Probabilité. Arbre pondéré.

ω1ω2ω3
A={ω1, ω2, ω3}
ω1ω2ω3A
P(A) = P(ω1) + P(ω2) + P(ω3)
A
P(A) = A
A B
A A
A
A∩B A B
A∪B A B
A B
P(A)=1−P(A)
P(A∪B) = P(A) + P(B)−P(A∩B)

1
3
2
3
1
3
2
3
1
3
2
3
(R,R) (R,B) (B,R) (B,B)
(B,R)

A
A A
B
B B
A∪B A ∪B
G P
1
2
1
2
V R J V R J
1
3
1
3
1
3
1
3
1
3
1
3
B T B T B T B T B T B T
1
2
1
2
1
2
1
2
1
2
1
2
1
2
1
2
1
2
1
2
1
2
1
2
GV B
GV T
GRB
GRT
GJ B
GJ T
P V B
P V T
P RB
P RT
P J B
P J T

(G,R,B)
P(G,R,B) = 1
2×1 3×1 2
=1
12
A={(G,J,B),(G,J,T ),(P,J,B),(P,J,T )}
p(A) = p(G,J,B) + p(G,J,T ),+p(P,J,B) + p(P,J,T )
=1
2×1 3×1 2+1 2×1 3×1 2+1 2×1 3×1 2+1 2×1 3×1 2+1 2×1 3×1 2
=1
3
C C =
{(G,J,B),(G,J,T ),(P,J,B),(P,J T )}
p(C) = 4 ×1
12
=1
3
p(A) = 1 −p(A)
= 1 −1
3
=2
3
B={(G,V,B),(G,R,B),(G,J,B),(P,V,B),(P,R,B),(P,J,B)}
p(B) = 6 ×1
12
=1
2
D
((P,V,B),(P,R,B),(P,J,B)
p(D) = 3 ×1
12
=1
4

A∪B
(G,J,B) (R,J,B)
p(A∩B) = 2 ×1
12
=1
6
p(A∪B) = p(A) + p(B)−p(A∩B)
=1
3+1 2−1 6
=4
6