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Home » Is Tatsumaki’s One-Punch Truly Unstoppable? Shocking Breakdown Inside! - jntua results

Is Tatsumaki’s One-Punch Truly Unstoppable? Shocking Breakdown Inside! - jntua results

📅 March 9, 2026 👤 scraface

Mar 09, 2026

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📌 \boxed{\dfrac{15}{64}}

📌 Pregunta: Una caja contiene 5 bolas rojas y 7 bolas azules. Si se seleccionan 4 bolas al azar sin reemplazo, ¿cuál es la probabilidad de que exactamente 2 sean rojas?

📌 Solución: Este es un problema de distribución hipergeométrica. Tenemos una población finita con dos grupos: 5 bolas rojas (éxitos) y 7 bolas azules (fallos), totalizando 12 bolas. Se extraen 4 bolas sin reemplazo.

📌 Queremos la probabilidad de que exactamente 2 sean rojas. La fórmula para la probabilidad hipergeométrica es:

📌 P(X = k) = \frac{\binom{K}{k} \binom{N-K}{n-k}}{\binom{N}{n}}

📌 $ N = 12 $ (total de bolas),

📌 $ K = 5 $ (bolas rojas),

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4! \times 3! = 24 \times 6 = \boxed{144}
Pregunta: Calcula \(\dbinom{12}{5} - \dbinom{12}{4}\).
Solución: Usando las propiedades de los coeficientes binomiales:
\dbinom{12}{5} = \dbinom{12}{7}, \quad \dbinom{12}{4} = \dbinom{12}{8}
Pero directamente:
\dbinom{12}{5} = 792, \quad \dbinom{12}{4} = 495
Así:
- 495 = \boxed{297}
\]Frage: Finde die Summe aller Winkel \( \theta \in [0^\circ, 360^\cir...
Beginne mit der gegebenen Gleichung:

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