hw3_soln

hw3_soln - Probability and Stochastic Processes: A Friendly...

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Unformatted text preview: Probability and Stochastic Processes: A Friendly Introduction for Electrical and Computer Engineers Roy D. Yates and David J. Goodman Problem Solutions : Yates and Goodman,2.2.1 2.2.2 2.2.4 2.2.6 2.3.2 2.3.6 2.3.9 and 2.3.11 Problem 2.2.1 (a) We wish to find the value of c that makes the PMF sum up to one. c4 1, implying c 4 7. ¢ c2 n 012 otherwise ¤ ¢¤ ¨¤ ¨ ¢ §¡ 1 is PN 1 47 ¢  0 27 ¨¤ ¢ ¨  PN ¤ © 1 ¢ ¢ ¦ PN 67 ¤ (b) The probability that N ¢ £¡ c n ¡¤ 0 PN n ¢ Therefore, ∑2 n c12 0 ¥¥ PN n ¢  © Problem 2.2.2 From Example 2.5, we can write the PMF of X and the PMF of R as ¢ ¤ ¢ £¡ ¢ PR r 14 r 0 34 r 2 0 otherwise ¢ x0 x1 x2 x3 otherwise ¤ ¢ ¢ 8 8 8 8 ¢ ¤ ¤ ¤ ¤ ¢ £¡ PX x 1 3 3 1 0 From the PMFs PX x and PR r , we can calculate the requested probabilities ¡ 1 8. ¨ ¡ PX 1 ¢  PR 2 ¢ £¡ 3 4. PX 2 7 8. ¤ ¢  1 (c) P R ¨ ¡ PX 0 3 ¢ £¡ (b) P X ¤ ¢ £¡ PX 0 ¤ 0 ¡ (a) P X ¢  ¢   Problem 2.2.4 (a) We choose c so that the PMF sums to one. ¢ §¡ PX 4 ¨ c 8 7c 8 1 8 7. ¤ ¢ Thus c c 4 ¨ x c 2 ¢ x ¢ ∑ PX (b) ¢ ¡ 1 8 74 ¢ PX 4 2 7 ¢  ¢ (c) PX 2  9 PX 4 8 ¢ ¡ 4 4 7 ¢ PX 72 ¢  (d) 8 ¢ £¡ PX 8 8 74 ¨ ¨ ¡ ¢ ! X 78 3 7 ¢ © P3 © Problem 2.2.6 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 2.3.2 (a) Each paging attempt is an independent Bernoulli trial with success probability p. The number of times K that the pager receives a message is the number of successes in n Bernoulli trials and has the binomial PMF " ¢ £¡ 0 p nk $¡ pk 1 k 01 otherwise ¢ n k ¥ ### %%%¥ ¥ PK k n (b) Let R denote the event that the paging message was received at least once. The event R has probability 1 p n p . For p ¡ ¢  ¢ " ¢   ¢   " To ensure that P R 0 95 requires that n ln 0 05 ln 1 n 1 86. Thus, n 2 pages would be necessary. 1 ¢ 0 0 8, we must have # PB " 1 0 " PB ¡ PR ¤ %¡ # & # ¢ & !  # & Problem 2.3.6 an interval of T minutes, buses arrive at the bus stop at a rate of 1 5 buses per minute. ¤ (a) From the definition of the Poisson PMF, the PMF of B, the number of buses in T minutes, is b! b 0 1 otherwise ¤'$ ¡¤ ¢ §¡ 2 minutes, the probability that three buses arrive in a two minute interval is 25 3! 0 0072 ( ¤'$ 2 5 3e # ¡¤ ¢ £¡ 10 minutes, the probability of zero busees arriving in a ten minute interval 10 5 0! ¢ ¤' $ e e 2 ($ ¢ ¡ ¢ PB 0 0 135 # ¢ PB 3 (c) By choosing T is T5 ¢ (b) Choosing T T 5 be 0 ### %%%¥ ¥ PB b (d) The probability that at least one bus arrives in T minutes is 0 ¢  PB ¢ " 23 0 minutes. # 2 1 " ¢  ( & 5 ln 100 1 e T5 &'$ & Rearranging yields T 1 0 99 # PB Problem 2.3.9 The packets are delay sensitive and can only be retransmitted d times. For t d , a packet is transmitted t times if the first t 1 attempts fail followed by a successful transmission on attempt t . Further, the packet is transmitted d times if there are failures on the first d 1 transmissions, no matter what the outcome of attempt d . So the random variable T , the number of times that a packet is transmitted, can be represented by the following PMF.  " " d 1 " )%%%¥ ¥ ¥ ### $¡ " $¡ " ¢ ¢ ¢ ¡ PT t p1 pt 1 t 12 1 pd 1 t d 0 otherwise Problem 2.3.11 (a) Let Sn denote the event that the Sixers win the series in n games. Similarly, Cn is the event that the Celtics in in n games. The Sixers win the series in 3 games if they win three straight, which occurs with probability 3 12 18 ¤ ¢¡¤ P S3 ¢ !  The Sixers win the series in 4 games if they win two out of the first three games and they win the fourth game so that 3 12 ¢ 0¡ ¤ 12 3 16 ¤ 3 2 ¡¤ P S4 ¢   The Sixers win the series in five games if they win two out of the first four games and then win game five. Hence, 4 12 ¢ 0¡ ¤ 12 3 16 ¤ 4 2 ¡¤ P S5 ¢   By symmetry, P Cn P Sn . Further we observe that the series last n games if either the Sixers or the Celtics win the series in n games. Thus,  P Sn P Cn ¢ !  ¨   n 2P Sn  ¢  ¢   PN ¢ Consequently, the total number of games, N , played in a best of 5 series between the Celtics and the Sixers can be described by the PMF ¢ ¢ ¤ ¤ ¢ ¤ ¢¡¤ ¢¡¤ ¢ ¡ ¤ ¢ ¡ PN n 2123 14 n3 3 4 21 12 38 n 4 24 125 38 n 5 2 0 otherwise (b) For the total number of Celtic wins W , we note that if the Celtics get w 3 wins, then the Sixers won the series in 3 w games. Also, the Celtics win 3 games if they win the series in 3,4, or 5 games. Mathematically,  ¨ ¨2    ¨ 1 w 3 P C5  P C4 w w ¢ P S3 P C3 ¢ w 012 3 ¥¥ ¢  ¢ PW  Thus, the number of wins by the Celtics, W , has the PMF shown below. ¤ ¢ ¢ ¢ ¤ ¨ ¤¨¤ ¤ ¢!  ¢!  ¢ !  ¢ ¤ ¤ ¢ §¡ PW w w0 w1 w2 12 w 3 otherwise ¢ P S3 18 P S4 3 16 3 16 P S5 1 8 3 16 3 16 0 (c) The number of Celtic losses L equals the number of Sixers’ wins WS . This implies PL l PWS l . Since either team is equally likely to win any game, by symmetry, PWS w PW w . This implies PL l PWS l PW l . The complete expression of for the PMF of L is ¢ §¡ ¡ ¢ ¢ ¢ ¢ ¤ ¤ ¤ ¤ ¢ §¡ 4 l0 l1 l2 l3 otherwise ¡ PW l 8 16 16 2 ¢ 3¡ ¢ §¡ ¢ §¡ PL l 1 3 3 1 0 ¢ §¡ ¡ ...
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This note was uploaded on 02/11/2012 for the course EEE 352 taught by Professor Ferry during the Spring '08 term at ASU.

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