Exercice 1 : Les probabilités Exercice 2 : Les probabilités

N
N
1
4
N
N →B10; 1
4
10−45
p(N= 5) = 10
51 45
×3
45
= 252 ∗243
1048576
=15309
262144
'0,0584
10−4(N64)
p(N64) = p(N= 0) + p(N= 1) + p(N= 2) + p(N= 3) + p(N= 4)
=10
01 40
×3
41
0 + 10
11 41
×3
49
+10
21 42
×3
48
+10
31 43
×3
47
+10
41 44
×3
46
=483327
524288
'0,9219
X
X
X
3
4
X
X →B5; 3
4
X(Ω) = {0; 1; 2; 3; 4; 5}

X
10−4
xi0 1 2 3 4 5
p(X=xi)5
03
40
×1
45
5
13
41
×1
44
... ... ... 5
53
45
×1
40
exacte 1
1024
15
1024
45
512
135
512
405
1024
243
1024
approche 0,0010 0,0146 0,0879 0,2637 0,3956 0,2373
p(X≥3)
p(X≥3) = 459
512
'0,8965
0,8965
E(X)
E(X) = n×p E(X) = 15
4E(X) = 3,75
X →B(10; 0,75)
p(X= 10) = 10
103 410
×1
40
=59049
1048576
'0,0563
X →B(6; 0,75)
p(X62) = 6
03 40
×1
46
+6
13 41
×1
45
+6
23 42
×1
44
=77
2048
'0,0376

X n
X
X →B(n; 0,6)
pnn
pn=p(X>1)
= 1 −p(X= 0)
= 1 −n
0(0,6)0×(0,4)n
= 1 −0,4n
1−0,4n>0,999 ⇐⇒ 0,4n<0,001
n= 7,0,4n'0,0016 >0,001 n= 8,0,4n'0,00066 <0,001
(un)
un=n2−3n+ 1 n∈N
∀n∈N, un=f(n)f(x) = x2−3x+ 1
f f [0; +∞[
∀x>0, f0(x) = 2x−3
∀x>1, f0(x)>0f[1; +∞[
(un)n>1
u0= 3 un+1 = 2u2
n+un+ 3 n∈N
∀n∈N, un+1 −un= 2u2
n+ 3 >0
(un)
un=3n+1
5nn∈N
∀n∈N, un>0un+1
un
=3n+2
5n+1 ×5n 3n+1
un+1
un
=3
5<1
(un)
un=2n
nn∈N∗
∀n∈N∗, un>0un+1
un
=2n
n+ 1
∀n∈N∗, n >1n+n>n+ 1 2n
n+ 1 >1
(un)

20%
un
u1= 1000 u2u3
u2=80
100u1+ 300 u2= 1100
u3=80
100u2+ 300 u3= 1180
n un+1 = 0,8un+ 300
80% 0,8un300
un+1 = 0,8un+ 300
vn= 1500 −un
(vn)
∀n∈N∗
vn+1 = 1500 −un+1
= 1500 −0,8un−300
= 1200 −0,8un
= 0,8×(1500 −un)
= 0,8vn
0,8v1= 500
n un= 1500 −500 ×(0,8)n−1
∀n∈N∗, vn= 500 ×(0,8)n−1
vn= 1500 −unn un= 1500 −500 ×(0,8)n−1
u10 '1432,89
N
un∈]1499,9; 1500,1[
N= 40
un∈]1499,99; 1500,01[
N= 50
lim
n→+∞
un= 1500