Question

Probability problem: let u1 and u2 be two urns each containing 6 balls identical to the hit such that u1 contain 2 red balls, 3 green, and one black; and u2 2 red balls, one green and 3 black. we draw at random and simultaneously 2 balls from u1 and put them in u2, then we draw two balls at random and simultaneously from u2 which we place in u1. what is the probability that the contents of each ballot box will remain unchanged after that? And what is the probability that the contents of the ballot boxes will be interchanged?

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Answer to a math question Probability problem: let u1 and u2 be two urns each containing 6 balls identical to the hit such that u1 contain 2 red balls, 3 green, and one black; and u2 2 red balls, one green and 3 black. we draw at random and simultaneously 2 balls from u1 and put them in u2, then we draw two balls at random and simultaneously from u2 which we place in u1. what is the probability that the contents of each ballot box will remain unchanged after that? And what is the probability that the contents of the ballot boxes will be interchanged?

Expert avatar
Rasheed
4.7
107 Answers
Pour résoudre ce problème, nous allons utiliser les règles de la probabilité.

Premièrement, déterminons la probabilité que le contenu de chaque urne reste inchangé.

Étape 1: Calcul de la probabilité que les boules tirées de u1 restent dans u1.

La probabilité de tirer une boule rouge de u1 est \frac{2}{6} .

Après avoir tiré une boule rouge de u1, il reste 5 boules dans u1 et 3 boules rouges dans u2.

La probabilité de tirer une deuxième boule rouge de u1 est donc \frac{3}{5} .

La probabilité que les deux boules tirées de u1 restent dans u1 est alors \frac{2}{6} \times \frac{3}{5} = \frac{6}{30} .

Étape 2: Calcul de la probabilité que les boules tirées de u2 restent dans u2.

La probabilité de tirer une boule rouge de u2 est \frac{2}{6} .

Après avoir tiré une boule rouge de u2, il reste 4 boules dans u2 et 2 boules rouges dans u1.

La probabilité de tirer une deuxième boule rouge de u2 est donc \frac{2}{4} .

La probabilité que les deux boules tirées de u2 restent dans u2 est alors \frac{2}{6} \times \frac{2}{4} = \frac{4}{24} .

Étape 3: Calcul de la probabilité globale que le contenu de chaque urne reste inchangé.

Les deux tirages sont indépendants, donc nous multiplions les deux probabilités précédentes pour obtenir la probabilité globale :

\frac{6}{30} \times \frac{4}{24} = \frac{1}{20} .

Donc, la probabilité que le contenu de chaque urne reste inchangé est de \frac{1}{20} .

Maintenant, calculons la probabilité que le contenu des urnes soit interchangé.

La probabilité que le contenu des urnes soit interchangé est complémentaire à la probabilité que le contenu de chaque urne reste inchangé. Donc,

Probabilité du contenu des urnes interchangé = 1 - Probabilité que le contenu de chaque urne reste inchangé.

Probabilité du contenu des urnes interchangé = 1 - \frac{1}{20} = \frac{19}{20} .

Réponse :

La probabilité que le contenu de chaque urne reste inchangé est de \frac{1}{20} .

La probabilité que le contenu des urnes soit interchangé est de \frac{19}{20} .

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