Indépendance Probabilités Conditionnelles

•A B
P r(A∩B) = P r(A)P r(B).
•Ai, i = 1, ..., n
J⊂ {1, ..., n}
P r 
\
i∈J
Ai
=Y
i∈J
P r(Ai).
Ai, i = 1, ..., n
Ai
Ω = {1,2,3,4,5,6,7,8}
1
8
A1={1,2,3,4}
A2={1,2,5,6}
A3={3,4,5,6}
P r(A1∩A2) = P r({1,2}) = 1
4=P r(A1)P r(A2)
P r(A1∩A3) = P r({3,4}) = 1
4=P r(A1)P r(A3)
P r(A2∩A3) = P r({5,6}) = 1
4=P r(A2)P r(A3)
P r(A1∩A2∩A3) = P r(∅) = 06=P r(A1)P r(A2)P r(A3) = 1
8.
B A
B A A
B
A
B

A
Ω′=A
3A
B2
B A 2
3P r(B) = 1 2
(Ω, P r)A
P(Ω)
P r(./A) [0,1]
P r(B/A) = P r(A∩B)
P r(A),∀B∈ P(Ω).
P r(B/A)B A
B A
P r(A∩B) = P r(B/A)P r(A).
P r(A)6= 0 A B
P r(B/A) = P r(B)
(A, P r(./A))
A1, ..., AnΩ
P r(B) =
n
X
i=1
P r(Ai)P r(B/Ai),∀B∈ P(Ω).
•
8
•
10
•P r(/S = 8) = 1/36
5/36 =1
5.
•P r(/S ≥10) = 2/36
6/36 =1
3.
B A1, A2, ..., An
P r(Ai)
P r(B/Ai)
B
P r(Ai/B)
A1, ..., AnΩ
P r(Ai)6= 0,∀i= 1, ..., n B
P r(Ai/B) = P r(Ai)P r(B/Ai)
Pn
j=1 P r(Aj)P r(B/Aj), i = 1,2, ..., n

A B
P r(A/B) = P r(A)P r(B/A)
P r(B).
•P r(A)A
•P r(A/B)A
F M D
C
P r(F/C) = P r(F)P r(C/F)
P r(C/F)P r(F) + P r(C/M)P r(M) + P r(C/D)P r(D),
1
3.3 4
1
3.3 4+1 3.2 5+1 3.1 5
=5
9.
X Y
Ω (X, Y )
Ω (X, Y )
X Y
A⊂X(Ω)
B⊂Y(Ω)
P r1(X∈A) = P r(X∈A, Y ∈R)
P r2(Y∈B) = P r(X∈R, Y ∈B).

X Y A ⊂X(Ω)
B⊂Y(Ω)
P r(X∈A, Y ∈B) = P r1(X∈A)P r2(Y∈B).
P r1P r2
(X, Y )P r
P r({F, F }) = P r({P, P }) =1
2P r({F, P }) = P r({P, F }) =0.
P r({F, F }) = P r({P, F }) = P r({F, P }) = P r({P, P }) = 1
4.
P r({P}) = P r({F}) = 1
2.
X A
X A
E(X/A) = EP r(./A)X
E(X/A) = 1
P r(A)X
ω∈A
X(ω)P r({ω}).
X A
{1,2,3}
A1
2
E(X/A) = 1
1
2
[1
6+2 6+3 6] = 2.