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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,
2.2.1 2.2.3 2.2.6 2.3.2 2.3.7 2.3.10 2.3.11 2.4.2
2.4.3 and 2.5.9
Problem 2.2.1 Solution
(a) We wish to fnd the value oF
c
that makes the PM± 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
≤
1 is
P
[
N
≤
1] =
P
[
N
= 0] +
P
[
N
= 1] = 4
/
7 + 2
/
7 = 6
/
7
(2)
Problem 2.2.3 Solution
(a) We must choose
c
to make the PM± 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)
(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)
1
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View Full DocumentProblem 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
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 Spring '08
 Duman

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