4.
Let X denotes the highest ranking achieved by a woman. Then
p
(
X
= 1) =
(
5
1
)
9!
10!
= 1
/
2,
p
(
X
= 2) =
(
5
1
)(
5
1
)
8!
10!
= 5
/
18,
p
(
X
= 3) =
(
5
2
)(
2
1
)(
5
1
)
7!
10!
= 5
/
34,
p
(
X
= 4) =
(
5
3
)(
5
1
)
3!6!
10!
= 5
/
84,
p
(
X
= 5) =
(
5
4
)(
5
1
)
4!5!
10!
= 5
/
252,
p
(
X
= 6) =
5!5!
10!
= 1
/
252.
For
i
= 7
,
8
,
9
,
10,
p
(
X
=
i
) = 0
.
7.
a) Let X denotes the max value to appear in the two rolls, then
S
x
=
{
1
,
2
,
3
,
4
,
5
,
6
}
,
b) Let Y denotes the min value to appear in the two rolls,, then
S
y
=
{
1
,
2
,
3
,
4
,
5
,
6
}
,
c) Let Z denotes the sum of the two rolls, then
S
z
=
{
2
,
3
,
4
,
5
,
6
,
7
,
8
,
9
,
10
,
11
,
12
}
.
d) Let A denotes the value of the first roll minus the value of the 2nd , then
S
A
=
{
5
,

4
,

3
,

2
,

1
,
0
,
1
,
2
,
3
,
4
,
5
}
.
8.
Assume the die is fair,
S
x
=
{
1
,
2
,
3
,
4
,
5
,
6
}
, then
p
(
X
= 1) = 1
/
36
, p
(
X
=
2) = 3
/
36
, p
(
X
= 3) = 5
/
36
, p
(
X
= 4) = 7
/
36
, p
(
X
= 5) = 9
/
36
, p
(
X
=
6) = 11
/
36
.
13.
Let
Y
1
∼
B
(1
,
0
.
3),
Y
2
∼
B
(1
,
0
.
6),
X
i
= 1000 with conditional probability
1/2,
X
i
= 500 with conditional probability 1/2. Let A1 denotes X1=1, X2=1;
A2 denotes X1=1,X2=0; A3 denotes X1=0,X2=1; A4 denotes X1=0,X2=0.
Then
p
(
X
= 0) =
p
(
X
= 0

A
4)
p
(
A
4) = (1

0
.
3)(1

0
.
6) = 0
.
28
,
p
(
X
= 500) =
p
(
X
= 500

A
2)
p
(
A
2) +
p
(
X
= 500

A
3)
p
(
A
3) = 1
/
2[0
.
3
*
(1

0
.
6) + (1

0
.
3)0
.
6] = 0
.
27
.