Epsilon 24 résolu par perplexity

Deux boules rouges et cinq boules blanches

Une urne contient deux boules rouges et cinq boules blanches. Deux joueurs Ali et Aloulou tirent tour à tour une boule de l’urne sans remise. Ali commence le premier. Quelle est la probabilité que Ali tire le premier une boule rouge ?

Perplexity a trouvé la bonne réponse !

To determine the probability that Ali is the first to draw a red ball, we need to consider the possible scenarios in which Ali draws a red ball before Aloulou does. Ali can win on his first, second, or third draw[1][2][3].

Here’s how we can break down the probabilities:

[latexpage]

Scenario 1: Ali wins on his first draw
- Ali draws a red ball on his first turn.
- Probability: $$P(R_1) = \frac{2}{7}$$ , where $$R_1$$ is the event that Ali draws a red ball on the first draw[1][2][3].
Scenario 2: Ali wins on his second draw
- Ali draws a white ball on his first turn, Aloulou draws a white ball on his turn, and then Ali draws a red ball.
- Probability: $$P(W_1 \cap W_2 \cap R_3) = \frac{5}{7} \cdot \frac{4}{6} \cdot \frac{2}{5} = \frac{40}{210}$$, where $$W_1$$ is the event that Ali draws a white ball on the first draw, $$W_2$$ is the event that Aloulou draws a white ball on the second draw, and $$R_3$$ is the event that Ali draws a red ball on the third draw[1][2][3].
Scenario 3: Ali wins on his third draw
- Ali draws a white ball, Aloulou draws a white ball, Ali draws a white ball, Aloulou draws a white ball, then Ali draws a red ball.
- Probability: $$P(W_1 \cap W_2 \cap W_3 \cap W_4 \cap R_5) = \frac{5}{7} \cdot \frac{4}{6} \cdot \frac{3}{5} \cdot \frac{2}{4} \cdot \frac{2}{3} = \frac{120}{2520} = \frac{2}{42} = \frac{1}{21}$$[1][2][3].

Now, we add the probabilities of each scenario to find the total probability that Ali wins:

$$
P(\text{Ali wins}) = P(R_1) + P(W_1 \cap W_2 \cap R_3) + P(W_1 \cap W_2 \cap W_3 \cap W_4 \cap R_5)
$$

$$
P(\text{Ali wins}) = \frac{2}{7} + \frac{40}{210} + \frac{1}{21}
$$

To add these fractions, we need a common denominator, which is 210:

$$
P(\text{Ali wins}) = \frac{2 \cdot 30}{7 \cdot 30} + \frac{40}{210} + \frac{1 \cdot 10}{21 \cdot 10}
$$

$$
P(\text{Ali wins}) = \frac{60}{210} + \frac{40}{210} + \frac{10}{210} = \frac{110}{210} = \frac{11}{21}
$$

Therefore, the probability that Ali is the first to draw a red ball is $$\frac{11}{21}$$.

Citations:
[1] https://www.macformath.net/math/proba/probacond/exemple_probacond.pdf
[2] https://www.bac.org.tn/wp-content/document-bac/Serie-Bac/Technique/Mathematique/S%C3%A9rie%20dexercices%20-%20Math%20-%20Probabilit%C3%A9s%20-%20Bac%20Technique.pdf
[3] http://www.panamaths.net/Documents/Exercices/SolutionsHTML/19/PROBA00010.htm
[4] http://maths.desfontaines.free.fr/IMG/pdf/proba_2_ex_5.pdf
[5] http://maths.desfontaines.free.fr/IMG/pdf/proba_2_ex_4.pdf
[6] https://www.meilleurenmaths.com/images/misyl/terminaleS/ProgrammeOfficiel/probabilites/probabilites-conditionnelles-exercice-2.pdf
[7] https://www.superprof.fr/ressources/maths/maths-terminale-s/statistique-aleatoire-pourcentage.html
[8] https://www.bibmath.net/ressources/index.php?action=affiche&quoi=bde%2Fproba%2Fva2-vadiscretefinie&type=fexo

Réponse de Perplexity: https://www.perplexity.ai/search/une-urne-contient-deux-boules-ut16igb8RleGDAaDPAkHdw?utm_source=copy_output