ENGRD 2700, Spring ’09
Homework 3 Solutions
Homework 3 Solutions
Problem 1
When two dice are rolled, there are 36 possible outcomes, which can be repre-
sented as ordered integer pairs (
a, b
) with 1
≤
a, b
≤
6.
(a)
We need to check if
P
(
A
∩
B
1
) =
P
(
A
)
P
(
B
1
). We have
A
=
{
(4
,
1)
,
(4
,
2)
,
(4
,
3)
,
(4
,
4)
,
(4
,
5)
,
(4
,
6)
}
,
B
1
=
{
(1
,
2)
,
(2
,
2)
,
(3
,
2)
,
(4
,
2)
,
(5
,
2)
,
(6
,
2)
}
,
A
∩
B
1
=
{
(4
,
2)
}
.
Since each of the 36 outcomes is equally likely to be realized, we have
P
(
A
) = 6
×
1
/
36 = 1
/
6
,
P
(
B
1
) = 6
×
1
/
36 = 1
/
6
,
P
(
A
∩
B
1
) = 1
/
36
.
Thus we see that
P
(
A
∩
B
1
) = 1
/
36 =
P
(
A
)
P
(
B
1
), which means that
A
and
B
1
are independent.
(b)
B
2
=
{
(1
,
2)
,
(2
,
1)
}
. Clearly,
A
∩
B
2
=
∅
, which means that
A
and
B
2
are
disjoint.
(c)
We have
P
(
A
) = 1
/
6
, P
(
B
2
) = 2
/
36 = 1
/
18
, P
(
A
∩
B
2
) =
P
(
∅
) = 0.
Therefore,
P
(
A
∩
B
2
) = 0
6
= 1
/
108 =
P
(
A
)
P
(
B
2
), so
A
and
B
2
are
not
inde-
pendent.
(d)
B
3
=
{
(3
,
6)
,
(6
,
3)
,
(4
,
5)
,
(5
,
4)
}
. So
A
∩
B
3
=
{
(4
,
5)
} 6
=
∅
, which means
that
A
and
B
3
are
not
disjoint.
(e)
We have
P
(
A
) = 1
/
6
, P
(
B
3
) = 4
/
36 = 1
/
9
, P
(
A
∩
B
3
) = 1
/
36. There-
fore,
P
(
A
∩
B
3
) = 1
/
36
6
= 1
/
54 =
P
(
A
)
P
(
B
3
), so
A
and
B
3
are
not
independent.
(f)
B
4
=
{
(1
,
6)
,
(6
,
1)
,
(2
,
5)
,
(5
,
2)
,
(3
,
4)
,
(4
,
3)
}
.
So
A
∩
B
4
=
{
(4
,
3)
} 6
=
∅
,
which means that
A
and
B
4
are
not
disjoint.
(g)
We have
P
(
A
) = 1
/
6
, P
(
B
4
) = 6
/
36 = 1
/
6
, P
(
A
∩
B
4
) = 1
/
36. Therefore,
P
(
A
∩
B
4
) = 1
/
36 =
P
(
A
)
P
(
B
4
), so
A
and
B
4
are independent.
(h)
If two events are not disjoint, then they may or may not be independent.
If two events are not independent, then they may or may not be disjoint. In
short, independence and disjointness are different things.
Problem 2
When a fair coin is flipped three times, there are 2
3
= 8 possible outcomes,
1