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I need to solve this question:

In a certain article manufacture, is known that one among ten articles is defective. What is the probability that a sample of four articles have:
(a) no one defective?
(b) exactly one defective?
(c) exactly two defectives?
(d) no more than two defectives?

My try:

(a)$$P(X = 0) \Leftrightarrow \frac{\binom{1}{0} \cdot \binom{9}{4}}{\binom{10}{4}} = 0.6$$
(b)$$P(X = 1) \Leftrightarrow \frac{\binom{1}{1} \cdot \binom{9}{3}}{\binom{10}{4}} = 0.4$$
(c)$$P(X = 2) \Leftrightarrow \frac{\binom{1}{2} \cdot \binom{9}{2}}{\binom{10}{4}} = 0$$
(d)$$P(X \leq 2) \Leftrightarrow P(X = 0) + P(X = 1) + P(X = 2) = 1$$

The pdf answer:
(a) 0,6561
(b) 0,2916
(c) 0,0486
(d) 0,9963

I really don't know what is wrong with my try. Please help.

Matheus Sousa
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    "In a certain article manufacture, it is known that one among ten is defective..." Do you believe that there are exactly ten items manufactured? Usually a factory makes thousands or tens of thousands of items... not just ten. Your answer is correct if there were exactly ten items created however it is implied that many more than ten were made and that the statement was that $10%$ of the items were defective. Assuming a "large" number were made, use the binomial distribution instead of the hypergeometric distribution. – JMoravitz Sep 29 '21 at 15:37
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    It helps when trying to figure out a book's interpretation if you can try factoring numbers. Here, $6561 = 3^8 = 9^4$ so $0.6561 = \frac{9^4}{10^4}$ which should heavily imply that the book used the binomial distribution here. – JMoravitz Sep 29 '21 at 15:47
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    You didn't say whether sampling was with or without replacement. – Michael Hardy Sep 29 '21 at 16:05
  • Answers look correct (d)=(a)+(b)+(c). – herb steinberg Sep 29 '21 at 16:09
  • @MichaelHardy so, I interpreted like, once you get the defective article, you don't replace it in the sample. The question doesn't say if has a replacement. – Matheus Sousa Sep 29 '21 at 16:19
  • @JMoravitz ok, I'll use binomial distribution here. But I don't understand why binomial is the right way. – Matheus Sousa Sep 29 '21 at 16:22
  • @JMoravitz, I mean I understood. As you said, if there's a limit in a population size, with these conditions, can be considered a Hypergeometric distribution. If I don't know the population size, its better I get the binomial distribution. Am I Right? – Matheus Sousa Sep 29 '21 at 16:27
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    When the population size is many times larger than the amounts we are interested in, we may choose to simplify the problem and approximate the solution by pretending that we are sampling with replacement, even if we weren't. The result is that we will get values that are incredibly close to correct, with far simpler arithmetic, and that doesn't depend on knowing exactly how many objects there were to sample from. – JMoravitz Sep 29 '21 at 16:58
  • Thanks @JMoravitz, it helps a lot. – Matheus Sousa Sep 29 '21 at 17:02

1 Answers1

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You didn't say whether sampling was with or without replacement.

What you did is correct if it was without replacement.

Try it the other way.

Michael Hardy
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    It is heavily implied it is without replacement. The issue is that it is ambiguous here is the population size. It is heavily implied that the population size is many orders of magnitude larger than the amount we are sampling and so we may model it as sampling with replacement (even though it isn't actually sampling with replacement). – JMoravitz Sep 29 '21 at 16:59