8
Solutions to HW1: MTH 540
2.5
a.
A
S
A
B
B
A
B
A
B
A
=
∩
=
∪
∩
=
∩
∪
∩
)
(
)
(
)
(
.
b.
A
A
B
B
B
A
B
B
A
B
=
∩
=
∩
∪
∩
=
∩
∪
)
(
)
(
)
(
)
(
.
c.
=
∩
∩
=
∩
∩
∩
)
(
)
(
)
(
B
B
A
B
A
B
A
0
/
.
The result follows from part a.
d.
)
(
)
(
B
B
A
B
A
B
∩
∩
=
∩
∩
=
0
/
.
The result follows from part b.
2.6
A
= {(1,2), (2,2), (3,2), (4,2), (5,2), (6,2), (1,4), (2,4), (3,4), (4,4),
(5,4), (6,4), (1,6), (2,6), (3,6), (4,6), (5,6), (6,6)}
C
= {(2,2), (2,4), (2,6), (4,2), (4,4), (4,6), (6,2), (6,4), (6,6)}
A
∩
B
= {(2,2), (4,2), (6,2), (2,4), (4,4), (6,4), (2,6), (4,6), (6,6)}
B
A
∩
= {(1,2), (3,2), (5,2), (1,4), (3,4), (5,4), (1,6), (3,6), (5,6)}
B
A
∪
= everything but {(1,2), (1,4), (1,6), (3,2), (3,4), (3,6), (5,2),
(5,4), (5,6)}
A
C
A
=
∩
2.11
a.
Since
P
(
S
) =
P
(E
1
) + … +
P
(E
5
) = 1, 1 = .15 + .15 +
.40 + 3
P
(E
5
).
So,
P
(E
5
) = .10 and
P
(E
4
) = .20.
b.