Probabilités - Normalesup.org

A, B Ω
A∩B6=∅A∩¯ B6=∅¯ A∩B6=∅¯ A∩¯ B6=∅
(a, b, c, d)∈]0,1[4PΩ
P(A|B) = a P (A|¯ B) = b P (B|A) = c P (B|¯ A) = d
x=P(A∩B)x+y+z+t= 1 x
P x y z t Ω
a b 
(1 −a)x−az = 0
bx +y+bz =b
(1 −c)x−cy = 0
dx +dy +z=d
ad(1 −b)(1 −c) = bc(1 −a)(1 −d)
r n r 6n
A:
B:
J1, rK J1, nKnrA
P(A)
P(B) = 1 −P(A)
N+ 1 N k k N −k
(n+ 1) n
N→+∞
k k/N
n(k/N)n
n πn=1
N+ 1
N
X
k=0 k
nn

Akk
P(An+1 |A1∩ · · · ∩ An) = P(A1∩ · · · ∩ An+1)
P(A1∩ · · · ∩ An)=πn+1 πn
πn−−−−−→
N→+∞Z1
0
tndt=1
n+ 1 
b r
d
n
pnp1=b
b+rp2=b b+r+d
r
b+r+b+d b+r+d
b
b+r=b b+r
n+ 1 k
b+dk
b+r+nd
pn=
n
X
k=0 n
kb+dk b+r+ndb b+rkr b+rn−k
kn
k=nn−1 k−1
pn=1
b+r+nd 
n
X
k=0 n
kbb b+rkr b+rn−k+
n−1
X
j=0 n−1
jbnd b+rb b+rjr b+rn−1−j
A1, . . . , An
Akp Ak+1
1−p
pnAnA1
0<p<1pnn
pnp1= 1 p2=p pn+1 =ppn+ (1 −p)(1 −pn)
X
E(X)26EX2
X n p ∈]0,1[
Y=1
X+ 1

kn
k=nn−1 k−1E(Y) = 1−(1 −p)n+1 p(n+ 1) 
Z
X ϕX:R→C
ϕX(u) = EeiuX 
ϕX2πC∞
ϕX(0) ϕ0
X(0) ϕ00 X(0)
X p
n p
X x0
P(X=x0) = 1
2πZ2π
0
ϕX(u)e−iux0du
ϕX=ϕY⇒X=Y
X Y
ϕX+Y=ϕXϕY