Sagot :
Réponse :
Explications étape par étape :
■ BONSOIR !
■ "il a perdu ses lunettes" --> Ivoirien ! ☺
■ nb d' issues possibles ?
3 Ivoiriens = III ; IIF ; IIA ; IFF ; IAA; IFA ;
FFF ; FFA ; FAA ;
AAA ; donc 10 issues possibles !
■ proba(3 Ivoiriens) = 5/12 x 4/11 x 3/10 = 1/22 = 10/220
proba(3 F) = 3/12 x 2/11 x 1/10 = 1/220
proba(3 A) = 4/12 x 3/11 x 2/10 = 1/55 = 4/220
p(2A et I ou F) = 3 x [ 4/12 x 3/11 x (5/10 + 3/10) ]
= 3 x 4/12 x 3/11 x 8/10 = 12/55 = 48/220
p(2F et I ou A) = 3 x [ 3/12 x 2/11 x (5/10 + 4/10) ]
= 3 x 3/12 x 2/11 x 9/10 = 27/220
p(2 I et F ou A) = 3 x [ 5/12 x 4/11 x (3/10 + 4/10) ]
= 3 x 5/12 x 4/11 x 7/10 = 7/22 = 70/220
p(IFA) = 6 x 5/12 x 3/11 x 4/10 = 3/11 = 60/220
TOTAL = (15+48+27+70+60) / 220 = 220/220 = 1 ♥
■ proba(au moins 1 F) = p(3 F) + p(2 F) + p(1 F)
= 1/220 + 27/220 + 3x3/12x9/11x8/10
= 1/220 + 27/220 + 108/220
= 136/220
= 34/55 .
autre méthode : p(≥ 1 F) = 1 - p(3 I) - p(3 A) - p(2A+I) - p(A+2I)
= 1 - 1/22 - 1/55 - 3/22 - 2/11
= 1 - 4/22 - 1/55 - 2/11
= 1 - 2/11 - 1/55 - 2/11
= 1 - 4/11 - 1/55
= 1 - 20/55 - 1/55
= 1 - 21/55
= 34/55 .
■ p(au plus 2 A) = 1 - p(3 A) = 1 - 1/55 = 54/55 .
■ p(2F + A) = 3 x 3/12 x 2/11 x 4/10 = 3/55 .