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Home » \textMean = \frac11010 = 11 - GetMeFoodie

\textMean = \frac11010 = 11 - GetMeFoodie

📅 April 30, 2026 👤 scraface

Apr 30, 2026

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📌 \left(\frac{1}{2}\right)^6 = \frac{1}{64}

📌 P(X = 4) = 15 \cdot \frac{1}{64} = \frac{15}{64}

📌 \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}}

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