Feuille d`exercices 3 - Université d`Orléans

B, C P (C)6= 0 P(B|B∪C)≥P(B)
P(B|B∪C)≥P(B|C)
U14 1234 U25
3 2 2 3 U1
U2
X
X X
10
1 10
E
E
X
E
E
E[X]V[X]σ[X]
99
100 E
n
k
k
X
n1
1G
G0
P(G=l) = P(G=−l)E[G]
G n = 4
P(G=l)G

X
X p
X{1, . . . , n}
A X =1A−1Ac
Y p =P(A)
X2Y−1
X, Y Z
{−1; +1}A=XY B =Y Z C =ZX
A B
B C
A C
ABC A, B
C
N N ≥2
k{1, . . . , n}bkk
A
X=1AX
p=Pn
k=1 bk(bk−1)
N(N−1) .
G
A B
A B
TAA TBB
1
3

X
Y
Z= max(X, Y )k∈ {1,...,6}P(Z≤k)
Z
n
M
20% 10
X
X
n= 500 P(X > 200)
X, Y
α β X +Y
α+β
X λ
n
P(X≥n)≤λn
n!
1
(n+k)! ≤1 n!
1
k!
(A1, . . . , A2n) 2n
p∈]0,1[ X=Pn
k=11AkY=P2n k=n+11AkZ=X+Y
P(Z=n)
n
X
k=0 n
k2
=2n
n.
X, Y
λ{X= 0}∩{Y= 0}⊂{X=Y}
∀x≥0, e−x≥1−x
P(X6=Y)≤2λ.
A B n
X A Y
B
X Y
P(X=Y)n

X
∀k∈N,∀n∈N, P (X > n +k|X > n) = P(X > k).
N
α
P(N=k) = exp(−α)αk
k!k∈N
p
S
S pα
s > 1X ζ s
∀n∈N∗P(X=n) = 1
ζ(s)
1
ns,
ζ(s) =
+∞
X
n=1
1
ns.
X ζ s
p∈NpN∗={np;n∈N∗}
P(X∈2N∗)p∈N∗P(X∈pN∗)
(p, q)
{X∈pN∗} {X∈qN∗}
Y1X
n, k ∈N∗P(Y=k|X=n)
Z=Y
XFZ
[0,1]
p, q p ≤q
P(Z=p
q)
φ(n) 1 n
n
ζ(s+ 1)
+∞
X
n=1
φ(n)
ns+1 =ζ(s).

(
+∞
X
n=1
1
n2)(
+∞
X
n=1
φ(n)
n3)(
+∞
X
n=1
φ(n)
n4) = Z3
2
x dx.
5ζ(4) = 2ζ(2)2
b n
s r
p
p b n r s
s r
p
i= 1,2,3 1−αi
i
αi
Aii Ri
i i = 1,2,3
i= 1,2,3
i
α1
10%
40% 80%
15%
0.90
0.80