Corrigé du devoir en temps libre n de Mathématiques 9

TSI2
no9
X k ∈N
P(X > n) =
+∞
X
k=n+1
P(X=k) =
+∞
X
k=n+1
pqk−1=p
+∞
X
k=0
qk+n
=pqn
+∞
X
k=0
qk=pqn
1−q=qn
P(X > n +k|X > n) = P{X > n +k}∩{X > n}
P(X > n)=P(X > n +k) P(X > n)
=qn+k
qn=qk=P(X > k)
X
∀k∈N,∀n∈N, P (X > n +k|X > n) = P(X > k) (1)
k
N∗
XN∗
p=P(X= 1) ∀n>0, G(n) = P(X > n)
XN∗{X > 0}
G(0) = P(X > 0) = 1
n, k ∈N
G(n+k) = P(X > n +k) = P{X > n +k}∩{X > n}
=P(X > n +k|X > n)×P(X > n)
=P(X > k)P(X > n) = G(n)G(k)
∀n, k ∈N, G(n+k) = G(n)G(k)

G(n+ 1) = G(n)G(1)
G(n)n∈N
G(1) = P(X > 1) = 1 −P(X= 1) = 1 −p
G(n)n∈N1−p
G(0) = 1
∀n∈N, G(n) = (1 −p)n
{X > n −1}={X > n} ∪ {X=n}
P(X > n −1) = P(X > n) + P(X=n)
P(X=n) = P(X > n −1) −P(X > n) = G(n−1) −G(n)
= (1 −p)n−1−(1 −p)n= (1 −p)n−11−(1 −p)
=p(1 −p)n−1
X
Ri=
p0< p < 1q= 1 −p X
{X= 2}
{X= 2}=R1∩R2
{X= 3}
{X= 3}= (R1∩R2∩R3)∪(R1∩R2∩R3)

TSI2
{X= 4}
{X= 4}= (R1∩R2∩R3∩R4)∪(R1∩R2∩R3∩R4)∪(R1∩R2∩R3∩R4)
Ri
P(X= 2) = P(R1)×P(R2) = p2
P(X= 3) = P(R1∩R2∩R3) + P(R1∩R2∩R3)
=P(R1)×P(R2)×P(R3) + P(R1)×P(R2)×P(R3)
= (p×q×p)+(q×p×p)=2p2q
P(X= 4) = P(R1∩R2∩R3∩R4) + P(R1∩R2∩R3∩R4) + P(R1∩R2∩R3∩R4)
= 3p2q2
{X=k}
{X=k}=
k−1
[
i=1
Rk∩Ri∩
k−1
\
j=1
j6=i
Rj
P(X=k) =
k−1
X
i=1
P
Rk∩Ri∩
k−1
\
j=1
j6=i
Rj
=
k−1
X
i=1
P(Rk)×P(Ri)×
k−1
Y
j=1
j6=i
P(Rj)
=
k−1
X
i=1
p2qk−2= (k−1)p2qk−2
k>2P(X=k) = (k−1)p2qk−2
A
A1=A2=
A= +∞
\
i=1
Ri!∪
+∞
[
k=1
Rk∩
+∞
\
j=1
j6=k
Rj
A
A={X∈N\ {1,2}} =
+∞
[
k=2
{X=k}

P(A) = 1 −P +∞
[
k=2
{X=k}!= 1 −
+∞
X
k=2
P(X=k)
= 1 −p2
+∞
X
k=2
(k−1)qk−2= 1 −p2
+∞
X
k=1
kqk−1
|x|<1
+∞
X
k=0
xk=1
1−x
+∞
X
k=1
kxk−1=1
(1 −x)2(∗)
P(A) = 1 −p2×1
(1 −q)2= 1 −p2 p2= 0
P(A) = 0
X
+∞
X
k=2
|kP (X=k)|=
+∞
X
k=2
k(k−1)p2qk−2
X
E(X) =
+∞
X
k=2
kP (X=k) =
+∞
X
k=2
k(k−1)p2qk−2
=p2
+∞
X
k=2
k(k−1)qk−2
(∗)
+∞
X
k=2
k(k−1)xk−2=2
(1 −x)3
X
E(X) = p2×2
(1 −p)3=2p2 p3
E(X) = 2
p

TSI2
T
T
q= 1 −p
p T
T
P(T=k)=(k−1)p2qk−2, k >2.
n∈N∗
P(T > n) = P +∞
[
k=n+1
{T=n}!=
+∞
X
k=n+1
P(T=n) = p2
+∞
X
k=n+1
(k−1)qk−2
|x|<1
f(x) =
+∞
X
k=n+1
xk−1=
+∞
X
k=0
xk+n=xn
+∞
X
k=0
xk=xn
1−x
f0(x) =
+∞
X
k=n+1
(k−1)xk−2=nxn−1(1 −x) + xn
(1 −x)2=xn−1(n+ (1 −n)x) (1 −x)2
P(T > n) = p2×nqn−1(1 −q) + qn
(1 −q)2=nqn−1p+qn
P(T > n) = qn−1(np +q)
p= 1/10
P(T > 60) = 9
1059
×60 ×1
10 +9 10
p= 1/10 P(T > 60) '0,014
p= 1/20
P(T > 60) = 19
2059
×60 ×1
20 +19 20