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THE UNIVERSITY OF HONG KONG
DEPARTMENT OF STATISTICS AND ACTUARIAL SCIENCE
STAT1301 Probability and Statistics I
Assignment 4 Solution
1. (a)
(
)
36
1
2
,
=
=
=
i
Y
i
X
P
for
6
,
5
,
4
,
3
,
2
,
1
=
i
(
)
18
1
,
=
+
=
=
j
i
Y
i
X
P
for
1
,...,
2
,
1
−
=
i
j
;
6
,
5
,
4
,
3
,
2
,
1
=
i
(b)
(
)
36
,
i
i
Y
i
X
P
=
=
=
for
6
,
5
,
4
,
3
,
2
,
1
=
i
(
)
36
1
,
=
=
=
j
Y
i
X
P
for
6
,...,
2
,
1
+
+
=
i
i
j
;
6
,
5
,
4
,
3
,
2
,
1
=
i
(c)
(
)
36
1
,
=
=
=
i
Y
i
X
P
for
6
,
5
,
4
,
3
,
2
,
1
=
i
(
)
18
1
,
=
=
=
j
Y
i
X
P
for
6
,...,
2
,
1
+
+
=
i
i
j
;
6
,
5
,
4
,
3
,
2
,
1
=
i
2.
,
(
)
(
2
1
1
,
2
2
2
1
1
x
x
p
p
x
X
x
X
P
+
−
=
=
=
)
,...
2
,
1
,
0
1
=
x
,
,...
2
,
1
,
0
2
=
x
3.
(
)
10
1
,
2
1
=
=
=
j
N
i
N
P
,
i
j
−
=
5
,...,
1
,
4
,
3
,
2
,
1
=
i
.
4. (a)
(
)
(
)
(
)
35
.
0
25
.
0
1
.
0
1
,
1
0
,
0
=
+
=
+
=
=
p
p
Y
X
P
(
)
(
)
2
.
0
0
,
1
=
=
>
p
Y
X
P
(b)
(
)
(
)
(
)
(
)
45
.
0
05
.
0
3
.
0
1
.
0
2
,
0
1
,
0
0
,
0
0
=
+
+
=
+
+
=
=
p
p
p
X
P
(
)
(
)
(
)
(
)
55
.
0
1
.
0
25
.
0
2
.
0
2
,
1
1
,
1
0
,
1
1
=
+
+
=
+
+
=
=
p
p
p
X
P
Marginal pmf of
X
:
( )
⎩
⎨
⎧
=
=
=
1
55
.
0
0
45
.
0
x
x
x
p
X
(
)
(
)
(
)
3
.
0
2
.
0
1
.
0
0
,
1
0
,
0
0
=
+
=
+
=
=
p
p
Y
P
(
)
(
)
(
)
55
.
0
25
.
0
3
.
0
1
,
1
1
,
0
1
=
+
=
+
=
=
p
p
Y
P
(
)
(
)
(
)
15
.
0
1
.
0
05
.
0
2
,
1
2
,
0
2
=
+
=
+
=
=
p
p
Y
P
Marginal pmf of
Y
:
( )
⎪
⎩
⎪
⎨
⎧
=
=
=
=
2
15
.
0
1
55
.
0
0
3
.
0
y
y
y
y
p
Y
(c) Possible values of
Y
X
+
: 0, 1, 2, 3
(
)
(
)
1
.
0
0
,
0
0
=
=
=
+
p
Y
X
P
(
)
(
)
(
)
5
.
0
2
.
0
3
.
0
0
,
1
1
,
0
1
=
+
=
+
=
=
+
p
p
Y
X
P
(
)
(
)
(
)
3
.
0
25
.
0
05
.
0
1
,
1
2
,
0
2
=
+
=
+
=
=
+
p
p
Y
X
P
(
)
(
)
1
.
0
2
,
1
3
=
=
=
+
p
Y
X
P
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