Correction - PC* Carnot Dijon

2014 −2015
Ω,T, P ) (A, B)∈ T 2d(A, B) = P(A∆B)
∀(A, B, C)∈ T 3, d(A, C)≤d(A, B) + d(B, C)
∀(A, B)∈ T 2,|P(A)−P(B)| ≤ d(A, B)
(X, Y )∈ T 2, X∆Y= (X∩Y)∪(X∩Y)X∩Y X ∩Y
d(X, Y ) = P(X∆Y) = P(X∩Y) + P(X∩Y)
d(A, C) = P(A∩C) + P(A∩C) = P(A∩C∩B) + P(A∩C∩B) + P(A∩
C∩B) + P(A∩C∩B)
P(A∩C∩B)≤P(C∩B)P(A∩C∩B)≤P(A∩B)
P(A∩C∩B)≤P(A∩B)P(A∩C∩B)≤P(C∩B)
d(A, C)≤P(C∩B) + P(A∩B) + P(A∩B) + P(C∩B)
P(C∩B) + P(A∩B) + P(A∩B) + P(C∩B) = P[(A∩B)∪(A∩B)] +
P[(C∩B)∪(C∩B)]
=P(A∆B) + P(B∆C) = d(A, B) + d(B, C)
d(A, C)≤d(A, B) + d(B, C)
C=∅d(A, ∅)≤d(A, B) + d(∅, B)
∀X∈ T , d(X, ∅) = P(∅∆X) = P(X)
P(A)≤d(A, B) + P(B)
A B P (B)≤d(A, B) + P(A)
|P(A)−P(B)| ≤ d(A, B)

2014 −2015
k∈[[1,52]] Akk
P(Ak)
52
k=1
P(Ak)
k∈[[1,52]] Bkk
P(Bk)
52
k=1
P(Bk)
Ω 52! C
Ak
C\ { }
(Ak) = 51! P(Ak) = (Ak)
(Ω) =1 52
k
52
k=1
P(Ak) = 1
Ak, k ∈[[1,52]]
B1=∅B2=∅
k≥3
k
k−1
k−1k−2
k−2k−1 49!
(Bk) = (k−1) ×(k−2) ×49!
P(Bk) = (Bk)
(Ω) =(k−1)(k−2) 52 ×51 ×50
52
k=1
P(Bk) =
52
k=3
P(Bk) =
52
k=3
(k−1)(k−2)
52 ×51 ×50 =
50
p=1
p(p+ 1)
52 ×51 ×50 =
50
p=1
p2+
50
p=1
p
52 ×51 ×50

2014 −2015
=1
52 ×51 ×5050 ×51 ×101 6+50 ×51 2=1 52101 6+1 2=1 3
Bk, k ∈[[1,52]]
D
3! = 6
P(D) = 1
3
2
a, b c Q(X) = aX2+bX +c
Q Q
Q
Ω = [[1,6]]3
(Ω) = 63= 216 A∈ P(Ω)
A(A)
216
S Q {(a, b, c)∈Ω/ b2−4ac > 0}
D Q {(a, b, c)∈Ω/ b2−4ac = 0}
C Q {(a, b, c)∈
Ω/b2−4ac < 0}S, D, C
4ac 6b
s(b), d(b), γ(b)
(a, c)b2>4ac b2= 4ac b2<4ac
a↓/c →1 2 3 4 5 6
1 4 8 12 16 20 24
2 8 16 24 32 40 48
3 12 24 36 48 60 72
4 16 32 48 64 80 96
5 20 40 60 80 100 120
6 24 48 72 96 120 144
γ(b) = 36 −s(b)−d(b)

2014 −2015
b b2s(b)d(b)γ(b)
1 1 0 0 36
2 4 0 1 35
3 9 3 0 33
4 16 5 3 28
5 25 14 0 22
6 36 16 1 19
(S) = 0 + 0 + 3 + 5 + 14 + 16 = 38 (D) = 0 + 1 + 0 + 3 + 1 = 5
(C) = 173
P(S) = 38
216 =19 108 P(D) = 5 216 P(C) = 173 216
P(C)
γ(b) (S, D, C)
P(C) = 1 −P(S)−P(D)
0,1,2, . . . , n k
E E
k a
E k E
k+ 1 b
0E
pnE n
k∈[[0, n]] AkE k
p0=P(A0) = 1
2
k P (Ak+1/Ak) = a P (Ak+1/Ak) = b
P(Ak+1/Ak) = 1 −b
P(Ak+1) = P(Ak+1/Ak)P(Ak) +
P(Ak+1/Ak)P(Ak)
P(Ak+1) = aP (Ak)+(1−b)(1−P(Ak)) pk+1 = (a+b−1)pk+1−b
x= (a+b−1)x+ 1 −b⇔x=1−b
2−a−b
pk+1 −1−b
2−a−b= (a+b−1)pk−1−b 2−a−b
k∈[[0, n −1]]
∀k∈[[0, n]], pk=1−b
2−a−b+ (a+b−1)kp0−1−b 2−a−b=1−b 2−a−b+
(a+b−1)kb−a
2(2 −a−b)
pn=1−b
2−a−b+ (a+b−1)nb−a 2(2 −a−b)

2014 −2015
b r n
n
k∈[[1, n]]
Bkk
Vkk
Vk=Bk∩n ∩
i=1,i̸=kBi=B1∩ ··· ∩ Bk−1∩Bk∩Bk+1 ∩ ··· ∩ Bn
A=n ∩
i=1 Vi
Vi
A
P(A) =
n
k=1
P(Vk)
k∈[[1, n]] P(Vk) = P(k−1 ∩
i=1 Bi)∩Bk∩(n
∩
j=k+1 Bj
Uk= (k−1 ∩
i=1 Bi)∩Bk
P(Vk) = P(B1)P(B2/B1)···P(Bk−1/k−2 ∩
j=1 Bj)P(Bk/k−1
∩
i=1 Bi)P(Bk+1/Uk)···P(Bn/Uk∩
n−1
∩
j=k+1 Bj)
k−1
b k
P(B1) = P(B2/B1) = ··· =P(Bk−1/k−2 ∩
j=1 Bj) = r
b+r
P(Bk/k−1 ∩
i=1 Bi) = b
b+r
k b −1
r
P(Bk+1/Uk) = P(Bk+1/Uk∩Bk) = ··· =P(Bn/Uk∩n−1 ∩
j=k+1 Bj) =
r
b+r−1
P(Vk) = r
b+rk−1b b+rr b+r−1n−k
=rn−1b
(b+r)k(b+r−1)n−k