Conditionnement et indépendance I Introduction aux probabilités

Ω
Ω
Ω
{ω}Ω
Ω
∅
A A A Ω
A B A ∩B
A B A ∪B
A B A ∩B=∅
A B A ⊂B
Ω = {ω1, ω2, . . . , ωn}n
PΩ
• {ωi}pi=P({ωi})
n
X
i=1
pi= 1
•A P (A){ωi}
A
Ω Ω
Ω [0,1]
•P(Ω) = 1
•A B P (A∪B) = P(A) + P(B)
A B
P(A) = 1 −P(A)P(∅) = 0
A⊂B P (A)≤P(B)
P(A∪B) = P(A) + P(B)−P(A∩B)
A1, A2, . . . , Ak
P(A1∪A2∪. . . ∪Ak) = .........

Ω = {ω1, ω2, . . . , ωn}
P({ω1}) = P({ω2}) = . . . =P({ωn}) = 1
n
A
P(A) = Card(A)
Card(Ω) =nombre d0issues favorables nombre total d0issues possibles
A
ΩP X
ΩRX
RΩX
X(Ω) = {x1, x2, . . . , xn}
(X=xi)X xi
XΩ
X X(Ω) = {x1, x2, . . . , xn}n
pi=P(X=xi) 1 ≤i≤n
valeurs xide X x1x2x3. . . xn
pi=P(X=xi)p1p2p3. . . pn
n
X
i=1
pi= 1
XΩx1, x2, . . . , xn
p1, p2, . . . , pn
•X E(X)E(X) =
n
X
i=1
pixi
•X V (X)V(X) =
n
X
i=1
pi(xi−E(X))2
•X σ(X)σ(X) = pV(X)

XΩa b
E(aX +b) = aE(X) + b
V(X) = E(X−E(X))2
V(aX +b) = a2V(X)
B A
A B PB(A)
PB(A) = P(A∩B)
P(B)
PB(A)A B
B PBΩ
A0≤PB(A)≤1PB(A) = 1 −PB(A)
P(B)6= 0 P(A∩B) = P(B)PB(A)