TD 3 : Conditionnement / Indépendance

p
p

b r
d+ 1
(Ω,P)n+1 A1, . . . , An+1
n≥1P(∩n
i=1Ai)>0
P
n+1
\
i=1
Ai=P(A1)P(A2|A1)P(A3|A1∩A2). . . P(An+1|A1∩A2∩ · · · ∩ An).
X
p+q=n,p≥1,q≥1
pq =(n+ 1)n(n−1)
6.
n+ 1 1 n+ 1
A
(i, j, k) 1 ≤i < j < k ≤n+ 1 A
{i, j, jk}P(A) = 1/6
P(A) = (n+ 1)n(n−1)−1X
p+q=n,p≥1,q≥1
pq.
k+ 1 i
(i−1)/k N N ≥2
N N −1
k
N=m+n m n n
m
(n+ 1)/(n+m+ 1) k
{1,2,3,4}
A={1,2}, B ={2,3}, C ={3,4}.
A B B C A C
A B C

A={1,...,6}×{1,2,5},
B={1,...,6}×{4,5,6},
C={(i, j)∈ {1,...,6}2:i+j= 9}.
P(A∩B∩C) = P(A)P(B)P(C)A B B C
A C
(Ω,P)
A B A{B{
n n ≥3
(Ω,P)
A B C A B ∪C
(Ω,P)
n A1, . . . , Ann≥2P(∪n
k=1Ak) =
1−Qn
k=1(1 −P(Ak))
Ωn n ≥2
6n n ≥0
n
n
0<p<1
1−p
n
A=B=
A B
n≥1 0 <p<1
Ω = {0,1}nP
P{(ε1, . . . , εn)}=pPn
i=1 εi(1 −p)n−Pn i=1 εi, εi∈ {0,1}.
Xi: (ε1, . . . , εn)7→ εi,1≤i≤n.
X1, . . . , Xn
(Ω,P)X Y
p∈]0,1[ q∈]0,1[
XY

(Ω,P)X1, . . . , Xnn
0< p < 1
S=Pn
i=1 Xi
(Ω,P)X Y
(m, p) (n, p)m n
p∈]0,1[ X+Y(m+n, p)
(Ω,P)X Y
λ > 0µ > 0X+Y
A
A
n N
Ω = {1, . . . , N}nPXi
i Yk
k Zk1k V
(Xi)1≤i≤n
{Yk=s}1≤k≤N(Xi)1≤i≤n
YkP{Yk= 0}
ZkYkZk(Zk)1≤k≤N
V(Zk)1≤k≤N
A B C
Ω
0 1 1
(1) (0,1) (0,0,1) (0,0,0,1)
Ω
ΩA B
(A∪B){
X A X = +∞A
k{X= 3k}X
X Y
B X Y