Probabilit´ es: exercice de base 4
Sources S ´ esamath
Seconde
SourcesS´esamathSeconde Probabilit´es: exercice de base 4
´ enonc´ e
On consid` ere deux ´ ev´ enements A et B tels que:
p(A) = 0, 5 p(B) = 0, 8 p(A ∩ B) = 0, 4 Calculer p A ∪ B
SourcesS´esamathSeconde Probabilit´es: exercice de base 4
correction
On calcule p(A ∪ B) sachant que P (A ∪ B) = p(A) + p(B) − p(A ∩ B)
SourcesS´esamathSeconde Probabilit´es: exercice de base 4
correction
On calcule p(A ∪ B) sachant que P (A ∪ B) = p(A) + p(B) − p(A ∩ B) P (A ∪ B) = 0, 5 + 0, 8 − 0, 4 = 0, 9
SourcesS´esamathSeconde Probabilit´es: exercice de base 4
correction
On calcule p(A ∪ B) sachant que P (A ∪ B) = p(A) + p(B) − p(A ∩ B) P (A ∪ B) = 0, 5 + 0, 8 − 0, 4 = 0, 9 A ∪ B est l’´ ev´ enement contraire de A ∪ B
SourcesS´esamathSeconde Probabilit´es: exercice de base 4
correction
On calcule p(A ∪ B) sachant que P (A ∪ B) = p(A) + p(B) − p(A ∩ B) P (A ∪ B) = 0, 5 + 0, 8 − 0, 4 = 0, 9 A ∪ B est l’´ ev´ enement contraire de A ∪ B donc p A ∪ B
= 1 − p(A ∪ B)
SourcesS´esamathSeconde Probabilit´es: exercice de base 4
correction
On calcule p(A ∪ B) sachant que P (A ∪ B) = p(A) + p(B) − p(A ∩ B) P (A ∪ B) = 0, 5 + 0, 8 − 0, 4 = 0, 9 A ∪ B est l’´ ev´ enement contraire de A ∪ B donc p A ∪ B
= 1 − p(A ∪ B) p A ∪ B
= 1 − 0, 9 = 0, 1
SourcesS´esamathSeconde Probabilit´es: exercice de base 4
