NTHU MATH 2810
Midterm Examination Solution
Oct 30, 2007
1.
(a) (
3pts
) Let
n
i
be the number of candies the
i
th child got, where
i
= 1
,
2
,
3
,
4
,
then
it must be satisfied that
n
i
≥
1
, i
= 1
,
2
,
3
,
4
,
and
n
1
+
n
2
+
n
3
+
n
4
= 10
.
(1)
The number of integer solutions for Eqn. (1) is
‡
10

1
4

1
·
= 84
.
(b) (
3pts
) For each of the 10 different books, there are four distinct (and independent)
choices of child it can be given to. So, the answer is 4
×
4
× · · · ×
4 = 4
10
.
2. (
8pts
) The three equations can be represented in terms of
p
1
, p
2
, p
3
,
and
p
4
as follows:
P
(
A
∪
B
) = 3
P
(
B
)
⇒
p
1
+
p
2
+
p
3
= 3(
p
1
+
p
3
)
,
(2)
P
(
A
∩
B
) = 0
.
4
P
(
A
∩
B
c
)
⇒
p
1
= 0
.
4
p
2
,
(3)
P
((
A
∪
B
)
c
) = 0
.
1
⇒
p
4
= 0
.
1
.
(4)
Furthermore, because (
A
∩
B
)
∪
(
A
∩
B
c
)
∪
(
A
c
∩
B
)
∪
(
A
c
∩
B
c
) = Ω (the whole sample space)
,
p
1
+
p
2
+
p
3
+
p
4
= 1
.
(5)
By solving Eqns. (2), (3), (4), and (5), we get
p
1
= 0
.
24
,
p
2
= 0
.
6
,
p
3
= 0
.
06
,
and
p
4
= 0
.
1
.
Therefore,
P
(
A
) =
p
1
+
p
2
= 0
.
84
.
3.
(a) (
4pts
) Let
E
be the event of picking a black ball the first draw. Then,
P
(
E

U
1
) =
3
5
and
P
(
E

U
2
) =
2
5
.
So, by the law of total probability,
P
(
E
)
=
P
(
E

U
1
)
·
P
(
U
1
) +
P
(
E

U
2
)
·
P
(
U
2
)
=
3
5
·
1
2
+
2
5
·
1
2
=
1
2
.
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 Fall '11
 ShaoWeiCheng
 Math, Probability, Probability theory, y1, red die

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