ISyE 2027
Probability with Applications
Fall 2011
R. D. Foley
Solution to Homework 5
October 19, 2011
1.
Solution:
(
1

3
n
)
n

1
= (
1

3
n
)
n
*
(
1

3
n
)
(
1
)
, the first part converges to
e

3
and the 2nd part converges to 1, so lim
n
→
∞
(
1

3
n
)
n

1
=
e

3
.
2.
Solution: (a)Bernoulli.
(b)Geometric.
(c)Binomial.
(d)Bernoulli.
(e)Geometric.
(f)None of the three.
3.
Solution: (a)
Pr
{
Y
=
k
}
=
1
3
,
k
=
0
1
2
,
k
=
1
1
6
,
k
=
2
(b)E
[
Y
] =
1
3
×
0
+
1
2
×
1
+
1
6
×
2
=
1
2
+
1
3
=
5
6
.
(c)E
[
Y
2
] =
1
3
×
0
2
+
1
2
×
1
2
+
1
6
×
2
2
=
1
2
+
2
3
=
7
6
.
(d)
Var
(
Y
) =
E
[
Y
2
]  (
E
[
Y
])
2
=
7
6

25
36
=
17
36
.
(e)
p
Var
(
Y
) =
√
17
6
.
(f)
c <
E
[
Y
]
, so c can be
5
6
, you would still be willing play.
4.
Solution: (a)
(
50
20
)
(
100
40
)
.
(b)1

(
50
40
)(
2
1
)
40
(
100
40
)
.
(c)
(
50
k
)(
2
2
)
k
(
50

k
40

2
k
)(
2
1
)
40

2
k
(
100
40
)
.
5.
Solution: (a)
μ
=
E
[
X
] =
1
3
×
9
10
+
7
×
1
10
=
1.
(b)E
[
X
2
] =
1
9
×
9
10
+
49
×
1
10
=
5,
σ
2
=
E
[
X
2
]  (
E
[
X
])
2
=
5

1
=
4,
so
σ
=
2.
(c)Pr
{
X
>
μ
+
3
σ
}
=
Pr
{
X
>
1
+
3
×
2
}
=
Pr
{
X
>
7
}
=
1
10
.
(d)No.
(e)Pr
{
X

μ

>
3
σ
}
=
Pr
{
X

1

>
6
}
=
Pr
{
X
>
7
}
=
1
10
.
(f)Pr
{
X
>
7
}
6
E
[
X
]
7
=
1
7
.
(g)Pr
{
X

μ

>
3
σ
}
6
Var
(
X
)
(
3
σ
)
2
=
σ
2
9
σ
2
=
1
9
.
6.
Solution:
Suppose
∑
∞
k
=
1
1
k
2
=
1
c
, then let Pr
{
X
=
k
}
=
c
k
2
, for
k
=
1, 2, 3,
· · ·
.
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 Fall '08
 Zahrn
 Harshad number, k2, R. D. Foley

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