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7.7 Probability, Possibility and Necessity
In the case where
F
is a Boolean sub-algebra of
P
(
X
)
, also probabilities
p
: F ₒ
[
0
,
1
]
can be taken into account. Once a dual-pair
(
N
, ˀ)
is given, it appears the
problem of which probabilities are
consistent
with
(
N
, ˀ)
.
A
C
A
C
The property
N
(
A
)
+
N
(
)
1, equivalent to
N
(
A
)
1
−
N
(
)
=
ˀ(
A
)
,
showsthatforall
A
∈ F
it is
N
(
A
)
ˀ(
A
),
provided the pair
(
N
, ˀ)
is dual. A probability
p
is said
consistent
with the dual pair
(
N
, ˀ)
if
N
(
A
)
p
(
A
)
ˀ(
A
)
,
for all
A
∈ F
.
In this hypothesis, if
N
(
A
)>
0, it is also
p
(
A
)>
0, and
ˀ(
A
)>
0, and if
ˀ(
A
)
=
0itis
N
(
A
)
=
p
(
A
)
=
0. That is,
•
If something is just a little bit necessary, it is probable and possible.
•
If something is not possible at all, it is neither necessary nor probable.
Analogously, if
p
(
A
)>
0itis
ˀ(
A
)>
0 although it could be
N
(
A
)
=
0.
That is,
•
If something is just a little bit probable, it is possible, but not necessarily necessary.
A
C
Notice that from
N
(
A
)
p
(
A
)
ˀ(
A
)
, or equivalently from 1
−
ˀ(
)
A
C
A
C
p
(
A
)
ˀ(
A
)
, follows 1
−
ˀ(
A
)
p
(
)
ˀ(
)
.
Example 7.7.1
Let is
X
={
1
,
2
,
3
}
, and
μ
=
0
.
7
|
1
+
1
|
2
+
0
.
5
|
3. Then, with
, and
N
(
A
)
ˀ
μ
(
)
=
(μ(
), μ
A
(
))
=
−
ˀ
μ
(
)
A
Sup
i
min
i
i
1
A
, we get:
μ
∈
X
ˀ
μ
(
)
=
μ(
)
=
.
ˀ
μ
(
)
=
μ(
)
=
ˀ
μ
(
)
=
μ(
)
=
.
1.
1
1
0
7,
2
2
1,
3
3
0
5.
ˀ
μ
(
{
1
,
2
}
)
=
max
(μ(
1
), μ(
2
))
=
1,
ˀ
μ
(
{
1
,
3
}
)
=
max
(
0
.
7
,
0
.
5
)
=
0
.
7,
ˀ
μ
(
{
2
,
3
}
)
=
1,
ˀ
μ
{
X
}=
1.
2.
N
μ
(
1
)
=
1
−
ˀ
μ
(
{
2
,
3
}
)
=
1
−
1
=
0,
N
μ
(
2
)
=
1
−
ˀ
μ
(
{
1
,
3
}
)
=
1
−
0
.
7
=
0
.
3,
N
μ
(
3
)
=
1
−
ˀ
μ
(
{
1
,
2
}
)
=
0
N
μ
(
{
1
,
2
}
)
=
1
−
ˀ
μ
(
3
)
=
0
.
5,
N
μ
(
{
1
,
3
}
)
=
1
−
ˀ
μ
(
2
)
=
0,
N
μ
(
{
2
,
3
}
)
=
1
−
ˀ
μ
(
1
)
=
0
.
3,
N
μ
(
X
)
=
1.
Hence, the consistent probabilities are given by triplets
p
(
1
)
,
p
(
2
)
,
p
(
3
)
in
[
0
,
1
]
such that
p
(
1
)
+
p
(
2
)
+
p
(
3
)
=
1, and verifying:
0
p
(
1
)
0
.
7, 0
.
3
p
(
2
)
1, 0
p
(
3
)
0
.
5
.
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