Introduction aux probabilités - Licence MIA 2e année

N
n= 10000 N
n= 10000 p= 10−50≤k≤n
P(N=k) = n
kpk(1 −p)n−k.
E(N) = np = 10000 ×10−5= 0.1
NB(n, p)n p
N λ =np = 0.1
P(N=k)'e−λλk
k!.
p
0.95 P(N= 0) ≥0.95
P(N= 0) = n
0p0(1 −p)n−0= (1 −p)n.
P(N= 0) ≥0.95 ⇔(1−p)n≥0.95 ⇔1−p≥0.951/n ⇔p≤pmin = 1−0.951/n = 1−0.9510−4.
pmin
(1 + α)m= 1 + mα +o(α)
pmin = 1 −(1 −0.05)10−4'1−(1 −10−4×0.05) = 5.10−6.
X
f(x) = α
√xx∈]0,1],
0,
f1
Z1
0
f(x)dx =αZ1
0
dx
√x=α√x
1/21
0
= 2α.
α= 1/2α f 1

P(−1≤X≤1/2) = Z1
2
−1
f(x)dx
=Z0
−1
0dx +Z1
2
0
dx
2√x
= 0 + √x1
2
0=r1
2=√2 2.
P(X= 1/2) = Z1
2
1
2
f(x)dx = 0.
P(X≥3/4) = Z+∞
3
4
f(x)dx
=Z1
3
4
dx
2√x+Z+∞
1
0dx
=√x1
3
4
+ 0 = 1 −r3
4= 1 −√3 2.
E(X) = Z+∞
−∞
xf(x)dx
=Z0
−∞
0×x dx +Z1
0
x dx
2√x+Z+∞
0
0×x dx
= 0 + 1
2Z1
0
√x dx + 0 = 1
2x3/2 3/21
0
=1
3.
V(X) = E(X2)−E(X)2
=Z+∞
−∞
x2f(x)dx −1
32
=Z0
−∞
0×x2dx +Z1
0
x2dx
2√x+Z+∞
0
0×x2dx −1
9
= 0 + 1
2Z1
0
x3/2dx + 0 = 1
2x5/2 5/21
0−1
9=1 5−1 9=4 45.

A=P(A) = 0.9T=
P(T)
P(T) = P(T|A)P(A) + P(T|Ac)P(Ac)
P(T|A) = 1/4
P(T|Ac) = 0
P(T) = 1
4×9 10 + 0 ×1 10 =9 40 = 0.225.
P(T|A) = 0
P(T|Ac) = 0.95
P(T) = 0 ×9
10 + 0.95 ×1 10 = 0.095.
P(A|Tc)
P(A|Tc) = P(Tc|A)P(A)
P(Tc)=(1 −P(T|A))P(A) 1−P(T)=(1 −0) ×0.9 1−0.095 =0.9 0.905 =180 181 '0.995.
B=B Bc
P(T|B) = P(T|A∩B)P(A|B) + P(T|Ac∩B)P(Ac|B) = 1
4×9 10 + 0 ×1 10 =9 40 = 0.225,
P(T|Bc) = P(T|A∩Bc)P(A|Bc) + P(T|Ac∩Bc)P(Ac|Bc) = 0 ×9
10 + 0.95 ×1 10 = 0.095.
P(A|Tc∩Bc) = P(Tc|A∩Bc)P(A|Bc)
P(Tc|Bc)=(1−P(T|A∩Bc))P(A|Bc) 1−P(T|Bc)=(1−0)×0.9 1−0.095 =0.9 0.905 =180 181 '0.995.

Gn
n G0= 0 Mn
•Gn=Gn−1+Mn
•Gn=Gn−1−Mn
Mn= 1 −Gn−11
•1G1=−1
•2G2=−3
•4G3=−7
•8G4= 1
N
N P (N=n)n≥1
N
MNMN
N
MN