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vaieri
1 month ago
3

4 out of 15 apples are rotten. They are taken out one by one at random and examined. The ones which are examined are not replace

d. What is the probability that the 9th one examined is the last rotten one?
Mathematics
1 answer:
Svet_ta [12.7K]1 month ago
3 0

Respuesta:

0.041

Explicación paso a paso:

Número de manzanas podridas = 4

Total de manzanas = 15

Manzanas buenas = 15 - 4 = 11

Para que la 9ª manzana examinada sea la última podrida, las primeras 8 deben contener 3 podridas y 5 buenas. Esto puede ocurrir de:

C(4,3) × C(11,5)

Recordando que C(n,r) = n! / [(n - r)! r!]

C(4,3) = 4! / (1! 3!) = 4

C(11,5) = 11! / (6! 5!) = 462

Elegir 8 manzanas de 15: C(15,8) = 15! / (7! 8!) = 6435

La 9ª manzana, siendo la última podrida, se elige entre las 7 restantes:

C(7,1) = 7! / (6! 1!) = 7

Por lo tanto, la probabilidad total es:

[C(4,3) × C(11,5)] / [C(15,8) × C(7,1)] = (4 × 462) / (6435 × 7) = 1848 / 45045 = 0.041

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