Construction d`une probabilité

Ω = {1,2, . . . , n}
{k}k
Ω = {1,2, . . . , n}
{1,2, . . . , k}k2
A, B Ω
A∩B6=∅, A ∩¯ B6=∅,¯ A∩B6=∅¯ A∩¯ B6=∅
(a, b, c, d)∈]0 ; 1[4PΩ
P(A|B) = a, P(A|¯ B) = b, P(B|A) = cP(B|¯ A) = d?
A B
max {0,P(A) + P(B)−1} ≤ P(A∩B)≤min {P(A),P(B)}

α∈RP ({k}) = αk
n
X
k=1
P ({k}) = P(Ω) = 1
α=2
n(n+ 1)
P α ∈R
P ({1,2, . . . , k}) = αk2
P(Ω) = 1 α= 1/n2
P ({k}) = P ({1, . . . , k})−P ({1, . . . , k −1}) = 2k−1
n2
P ({k}) = 2k−1
n2
P ({1,2, . . . , k}) =
k
X
i=1
2i−1
n2=k2 n2
P
x= P(A∩B), y = P(A∩¯ B), z = P( ¯ A∩B)t= P( ¯ A∩¯ B)
x, y, z, t ≥0
x+y+z+t= P(A) + P( ¯ A) = 1
x, y, z, t
PΩ
A∩B A ∩¯ B¯ A∩B
¯
A∩¯ B x, y, z
t
x, y, z, t ≥0
P(A|B) = a, P(A|¯ B) = b, P(B|A) = cP(B|¯ A) = d
P(A) = x+yP(B) = x+z
P(A|B) = a x =a(x+z)
y=b(1 −(x+z)), x =c(x+y)z=d(1 −(x+y))
(1 −a)x−az = 0
bx +y+bz =b
(1 −c)x−cy = 0
dx +dy +z=d
x=abc
a(1 −c) + bc, y =ab(1 −c) a(1 −c) + bc z=(1 −a)bc a(1 −c) + bc
ad(1 −b)(1 −c) = bc(1 −a)(1 −d)
(x, y, z)x, y, z ≥0x+y+z≤1
t≥0x+y+z+t= 1
ad(1 −b)(1 −c) = bc(1 −a)(1 −d)
abcd
1−1
b1−1 c=1−1 a1−1 d

A∩B⊂AP(A∩B)≤P(A) P(A∩B)≤P(B)
P(A∩B)≤min {P(A),P(B)}
P(A∩B)≥0 P(A∪B)≤1
P(A∪B) = P(A) + P(B)−P(A∩B)
P(A∩B)≥P(A) + P(B)−1
max {0,P(A) + P(B)−1} ≤ P(A∩B)