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A box with 15 spare parts for a type of machine contains 10 in good condition and 5 defective. Three parts are drawn randomly without replacement. ¿What is the probability that there are at least two in good condition?

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In order to get at least two spare parts in good condition, one must pick two or three spare parts in good condition. The probability of first selecting two parts in good condition, and then one in bad condition, equals:

$$\frac{10}{15} \cdot \frac{9}{14} \cdot \frac{5}{13}$$

Since there are three ways to select the spare part in bad condition (on the first, on the second and on the third turn), the total probability equals:

$$\frac{5}{15} \cdot \frac{10}{14} \cdot \frac{9}{13} + \frac{10}{15} \cdot \frac{5}{14} \cdot \frac{9}{13} + \frac{10}{15} \cdot \frac{9}{14} \cdot \frac{5}{13} = \frac{1350}{2730}$$

The probability of selecting three spare parts in good condition equals:

$$\frac{10}{15} \cdot \frac{9}{14} \cdot \frac{8}{13} = \frac{720}{2730}$$

Adding this all up, we get:

$$\frac{1350}{2730} + \frac{720}{2730} = \frac{2070}{2730} \approx 0.758$$

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