Probabilité conditionnelle - Institut de Mathématiques de Bordeaux

F P (F) = 31.1
74.7'0.42
F∩S
P(F∩S) = 13.8
74.7'0.18
p=13.8
31.1'0.44
p=P(F∩S)
P(F)F
F
(Ω,P(Ω), P )
A P (A)>0B B
A B A PA(B)
PA(B) = P(B∩A)
P(A).
PA(A)=1
PA(Ω,P(Ω))

PAP(Ω) R+
PA(Ω) = P(Ω∩A)
P(A)=P(A) P(A)= 1 B C B∩A C∩A
PA(B∪C) = P((B∪C)∩A)
P(A)=P((B∩A)∪(C∩A)) P(A)=P(B∩A) P(A)+P(C∩A) P(A)=PA(B) + PA(C).
A B P (A∩B)>0 (PA)B=P(A∩B)
PAP(A)≥P(A∩B)>0PA(B)>0
E(PA)B(E) = PA(B∩E)
PA(B)=P(A∩(B∩E)) P(A)
P(A)
P(A∩B)=P(A∩B)(E)
A B (Ω,P(Ω), P )P(A)>0B
A PA(B) = P(B)
A PA(B) = P(B)
P(B∩A) = P(B)P(A)
(Ω,P(Ω), P )
A B P (A∩B) = P(A)P(B)
N A1ANk
N1≤i1<· · · < ik≤N P (Ai1∩. . . ∩Aik) = P(Ai1)· · · P(Aik)
A
B P (A∩B)=06=P(A)P(B)
N k = 2
A
B C
A B C
A B A B
P(A∩B) = P(A)−P(A∩B) = P(A)−P(A)P(B) = P(A)(1 −P(B)) = P(A)P(B)
(Ω,P(Ω), P )A B P (A)>0P(A∩B) =
P(A)PA(B)
A
(Ω,P(Ω), P ) (Ai)1≤i≤n
PT1≤i≤n−1Ai>0
P
\
1≤i≤n
Ai
=P(A1)PA1(A2)P(A1∩A2)(A3)· · · P(A1∩...∩An−1)(An).

k n −1PT1≤i≤kAi≥PT1≤i≤n−1Ai>0
n(Hn) (Ai)1≤i≤n
PT1≤i≤n−1Ai>0PT1≤i≤nAi=P(A1)PA1(A2)P(A1∩A2)(A3)· · · P(A1∩...∩An−1)(An)
(H2) (Hn) (Hn+1)n≥2
PT1≤i≤nAi>0
P
\
1≤i≤n+1
Ai
=P(A1∩. . . ∩An)P(A1∩...∩An)(An+1).
(Hn)P(A1∩. . . ∩An−1)≥P(A1∩. . . ∩An)>0 (Hn+1)
(H2) (Hn) (Hn+1)n≥2 (Hn)n≥2
(Ai)1≤i≤nS1≤i≤nAi= Ω
(Ω,P(Ω), P ) (Ai)1≤i≤n
A
P(A) =
n
X
i=1
P(Ai)PAi(A).
P(A) = P(A∩(S1≤i≤nAi)) = Pn
i=1 P(A∩Ai)
@@@@@
@
A
B
C
P(A)
P(B)
P(C)
PPPPP
P
A∩E
A∩E
PA(E)
PA(E)
PPPPP
P
B∩E
B∩E
PB(E)
PB(E)
PPPPP
P
C∩E
C∩E
PC(E)
PC(E)
P(A∩E) = P(A)PA(E)
P(E) = P(A)PA(E) + P(B)PB(E) + P(C)PC(E)
PA(E) + PA(E)=1

(H1)
(H2)
(H3)
k
PA(B)A B
A
B
(Ω,P(Ω), P )A1A2
PA1(A2) = PA2(A1)P(A2)
P(A1).
P(A1)PA1(A2) = P(A1∩A2) = P(A2)PA2(A1)
(Ω,P(Ω), P ) (Ai)1≤i≤n
i P (Ai)>0A i
PA(Ai) = P(Ai)PAi(A)
n
P
j=1
P(Aj)PAj(A)
.
PA(Ai) = P(Ai)PAi(A)
P(A)
M=
P0=PM(P0) = 10−3PM(P0)=2×10−3
U1U2U1U2
UiUiN B
P(U1) = P(U2)=1/2
PU1(N) = 7
7+3 PU2(N) = 9 9+11
P(N) = 1
2(7 10 +9 20 ) = 0.575
PUi
(Ω,P(Ω), P )PUiUi
Ωi1≤i≤n
Ω1p1
2≤k≤n(ω1, . . . , ωk−1)∈Ω1× · · · × Ωk−1Ωkp(ω1,...,ωk−1)
k
(Ω,P(Ω), P )
P({ω1} × Ω2× · · · Ωn) = p1({ω1})
P({ω1}×···×{ωk−1}×Ωk×···×Ωn)({ω1}×···×{ωk} × Ωk+1 × · · · × Ωn) = p(ω1,...,ωk−1)
k({ωk})
Ω=Ω1×· · ·×ΩnP P ({ω1}×· · ·×{ωn}) = p1({ω1})p(ω1)
2({ω2})· · · p(ω1,...,ωn−1)
n({ωn})

k p(ω1,...,ωk−1)
k(ω1, . . . , ωk−1)k
PΩ1× · · · × Ωk−1× {ωk} × Ωk+1 · · · × Ωn
Ω1× · · · × Ωj−1× {ωj} × Ωj+1 · · · × Ωnk6=j
PΩ
P(Ω) P(E) = P
ω∈E
P({ω})P
ω∈Ω
P({ω})=1
k p(ω1,...,ωk−1)
k
P
ω∈Ω
P({ω}) = P
ω1∈Ω1
p1({ω1}) P
ω2∈Ω2
p(ω1)
2({ω2}). . . P
ωn∈Ωn
p(ω1,...,ωn−1)
n({ωn})!. . .!= 1
(1) 
Ω1={1,2}p1({1}) = p1({2}) = 1/2
Ω2={n, b}p(1)
2({n}) = 7
10 p(2)
2({n}) = 9
20 Ω = {(1, n),(2, n),(1, b),(2, b)}P({(1, n)}) =
1
2
7
10 Ui={(i, n),(i, b)}i= 1 2 N={(1, n),(2, n)}P(Ui) = 1/2
PU1(N) = 7
10
a k
p2a
P(n) 2a n
n
n+k
P(n) = pP (n+k) + (1 −p)P(n−k).
a k ul=P(lk)
ul+2 −1
pul+1 +1−p pul= 0
u0=P(0) = 0 u2a
k=P(2a) = 1
x2−1
px+1−p p= 0
1−p
p
p6= 1/2p1/2ul=
α+β1−p
plα β u0= 0 u2a
k= 1
P(a) = ua/k =1−(1−p
p)a/k
1−(1−p
p)2a/k =1
1+(1−p
p)a/k ,
k=a p
p= 1/2x2−1
px+1−p p= 0
ul=α+lβ ul=lk
2a
P(a) = ua/k = 1/2k