Probability and Stochastic Processes:
A Friendly Introduction for Electrical and Computer Engineers
Edition 2
Roy D. Yates and David J. Goodman
Problem Solutions
: Yates and Goodman,
2.6.5 2.7.2 2.7.4 2.7.6 2.7.7 2.8.5 2.8.9 2.8.11
2.9.7 and 2.10.5
Problem 2.6.5 Solution
(a) The source continues to transmit packets until one is received correctly. Hence, the
total number of times that a packet is transmitted is
X
=
x
if the first
x

1 trans
missions were in error. Therefore the PMF of
X
is
P
X
(
x
) =
q
x

1
(1

q
)
x
= 1
,
2
, . . .
0
otherwise
(1)
(b) The time required to send a packet is a millisecond and the time required to send an
acknowledgment back to the source takes another millisecond. Thus, if
X
transmis
sions of a packet are needed to send the packet correctly, then the packet is correctly
received after
T
= 2
X

1 milliseconds. Therefore, for an odd integer
t >
0,
T
=
t
iff
X
= (
t
+ 1)
/
2. Thus,
P
T
(
t
) =
P
X
((
t
+ 1)
/
2) =
q
(
t

1)
/
2
(1

q
)
t
= 1
,
3
,
5
, . . .
0
otherwise
(2)
Problem 2.7.2 Solution
Whether a lottery ticket is a winner is a Bernoulli trial with a success probability of 0
.
001.
If we buy one every day for 50 years for a total of 50
·
365 = 18250 tickets, then the number
of winning tickets
T
is a binomial random variable with mean
E
[
T
] = 18250(0
.
001) = 18
.
25
(1)
Since each winning ticket grosses $1000, the revenue we collect over 50 years is
R
= 1000
T
dollars. The expected revenue is
E
[
R
] = 1000
E
[
T
] = 18250
(2)
But buying a lottery ticket everyday for 50 years, at $2.00 a pop isn’t cheap and will cost
us a total of 18250
·
2 = $36500. Our net profit is then
Q
=
R

36500 and the result of our
loyal 50 year patronage of the lottery system, is disappointing expected loss of
E
[
Q
] =
E
[
R
]

36500 =

18250
(3)
1
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Problem 2.7.4 Solution
Given the distributions of
D
, the waiting time in days and the resulting cost,
C
, we can
answer the following questions.
(a) The expected waiting time is simply the expected value of
D
.
E
[
D
] =
4
X
d
=1
d
·
P
D
(
d
) = 1(0
.
2) + 2(0
.
4) + 3(0
.
3) + 4(0
.
1) = 2
.
3
(1)
(b) The expected deviation from the waiting time is
E
[
D

μ
D
] =
E
[
D
]

E
[
μ
d
] =
μ
D

μ
D
= 0
(2)
(c)
C
can be expressed as a function of
D
in the following manner.
C
(
D
) =
90
D
= 1
70
D
= 2
40
D
= 3
40
D
= 4
(3)
(d) The expected service charge is
E
[
C
] = 90(0
.
2) + 70(0
.
4) + 40(0
.
3) + 40(0
.
1) = 62 dollars
(4)
Problem 2.7.6 Solution
Since our phone use is a geometric random variable
M
with mean value 1
/p
,
P
M
(
m
) =
(1

p
)
m

1
p
m
= 1
,
2
, . . .
0
This is the end of the preview.
Sign up
to
access the rest of the document.
 Spring '08
 Duman
 Probability theory, Expected Profit

Click to edit the document details