S07 EEE 350 Random Signal Analysis
Homework 4 Solutions
Feb. 15th, 2007
Problem Solutions
: Yates and Goodman,
1.10.1 1.10.2 2.2.1 2.2.3 2.2.6 and 2.2.9
Problem 1.10.1 Solution
From the problem statement, we can conclude that the device components are confgured
in the ±ollowing way.
W
1
W
2
W
5
W
3
W
4
W
6
To fnd the probability that the device works, we replace series devices 1, 2, and 3, and
parallel devices 5 and 6 each with a single device labeled with the probability that it works.
In particular,
P
[
W
1
W
2
W
3
] = (1

q
)
3
,
(1)
P
[
W
5
∪
W
6
] = 1

P
[
W
c
5
W
c
6
] = 1

q
2
.
(2)
This yields a composite device o± the ±orm
1q
2
1q
(
)
1q
3
The probability
P
[
W
0
] that the two devices in parallel work is 1 minus the probability that
neither works:
P
±
W
0
²
= 1

q
(1

(1

q
)
3
)
.
(3)
Finally, ±or the device to work, both composite device in series must work.
Thus, the
probability the device works is
P
[
W
] = [1

q
(1

(1

q
)
3
)][1

q
2
]
.
(4)
Problem 1.10.2 Solution
Suppose that the transmitted bit was a 1. We can view each repeated transmission as an
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 Spring '08
 Duman
 Probability

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