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📅 February 27, 2026 👤 scraface

Feb 27, 2026

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\binom{6}{4} = \frac{6!}{4!2!} = \frac{720}{24 \cdot 2} = 15
\left(\frac{1}{2}\right)^6 = \frac{1}{64}
Entonces:
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 sele...
Solución: Este es un problema de distribución hipergeométrica. Tenemos...
Queremos la probabilidad de que exactamente 2 sean rojas. La fórmula p...
P(X = k) = \frac{\binom{K}{k} \binom{N-K}{n-k}}{\binom{N}{n}}
Donde:

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