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Hw1_02_booksol

Hw1_02_booksol - Probability and Stochastic Processes...

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Probability and Stochastic Processes Homework 1 Solutions (Book Problems) Problem Solutions : Yates and Goodman,1.4.1 1.4.4 1.4.5 1.6.7 1.7.10 and 1.9.5 Problem 1.4.1 From the table we look to add all the disjoint events that contain H 0 to express the probability that a caller makes no hand-offs as P [ H 0 ] = P [ LH 0 ]+ P [ BH 0 ] = 0 . 1 + 0 . 4 = 0 . 5 In a similar fashion we can express the probability that a call is brief by P [ B ] = P [ BH 0 ]+ P [ BH 1 ]+ P [ BH 2 ] = 0 . 4 + 0 . 1 + 0 . 1 = 0 . 6 The probability that a call is long or makes at least two hand-offs is P [ L H 2 ] = P [ LH 0 ]+ P [ LH 1 ]+ P [ LH 2 ]+ P [ BH 2 ] = 0 . 1 + 0 . 1 + 0 . 2 + 0 . 1 = 0 . 5 Problem 1.4.4 consequence of part 4 of Theorem 1.4. (a) Since A A B , P [ A ] P [ A B ] . (b) Since B A B , P [ B ] P [ A B ] . (c) Since A B A , P [ A B ] P [ A ] . (d) Since A B B , P [ A B ] P [ B ] . Problem 1.4.5 (a) For convenience, let p i = P [ FH i ] and q i = P [ VH i ] . Using this shorthand, the six unknowns p 0 , p 1 , p 2 , q 0 , q 1 , q 2 fill the table as H 0 H 1 H 2 F p 0 p 1 p 2 V q 0 q 1 q 2 However, we are given a number of facts: p 0 + q 0 = 1 / 3 p 1 + q 1 = 1 / 3 p 2 + q 2 = 1 / 3 p 0 + p 1 + p 2 = 5 / 12
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