BCPST1B Exemples préliminaires : I) Probabilité conditionnelle

1B
(Ω,P(Ω), P )A
PA:P(Ω) →RB7→ P(A∩B)
P(A)(Ω,P(Ω))
PAA A
•B A ∩B\A P (A∩B)6P(A)
06P(A∩B)
P(A)61
PA:P(Ω) →[0,1]
•PA(Ω) = P(A∩Ω)
P(A)=P(A) P(A)
PA(Ω) = 1
•B1B2B1∩B2=∅
PA(B1∪B2) = · · · =PA(B1) + PA(B2)
•P PA
PA(B∪C) = PA(B) + PA(C)−PA(B∩C)
•PA(A)=1
•PA(B) = PB /A
•PB /A B /A
•P(A∩B) = P(A)×PA(B)

Ω
@@@
@
A
A
HHH
H
B
B
HHH
H
B
B
P(A)
PA(B)
•P PA(B) = card(A∩B)
card(A)
Ω = {1,2,3,4,5,6}
A A ={2,4}
B B ={1,3,5}
PAPB
ω
P({ω})1
6
1
6
1
6
1
6
1
6
1
6
PA({ω})
PB({ω})
9 6
PΩ 15
A1
A2
A3
card (Ω) =................. card (A1) =................. card (A1∩A2) =.................
card (A3) =................. card (A1∩A3) =................. card (A1∩A2∩A3) =.................
P(A1) =................. P(A1∩A2) =.................
PA1∩A2(A3) =................. P(A1∩A2∩A3) =.................
PA1(A2) =................. PA1∩A3(A2) =.................
PA1(A2) =................. PA1(A2) =.................

PA1∩A2(A3) =................. PA1∩A2(A3) =.................
P(A2) =................. P(A2∩A3) =.................
PA1∩A3(A2) =................. P(A3) =.................
PA2(A1) =................. PA1∩A3(A2) =.................
P(A∩B∩C) = P(A)PA(B∩C) = P(A)PA(B) (PA)B(C)
P(A)6= 0 PA(B)6= 0
(PA)B(C) =.......................................................
PA∩B(C) =.......................................................
(PA)B(C) = PA∩B(C)
A B P (A∩B)6= 0
P(A∩B∩C) = P(A)PA(B)PA∩B(C)
A∩B⊂A P (A∩B)6= 0 P(A)6= 0
A1, . . . , Akk k >2P(A1∩. . . ∩Ak−1)>0
P(A1∩. . . ∩Ak) = P(A1)×PA1(A2)× · · · × PA1∩...∩Ak−1(Ak)

(Ω,P(Ω), P ){A1, . . . , Ak}k k >1
{A1, . . . , Ak}P
{A1, . . . , Ak} ∀i∈[[1, k]], P (Ai)6= 0
(Ω,P(Ω), P ) (A1, . . . , Ak)k
À(A1, . . . , Ak)B
P(B) = P(A1∩B) + P(A2∩B) + · · · +P(Ak∩B)
P(B) =
k
X
i=1
P(Ai∩B)
Á(A1, . . . , Ak)P B
P(B) = P(A1)×PA1(B) + P(A2)×PA2(B) + · · · +P(Ak)×PAk(B)
P(B) =
k
X
i=1
P(Ai)×PAi(B)
B= (A1∩B)∪ · · · ∪ (Ak∩B) (Ai∩B)
P(B) = P(A1∩B) + · · · +P(Ak∩B)

P(A)6∈ {0,1}(A, A)
P
P(B) = P(A)PA(B) + P(A)PA(B)
Ê
Ën∈N∗n
U1
U2
nUn
N
Ukk k(k−1)
k∈[[1, n]] Ukk
k∈[[1, n]] Uk
k∈[[1, n]]
k
U1U2nUn
N=1+2+· · · +n
N=n(n+ 1)
2
Ukk k(k−1)
k∈[[1, n]] Ukk
N k Uk
P(Uk) = k
NP(Uk) = 2k n(n+ 1)