) The probability remains unchanged because the output of the ma- chine being independent of the past behavior, the order in which it prints the characters doesn’t matter.
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