TD 1 1. Génotype 2. Couple 3. Une urne 4. Examen

2
4
X1X2X3X4
90%
3 1 2 3
(X, Y )X
Y
(X, Y )
1
X Y X Y
X Y
X Y
×
5×(X+Y)
Z Z

40
(X, Y ) IN∗×IN∗
IP(X=i, Y =j) = a−ibj(i, j)∈IN∗×IN∗,
a, b > 0P∞
i=nxi=xn/(1 −x)n∈IN |x|<1
a b
X Y X Y
IP(X=Y)
Z=X−Y
X1, X2, . . .
0 1 IP(X1= 0) = p, 0<p<1N
{0,1,2}IP(N= 0) = IP(N= 1) = IP(N= 2) N(Xi)i∈IN
S
S=
N
X
i=1
Xi
S= 0 N= 0
(N, S)S
IP(S|N= 1) IP(S|N= 2)
IE(S|N= 0) IE(S|N= 1) IE(S|N= 2)
IE(S)
N{1, . . . , n}
IE(S)
N P ois(λ) IE(S)

(X, Y )N
pkn = IP(X=k, Y =n) = λke−λαn(1 −α)k−n
n!(k−n)! {0≤n≤k}(k, n)
λ > 0 0 ≤α≤1
X
Y X Y
Z=X−Y
(m, n)∈N2IP(Y=m|Z=n)Y Z
2,2
1/2
n
Bin(n, 1/2)
a b

(X1, . . . , Xn, . . .)Xi∼Ber(1/2)
3
(1,0,0)
(0,1,0)
(0,0,1)
9 10
1 2 1 2
3 10 7
10
8 10
2 10
(Xt)t∈N
3
(Xt)t∈N
(Xt)t∈NX0= 0
1!
M= 0 1
1 0 !
5
µ
MnX0→µ

◦e1
◦e2◦e3
◦ ◦
X1, . . . , Xn, . . .
2e1e1
IP(X3=e3|X1=e3)
◦
Xt∈E={(1,0)T,(0,1)T}IP(X0=
(0,1)T)=1 M= 0,9 0,2
0,1 0,8!
(X0, . . . , X3) = ((0,1)T,(0,1)T,(1,0)T,(1,0)T) (X2, X3) =
((1,0)T,(1,0)T)X3= (1,0)T
X2= (1,0)TX3= (1,0)T
(Xt)t∈N
Xt∈E={A, G, C, T }Xt
Xt