hw3soln_06 - ECE302 Spring 2006 HW3 Solutions February 2,...

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Unformatted text preview: ECE302 Spring 2006 HW3 Solutions February 2, 2006 1 Solutions to HW3 Note: Most of these solutions were generated by R. D. Yates and D. J. Goodman, the authors of our textbook. I have added comments in italics where I thought more detail was appropriate. I have also largely rewritten the solutions to problems 2.10.1, 2.10.2, and 2.10.3. Problem 2.2.1 The random variable N has PMF P N ( n ) = braceleftbigg c (1 / 2) n n = 0 , 1 , 2 , otherwise . (a) What is the value of the constant c ? (b) What is P [ N 1]? Problem 2.2.1 Solution (a) We wish to find the value of c that makes the PMF sum up to one. P N ( n ) = braceleftbigg c (1 / 2) n n = 0 , 1 , 2 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 2.2.2 For random variables X and R defined in Example 2.5, find P X ( x ) and P R ( r ). In addition, find the following probabilities: (a) P [ X = 0] (b) P [ X < 3] (c) P [ R > 1] Problem 2.2.2 Solution From Example 2.5, we can write the PMF of X and the PMF of R as P X ( x ) = 1 / 8 x = 0 3 / 8 x = 1 3 / 8 x = 2 1 / 8 x = 3 otherwise P R ( r ) = 1 / 4 r = 0 3 / 4 r = 2 otherwise (1) From the PMFs P X ( x ) and P R ( r ), we can calculate the requested probabilities ECE302 Spring 2006 HW3 Solutions February 2, 2006 2 (a) P [ X = 0] = P X (0) = 1 / 8. (b) P [ X < 3] = P X (0) + P X (1) + P X (2) = 7 / 8. (c) P [ R > 1] = P R (2) = 3 / 4. Problem 2.2.3 The random variable V has PMF P V ( v ) = braceleftbigg cv 2 v = 1 , 2 , 3 , 4 , otherwise . (a) Find the value of the constant c . (b) Find P [ V { u 2 | u = 1 , 2 , 3 , } ]. (c) Find the probability that V is an even number. (d) Find P [ V > 2]. Problem 2.2.3 Solution (a) We must choose c to make the PMF of V sum to one. 4 summationdisplay 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) ECE302 Spring 2006 HW3 Solutions February 2, 2006 3 Problem 2.2.4 The random variable X has PMF P X ( x ) = braceleftbigg c/x x = 2 , 4 , 8 , otherwise . (a) What is the value of the constant c ? (b) What is P [ X = 4]? (c) What is P [ X < 4]? (d) What is P [3 X 9]? Problem 2.2.4 Solution (a) We choose c so that the PMF sums to one. summationdisplay x P X ( x ) = c 2 + c 4 + c 8 = 7 c 8 = 1 (1) Thus c = 8 / 7....
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This note was uploaded on 10/19/2008 for the course ECSE 4500 taught by Professor Woods during the Fall '08 term at Rensselaer Polytechnic Institute.

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hw3soln_06 - ECE302 Spring 2006 HW3 Solutions February 2,...

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