Third Homework Set — Solutions
Chapter 2
Problem 37
(a) There are
(
10
5
)
selections for the final exam. The number of selec
tions that allow the student to solve all problems is
(
7
5
)
, so that
the desired probability is
(
7
5
)
(
10
5
)
= 0
.
08333.
(b) There are
(
7
4
)
·
(
3
1
)
selections that’ll let the student solve exactly
four problems, so that the probability of solving at least four prob
lems is
(
7
5
)
+
(
7
4
)
·
(
3
1
)
(
10
5
)
=
1
2
.
Problem 43
(a) There are
n
! ways to arrange
n
people in a line. There are 2(
n

1)!
ways to arrange them such that
A
and
B
are next to each other.
Hence, the probability of
A
and
B
being next to each other is
2(
n

1)!
n
!
=
2
n
.
(b) If
n
= 2, then
A
and
B
will always be next to each other. Now,
assume that
n >
3. After
A
picks a seat, there are
n

1 seats left,
two of which are next to
A
, so that the desired probability is
2
n

1
.
Problem 50 The probability that you have five spades and your partner has the
remaining eight spades is
(
13
5
)(
39
8
)(
8
8
)(
31
5
)
(
52
13
,
13
,
26
)
= 2
.
6084
·
10

6
.
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 Probability, Trigraph, Prime number, EMC E4, E1 E2 E3

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