（
1
）
Solution:
.
(2) Solution:
(a)
m
a
x
(()
,()
)
(
)
PA PB
PA B
≤∪
; and any probability should be a number between 0 and 1.
0.7
(
)
1
≤
∪≤
(
) m
i
n
)
∩≤
; and the Bonferroni’s inequality gives the lower boundary of
()
∩
.
()1
(
)
+
−≤
∩
. So
0.1
(
)
0.4
≤
(b)The Boole’s inequality gives the upper boundary of
∪
.
(
)
PA B PA PB
+
,
so
0.6
(
)
0.8
≤
A1
A2
A3
A1
A2
A3
When A1, A2, A3 are disjoint from each other,
obviously,
(1
2
3
)
)
(2
)
(3
)
PA A
A
PA PA
PA
∪∪
=
+
+
When A1, A2 and A3 are not disjoint from each
other,
2
3
)
)
)
)
A
<
+
+
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View Full DocumentSimilar to part (a), but here the boundary given by Bonferroni’s inequality is trivial
(
0.2
(
)
PA B
−<
∩
)
0(
)
0
.
2
≤
∩≤
(c)The Boole’s inequality is
()
()1
(
)
PA PB
+−
≤
∩
.Only when
()1 0
>
,
the Boole’s inequality provides the nontrivial information. Otherwise, the lower boundary
of
∩
got from Boole’s inequality is just a nonpositive number which is already known
from the definition of probability.
(d)The Bonferroni’s inequality is
(
)
PA B PA PB
∪≤
+
. Only when
() 1
+<
,
the Bonferroni’s inequality provides the nontrivial information.
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 Spring '08
 GerardoHernandezdelValle
 Zagreb, Highways in Croatia, The Final, A1 A2 A3, Belgrade bypass

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