BANQUE PROBABILITÉS

X
Y
X
Y
X
Y
p=2
10 =1 5
4
5
X
(5,1
5)
X(Ω) = J0,5K∀k∈J0,5KP(X=k) = 5
k(1
5)k(4 5)5−k
E(X) = 5 ×1
5= 1 V(X) = 5 ×1 5×1−1 5=4 5= 0,8
Y= 2X−3(5 −X)Y= 5X−15
Y(Ω) = {5k−15 k∈J0,5K}
∀k∈J0,5KP(Y= 5k−15) = P(X=k) = 5
k(1
5)k(4 5)5−k
Y= 5X−15 E(Y) = 5E(X)−15 = 5 −15 = −10
Y= 5X−15 V(Y) = 25V(X) = 25 ×4
5= 20
X
X(Ω) = J0,2K
A
ΩAcard Ω = 10
5
k∈J0,2K
(X=k)k(5 −k)

2
k  8 5−k
∀k∈J0,2KP(X=k) = 2
k×8 5−k
10
5
Y= 5X−15
Y(Ω) = {5k−15 k∈J0,2K}
∀k∈J0,2KP(Y= 5k−15) = P(X=k) = 2
k×8 5−k
10
5

∀q∈N∗
k>qk
qxk−q∀x∈]−1,1[
+∞
k=qk
qxk−q=1 (1 −x)q+1 .
p∈]0,1[ r∈N∗
t= 0
t= 1
p
r
X
X
X
X(Ω) = Jr, +∞J
n∈Jr, +∞J
(X=n)n
n−1 (r−1)
((n−1) −(r−1)) n
(n−1) n−1
r−1
P(X=n) = n−1
r−1pr−1(1 −p)(n−1)−(r−1) ×p
∀n∈Jr, +∞JP(X=n) = n−1
r−1pr(1 −p)n−r
n>r
nP (X=n)
n∈Jr, +∞J
nP (X=n) = nn−1
r−1pr(1 −p)n−r=n(n−1)! (n−r)!(r−1)!pr(1 −p)n−r=rn! (n−r)!r!pr(1 −p)n−r
nP (X=n) = rn
rpr(1 −p)n−r
n>r
nP (X=n) = rpr
n>rn
r(1 −p)n−r
p∈]0,1[ (1 −p)∈]0,1[
n>r
nP (X=n)E(X)
E(X) =
+∞
n=r
nP (X=n) = rpr
+∞
n=rn
r(1 −p)n−r=rpr
(1 −(1 −p))r+1
E(X) = r
p

a∈]0,+∞[
(X, Y )N2
∀(j, k)∈N2P(X=j, Y =k) =
(j+k)1
2j+k
ej!k!
X Y
X Y
E2X+Y
∀x∈Rxn
n!
+∞
0
xn
n!=x
Y(Ω) = N
k∈N
P(Y=k) =
+∞
j=0
P(X=j∩Y=k) =
+∞
j=0
(j+k)1
2j+k
j!k!
j>0
j1
2j+k
j!k!=1
2k+1
k!
j>1
1
2j−1
(j−1)! 
j>0
j1
2j+k
j!k!
+∞
j=0
j1
2j+k
j!k!=1
2k+1
k!
+∞
j=1
1
2j−1
(j−1)! =1
2k+1
k!
1
2=1
2k+1
k!√e(∗). ⋆
j>0
k1
2j+k
j!k!=
k1
2k
k!
j>0
1
2j
j!
j>0
k1
2j+k
j!k!
+∞
j=0
k1
2j+k
j!k!=
k1
2k
k!
+∞
j=0
1
2j
j!=
k1
2k
k!
1
2=
k1
2k
k!√e(∗∗). ⋆⋆
P(Y=k) = 1
2k+1
k!√e+
k1
2k
k!√e=
(1
2+k)1 2k
k!√e
X Y
X(Ω) = N∀j∈NP(X=j) =
(1
2+j)1 2j
j!√e

X Y
P(X= 0, Y = 0) = 0 P(X= 0)P(Y= 0) ̸= 0
∀(j, k)∈N2aj,k = 2j+kP(X=j, Y =k)
aj,k =j+k
j!k!=j j!k!+k j!k!
∀k∈N
j>0
j
j!k!=1 k!
j>1
1
(j−1)!
+∞
j=0
j
j!k!=1 k!
+∞
j=1
1
(j−1)! =1 k!
j>0
k
j!k!=k k!
j>0
1
j!
+∞
j=0
k
j!k!=k k!
j=0
1
j!=k k!
k>0
1
k!
k>0
k
k!=
k>1
1
(k−1)!
+∞
k=0
1
k!=
+∞
k=0
k
k!=
(aj,k)(j,k)∈N2
E2X+YE2X+Y= 2