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ECE 310, Spring 2005, HW 5 solutions
Problem E4.1
p
3
= 1

P
(
{
ω
1
,ω
2
}
) = 0
.
35
p
1
= 1

P
(
{
ω
2
,ω
3
}
) = 0
.
15
p
2
= 1

p
1

p
3
= 0
.
5
P
(
{
ω
1
,ω
3
}
) =
p
1
+
p
3
= 0
.
5
Problem E4.4
Let
A
be the event that less than two errors occurred:
P
(
A
) =
1
X
k
=0
±
5
k
¶
(0
.
1)
k
(0
.
9)
5

k
= (0
.
9)
5
+ 5(0
.
1)(0
.
9)
4
≈
0
.
92
Problem E4.10
p
c
=
∑
2
k
=0
(
100
k
)
(0
.
01)
k
(0
.
99)
100

k
≈
0
.
92
The most probable number of errors is
b
(101)(0
.
01)
c
= 1
Problem E4.16
Lifetime
L
should be modelled with geometric distribution; we know
β
=
m
m
+1
=
10
11
, thus:
P
(
L
=
n
) = (1

10
11
)(
10
11
)
n
=
10
n
11
n
+1
,
n
∈
N
Problem E4.19
We know that
T
has geometric distribution:
P
(
T
=
k
) = (1

β
)
β
k

1
, k
= 1
,
2
,...
For
k
= 1, we know
P
(
T
= 1) =
P
(
{
5
,
6
}
) =
1
3
, so: (1

β
) =
1
3
⇒
β
=
2
3
⇒
P
(
T
=
k
) =
1
3
(
2
3
)
k

1
=
2
k

1
3
k
, k
= 1
,
2
,...
Problem E4.20
We must use
Poisson
(100). Hence:
P
(
s
) =
∑
120
k
=80
e

100 100
k
k
!
≈
0
.
96
Problem E4.21
a) Use the Poisson model. Let
A
be the desired event, then:
P
(
A
) =
e

γT
(
γT
)
0
0!
=
e

γT
b) Using the result of section 4.4.4:
k
*
=
b
γT
c
P
k
*
=
e

γT
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This homework help was uploaded on 09/26/2007 for the course ECE 3100 taught by Professor Haas during the Spring '05 term at Cornell University (Engineering School).
 Spring '05
 HAAS
 Probability theory, AirTrain Newark, Geometric distribution, Problem E4.16 Lifetime

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