SOLUTIONS FOR HWK # 2
1.37
The sample space consists of all possible ordered sample of size 4 from 10
digits.
N
= 10
4
= 10
,
000.
(a) Each permutation of the
different
digits 6, 7, 8, 9 is favorable.
Therefore
N
(
A
) =
P
(4) = 4! = 24;
P
(
A
) =
24
10
,
000
= 0
.
0024.
(b)
We consider
distinguishable
permutations of three kinds of digits; here
n
6
=
1
, n
7
= 1
, n
8
= 2
, n
= 4. Therefore
N
(
A
) =
4!
1!1!2!
=
24
2
= 12;
P
(
A
) =
12
10
,
000
.
(c)
n
7
= 2
, n
8
= 2
, n
= 4;
N
(
A
) =
4!
2!2!
=
24
4
= 6;
P
(
A
) =
6
10
,
000
.
(d)
n
7
= 1
, n
8
= 3;
N
(
A
) =
4!
1!3!
=
24
6
= 4;
P
(
A
) =
4
10
,
000
.
1.311(c)
The series stops at the sixth game in the following two (mutually
exclusive) cases:
C
1
= ”
AL
wins 4 games in the first 5 games and wins the 6th game”;
C
2
= ”
NL
wins 4 games in the first 5 games and wins the 6th games.
Let us mark a game by ”AL” (resp. ”NL”) if it won by AL (resp. by NL). In case
C
1
the 6th game is chosen for ”AL” and from the previous 5 games 3 ones have to be
chosen for the ”AL” and 2 games for ”NL”. So in this case the number of choices is
(
5
3
)
=
(
5
2
)
=
5!
·
4
2!3!
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 '08
 staff
 Math, Playing card, US standard clothing size

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