Final answer:
The question involves calculating the probability of randomly selecting none of the defective machine parts from a lot using hypergeometric distribution. The formula for this probability takes into account the total number of items, the number of defective and non-defective items, and the sample size.
Step-by-step explanation:
The subject of this question is using probability in the context of hypergeometric distribution to calculate the chance of a certain outcome when selecting from a set with known quantities of 'defective' and 'non-defective' items. Specifically, we have a lot of 25 machine parts with 6 defective ones. We sample 10 parts and want to find the probability that none are defective.
To calculate this, we can use the hypergeometric probability formula:
P(X = k) = \( \frac{{\choose{n1}{k} \choose{n2}{n-k}}}{{\choose{N}{n}}} \)
where:
- N is the total number of items (25 machine parts).
- n1 is the number of defective items (6 parts).
- n2 is the number of non-defective items (25 - 6 = 19 parts).
- n is the sample size (10 parts).
- k is the number of defective items we are interested in finding within the sample (0 parts).
By plugging in our values we get:
P(0 defective) = \( \frac{{\choose{6}{0} \choose{19}{10}}}{{\choose{25}{10}}} \)
After calculating the combinations and dividing, we get the probability of selecting none of the defective parts when sampling 10 from the lot. This calculation gives us the insight into the quality of the shipment assuming that the sampling process is random and without replacement.