hw3-solution

hw3-solution - Probability and Stochastic Processes: A...

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Probability and Stochastic Processes: A Friendly Introduction for Electrical and Computer Engineers Edition 2 Roy D. Yates and David J. Goodman Problem Solutions : Yates and Goodman, 2.2.1 2.2.3 2.2.6 2.3.2 2.3.7 2.3.10 2.3.11 2.4.2 2.4.3 and 2.5.9 Problem 2.2.1 Solution (a) We wish to fnd the value oF c that makes the PM± sum up to one. P N ( n ) = ± c (1 / 2) n n = 0 , 1 , 2 0 otherwise (1) ThereFore, 2 n =0 P N ( n ) = c + c/ 2 + c/ 4 = 1, implying c = 4 / 7. (b) The probability that N 1 is P [ N 1] = P [ N = 0] + P [ N = 1] = 4 / 7 + 2 / 7 = 6 / 7 (2) Problem 2.2.3 Solution (a) We must choose c to make the PM± oF V sum to one. 4 X v =1 P V ( v ) = c (1 2 + 2 2 + 3 2 + 4 2 ) = 30 c = 1 (1) Hence c = 1 / 30. (b) Let U = { u 2 | u = 1 , 2 , . . . } so that P [ V U ] = P V (1) + P V (4) = 1 30 + 4 2 30 = 17 30 (2) (c) The probability that V is even is P [ V is even] = P V (2) + P V (4) = 2 2 30 + 4 2 30 = 2 3 (3) (d) The probability that V > 2 is P [ V > 2] = P V (3) + P V (4) = 3 2 30 + 4 2 30 = 5 6 (4) 1
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Problem 2.2.6 Solution The probability that a caller fails to get through in three tries is (1 - p ) 3 . To be sure that at least 95% of all callers get through, we need (1
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hw3-solution - Probability and Stochastic Processes: A...

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