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# soln3 - CMPE 107 Homework 4 Problem Solutions Yates and...

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CMPE 107 Homework 4 October 11, 2001 Problem Solutions : Yates and Goodman,2.2.1 2.2.3 2.2.4 2.2.4 1.6.5 2.2.6 2.2.7 2.2.9 2.3.1 1.8.2 1.8.4 and 1.8.5 Contact TA with questions Problem 2.2.1 (a) We wish to find the value of c that makes the PMF sum up to one. P N ( n ) = c ( 1 / 2 ) n n = 0 , 1 , 2 0 otherwise 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 Problem 2.2.3 (a) We must choose c to make the PMF of V sum to one. 4 v = 1 P V ( v ) = c ( 1 2 + 2 2 + 3 2 + 4 2 ) = 30 c = 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 (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 (d) The probability that V > 2 is P [ V > 2 ] = P V ( 3 )+ P V ( 4 ) = 3 2 30 + 4 2 30 = 5 6 1

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Problem 2.2.4 (a) We choose c so that the PMF sums to one. x P X ( x ) = c 2 + c 4 + c 8 = 7 c 8 = 1 Thus c = 8 / 7. (b) P [ X = 4 ] = P X ( 4 ) = 8 7 · 4 = 2 7 (c) P [ X < 4 ] = P X ( 2 ) = 8 7 · 2 = 4 7 (d) P [ 3 X 9 ] = P X ( 4 )+ P X ( 8 ) = 8 7 · 4 + 8 7 · 8 = 3 7 Problem 2.2.4 (a) We choose c so that the PMF sums to one. x P X ( x ) = c 2 + c 4 + c 8 = 7 c 8 = 1 Thus c = 8 / 7. (b) P [ X = 4 ] = P X ( 4 ) = 8 7 · 4 = 2 7 (c) P [ X < 4 ] = P X ( 2 ) = 8 7 · 2 = 4 7 (d) P [ 3 X 9 ] = P X ( 4 )+ P X ( 8 ) = 8 7 · 4 + 8 7 · 8 = 3 7 2
Problem 1.6.5 For a sample space S = { 1 , 2 , 3 , 4 } with equiprobable outcomes, consider the events A 1 = { 1 , 2 } A 2 = { 2 , 3 } A 3 = { 3 , 1 } Each event A i has probability 1 / 2. Moreover, each pair of events is independent since P [ A 1 A 2 ] = P [ A 2 A 3 ] = P [ A 3 A 1 ] = 1 / 4 However, the threee events A 1 , A 2 , A 3

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soln3 - CMPE 107 Homework 4 Problem Solutions Yates and...

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