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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.
 Spring '08
 Ferry

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