TD3. Probabilités conditionnelles

e
A B
pA(B)
I U
I U
Ω
A Hn
n pnHn
Hn
Hn
P(A)
U1U2p1p2
π U1
U1
15%
95% 10%
Ω = [0,1[ \D p
[0,1[ m∈N
Am={x∈Ω, am(x) = 1},
am(x)m∈ {0,1}xD
p(D) = 0

x∈Ω
x∈Am⇐⇒ x=N
2m−1+1 2m+y 2mN∈ {0, . . . , 2m−1−1}, y ∈Ω.
Am2m−1
p(Am) = 1/2
n < m An∩Am2m−2
1/2mp(An∩Am) = p(An)p(Am)
Ann∈N∗
R1R20< p < 1
q= 1 −p
p=P( ) = 1/2 = P( ) = q
n∈NAnn n + 1 n+ 2
R1R2R1
Ω
(A3n+1)n∈N
(An)n∈N
lim supn∈N(A3n+1)R1R2R1
p(An)Pn∈Np(A3n+1)