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í��ë¥ - Problem Solutions Chapter 2 Problem 2.2.1 Solution(a We wish to nd the value of c that makes the PMF sum up to one PN(n = Therefore

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Problem Solutions – Chapter 2 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 ) = ± 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 1is P [ N 1] = P [ N = 0] + P [ N = 1] = 4 / 7 + 2 / 7 = 6 / 7 (2) 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 0 otherwise P R ( r ) = 1 / 4 r = 0 3 / 4 r = 2 0 otherwise (1) From the PMFs P X ( x ) and P R ( r ) , we can calculate the requested probabilities (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 Solution (a) We must choose c to make the PMF 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) 37
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(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 > 2is P [ V > 2] = P V ( 3 ) + P V ( 4 ) = 3 2 30 + 4 2 30 = 5 6 (4) Problem 2.2.4 Solution (a) We choose c so that the PMF sums to one. X x P X ( x ) = c 2 + c 4 + c 8 = 7 c 8 = 1 (1) Thus c = 8 / 7. (b) P [ X = 4] = P X ( 4 ) = 8 7 · 4 = 2 7 (2) (c) P [ X < 4] = P X ( 2 ) = 8 7 · 2 = 4 7 (3) (d) P [3 X 9] = P X ( 4 ) + P X ( 8 ) = 8 7 · 4 + 8 7 · 8 = 3 7 (4) Problem 2.2.5 Solution Using B (for Bad) to denote a miss and G (for Good) to denote a successful free throw, the sample tree for the number of points scored in the 1 and 1 is ± ± ± ± ± ± B 1 p G p ± ± ± ± ± ± B 1 p G p Y = 0 Y = 1 Y = 2 From the tree, the PMF of Y is P Y ( y ) = 1 py = 0 p ( 1 p ) y = 1 p 2 y = 2 0 otherwise (1) 38
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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 p ) 3 0 . 05. This implies p = 0 . 6316. Problem 2.2.7 Solution In Problem 2.2.6, each caller is willing to make 3 attempts to get through. An attempt is a failure if all n operators are busy, which occurs with probability q = ( 0 . 8 ) n . Assuming call attempts are independent, a caller will suffer three failed attempts with probability q 3 = ( 0 . 8 ) 3 n . The problem statement requires that ( 0 . 8 ) 3 n 0 . 05. This implies n 4 . 48 and so we need 5 operators. Problem 2.2.8 Solution From the problem statement, a single is twice as likely as a double, which is twice as likely as a triple, which is twice as likely as a home-run. If p is the probability of a home run, then P B ( 4 ) = pP B ( 3 ) = 2 B ( 2 ) = 4 B ( 1 ) = 8 p (1) Since a hit of any kind occurs with probability of .300, p + 2 p + 4 p + 8 p = 0 . 300 which implies p = 0 . 02. Hence, the PMF of B is P B ( b ) = 0 . 70 b = 0 0 . 16 b = 1 0 . 08 b = 2 0 . 04 b = 3 0 . 02 b = 4 0 otherwise (2) Problem 2.2.9 Solution (a) In the setup of a mobile call, the phone will send the “SETUP” message up to six times. Each time the setup message is sent, we have a Bernoulli trial with success probability p .Of course, the phone stops trying as soon as there is a success. Using r to denote a successful response, and n a non-response, the sample tree is ± ± ± ± ± r p n 1 p ± ± ± ± ± r p n 1 p ± ± ± ± ± r p n
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This note was uploaded on 12/12/2011 for the course EE EE980 taught by Professor Ji during the Spring '11 term at Akademia Rolnicza w Wrocławiu.

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í��ë¥ - Problem Solutions Chapter 2 Problem 2.2.1 Solution(a We wish to nd the value of c that makes the PMF sum up to one PN(n = Therefore

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