corrigé

◦
D1
360 −120 = 240 D1
D1A=D1∩D2
P(A) = (A)
(Ω) =240 120000 =2 1000 = 0,002 Ω
B=D1∩D2
D1∪D2
P(B) = 1 −P(D1∪D2) = 1 −(P(D1) + P(D2)−P(D1∩D2))
P(D1) = (D1)
(Ω) =360 120000 =3 1000 = 0,003
P(D2) = (D2)
(Ω) =600 120000 =6 12000 =1 2000 = 0,005
P(D1∩D2) = (D1∩D2)
(Ω) =120 120000 =1 1000 = 0,001
P(B) = 1 −(0,003 + 0,005 −0,001) = 1 −0,007 = 0,993
X(Ω) = {2,45 ; 4,05 ; 6,45 ; 8,05}P(X= 2,45) = P(B) = 0,993
P(X= 4,05) = P(D1∩D2) = P(A) = 0,002
P(X= 8,05) = P(D1∩D2) = 0,001
P(X= 6,45) = 1 −(P(X= 2,45) + P(X= 4,05) + P(X= 8,05)) = 1 −(0,993 + 0,002 + 0,001)
= 1 −0,996 = 0,004
X
xi
P(X=xi)
E(X) =
4
∑
i=1
xiP(X=xi) = 2,45 ×0,993 + 4,05 ×0,002 + 6,45 ×0,004 + 8,05 ×0,001 = 2,4748
e
•10 −1 = 9
•5−1 = 4
• −1n
1 + 3 + n=n+ 4 Ω
•P(X= 9) = (X= 9)
(Ω) =1 n+ 4 n+ 4
•P(X= 4) = (X= 4)
(Ω) =3 n+ 4 n+ 4

•P(X=−1) = (X=−1)
(Ω) =n n+ 4 n n + 4
X
xi−1
P(X=xi)1
n+ 4
3
n+ 4
n
n+ 4
E(X) =
3
∑
i=1
xiP(X=xi) = 9 ×1
n+ 4 + 4 ×3 n+ 4 −1×n n+ 4 =9 + 12 −n n+ 4 =21 −n n+ 4
E(X) = 0 ⇔21 −n
n+ 4 = 0
E(X)=0 ⇔21 −n= 0 ⇔n= 21
•10 −m1
20 n= 16
•5−m3
20
• −m16
20 =4 5
X
xi10 −m5−m−m
P(X=xi)1
20
3
20
4
5
E(X) =
3
∑
i=1
xiP(X=xi) = (10 −m)×1
20 + (5 −m)×3 20 −m×4 5
=10 −m+ 3(5 −m)−16m
20 =10 −17m+ 15 −3m 20 =25 −20m 20 =5−4m 4
E(X) = 0 ⇔5−4m
4= 0 ⇔5−4m= 0 ⇔4m= 5 ⇔m=5 4= 1,25
X
xi10 −m5−m−m
P(X=xi)1
n+ 4
3
n+ 4
n
n+ 4
E(X) = (10 −m)×1
n+ 4 + (5 −m)×3 n+ 4 −m×n n+ 4
=10 −m+ 3(5 −m)−nm
n+ 4 =10 −m+ 15 −3m−nm n+ 4 =25 −4m−nm n+ 4
E(X)=0 ⇔25 −4m−nm
n+ 4 = 0
E(X) = 0 ⇔25 −4m−nm = 0
n m nm + 4m= 25 ⇔m(n+ 4) = 25

1 + x6x2+x3f(x)6g(x)
CfCgS= [1 ; +∞[
f(x)−g(x) = 1 + x−(x2+x3) = 1 + x−x2(1 + x)x2+x3x2
(1+x)f(x)−g(x) = (1+x)−x2(1+x) = (1+x)(1−x2)
f(x)−g(x) = (1+x)(1+x)(1−x) 1−x2= (1+x)(1−x)f(x)−g(x) = (1+x)2(1−x)
(1 + x)260f(x)−g(x) = (1 + x)2(1 −x)
(1 −x)
1 + x6x2+x3⇔f(x)6g(x)⇔f(x)−g(x)60⇔1−x60
1 + x6x2+x3⇔16x
1 + x6x2+x3S= [1 ; +∞[
N0,35N
6×3100 + 500 ×52 + 2000 ×12 = 68600
0,35N > 68600 ⇔N > 68600
0,35 ⇔N > 196000
•3100 −0,35 = 3099,35
•52 −0,35 = 51,65
•12 −0,35 = 11,65
• −0,35
G P (G= 3099,65) = 6
280000 =3 14000
P(G= 51,65) = 500
280000 =1 560

P(G= 11,65) = 2000
280000 =1 140
P(G=−0,35) = 280000 −6−500 −2000
280000 =277494 280000 =19821 20000
G
xi3099,65 51,65 11,65 −0,35
P(G=xi)3
14000
1
560
1
140
19821
20000
E(G) =
4
∑
i=1
xiP(G=xi) = 3099,65×3
140000 +51,65×1 560 +11,65×1 140 −0,35×19821 20000 =−0,105