Correction feuille TD 3 : probabilités conditionnelles, indépendance

S D
P(S|D) = P(S∩D)
P(D)=10/36 30/36 =1 3
P S
P(P|S) = P(S|P)P(P)
P(S)=1/2×1/4 1/2×1/4+1/6×3/4=1 2
{1,2,3,4}
A={1,2}, B ={2,3}, C ={1,3}.
A B B C A C
A B C
P(A) = P(B) = P(C) = 1
2;P(A∩B) = 1 4:P(A∩C) = 1 4;P(B∩C) = 1 4:P(A∩B∩C)=0
(Ω,P)X Y
p∈]0,1[ q∈]0,1[

XY
Z=XY 0 1 Z1X Y 1
P(Z= 1) = P((X= 1) ∩(Y= 1)) = pq
X Y
P(Z= 0) = 1 −pq
XY pq
X
Xx1 2
P(X=x) 3/5 2/5
(X, Y )
X Y 1 2
11
3
4
15
24
15
2
15
Z XY
Z=XY z1 2 4
P(Z=z) 1/3 8/15 2/15
A F, M, H
P(A) = P(A|F)P(F) + P(A|M)P(M) + P(A|H)P(H) = 0.05 ∗0.2+0.15 ∗0.5+0.3∗0.3 = 0.0166
AcA
P(F|Ac) = P(Ac|F)P(F)
P(Ac)=0.95 ∗0.2 1−0.0166

I D
P(I|D)1×0.05
1×0.05 + 0.2×0.95 =5 24
5/24 >0.2
AGCFG
F
GG, GF, F G, F F
1/4P(CF|AG)
P(F|G)
P(CF|AG) = P(CF∩AG)
P(AG)=1/4 1/2=1 2
P(F|G) = P(F∩G)
P(G)=1/2 3/4=2 3
p
F F F G
GG TF
F G TF
P(F G|TF) = P(TF|F G)P(F G)
P(TF)
P(TF)
P(TF) = P(TF|GG)P(GG) + P(TF|F G)P(F G) + P(TF|F F )P(F F ) = 1
2p+1 4

P(F G|TF) =
1
2p
1
2p+1 4
A={1,...,6}×{1,2,5},
B={1,...,6}×{4,5,6},
C={(i, j)∈ {1,...,6}2:i+j= 9}.
P(A∩B∩C) = P(A)P(B)P(C)A B B
C A C
P(A) = 1
2;P(B) = 1 2;P(C) = 1 9
A∩B={1,...,6}×{5}P(A∩B) = 1
6
B∩C={(5,4),(4,5),(3,6)}
P(B∩C) = 1
12 A∩C={(4,5)}
P(A∩C) = 1
36 A∩B∩C={(4,5)}
P(A∩B∩C) = 1
36
(Ω,P)A B C
A B ∪C
A∩(B∪C) = (A∩B)∪(A∩C)
A B C A B
A C
P(A∩(B∪C)) = P((A∩B)∪(A∩C))
=P(A∩B) + P(A∩C)−P((A∩B)∩(A∩C))
=P(A)P(B) + P(A)P(C)−P(A∩B∩C)
=P(A)P(B) + P(A)P(C)−P(A)P(B)P(C)
A B A C A, B C
P(A)P(B∪C) = P(A) [P(B) + P(C)−P(B∩C)]
=P(A) [P(B) + P(C)−P(B)P(C)]
=P(A)P(B) + P(A)P(C)−P(A)P(B)P(C)
B C
n

0< p < 1
1−p
p(k)
cp(k) e
p(k+1)
c=pp(k) c+ (1 −p)p(k) e
p(k+1)
e= (1 −p)p(k) c+pp(k) e
k+ 1
k k + 1
k k + 1
p(k+1)
cp(k+1) e
p(k+1)
e= 1 −p(k+1) c
~p(k+1) =T ~p(k)
~p(k)= p(k)
c
p(k)
e!
T2×2
~p(k+1) =p1−p
1−p p  p(k)
c
p(k)
e!
n
k= 0
~p(n)=Tn1
0
A
A
G A G −6,−5,...,−1,1, . . . 6
A1/12
X
Y
%
% %
X Y
X Y faible interm.´elev´e
homme 0.16 0.16 0.08
femme 0.24 0.24 0.12