Math 105, Fall 2004 – Solutions to Prelim 2
1.
(a)
Henri has to take 3 bottles from 5+7=12, and the order in which they are taken doesn’t
matter. Therefore, there are
12
3
=
12!
9! 3!
=
12
·
11
·
10
·
9!
9! 3!
=
12
·
11
·
10
3!
=
1320
6
= 220
possible choices.
(b)
Henri has to choose 1 bottle from the 5 bottles of red and 2 bottles from the 7 bottles
of white. By the multiplication principle, there are
5
1
·
7
2
= 5
·
21 = 105
ways of doing this.
2.
Note that
S
=
{
0
,
1
,
2
, . . . ,
98
,
99
}
is the sample space for this problem.
(a)
E
=
{
0
,
1
,
2
,
3
,
4
,
5
,
6
,
7
,
8
,
9
,
10
,
11
,
12
,
13
,
14
,
15
,
16
,
17
,
18
,
19
}
.
F
=
{
9
,
18
,
27
,
36
,
45
,
54
,
63
,
72
,
81
,
90
}
.
Assuming equally likely outcomes, we have:
P
(
E
) =
n
(
E
)
n
(
S
)
=
20
100
=
1
5
and
P
(
F
) =
n
(
F
)
n
(
S
)
=
10
100
=
1
10
.
(b)
Assuming equally likely outcomes,
P
(
E
∩
F
) =
n
(
E
∩
F
)
n
(
s
)
=
n
(
{
9
,
18
}
)
100
=
2
100
=
1
50
.
So by definition of conditional probability, we have:
P
(
E

F
) =
P
(
E
∩
F
)
P
(
F
)
=
1
/
50
1
/
10
=
1
5
.
Hence
P
(
E
) =
1
5
=
P
(
E

F
), and so
E
and
F
are independent events.