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This was a homework assignment I was asked to do:

"A lot of $n$ contains $k$ defectives, and $m$ are selected randomly and inspected. How should the value of $m$ be chosen so that the probability that at least one defective item turns up is $0,90$? Determine the answer when $n=1000$ and $k=10$."

I tried solving this by denoting A as the event that the number of defectives are found is larger than or equal than to $1$, and B that zero defectives are found. Then I figured that $P(A) = 1 - P(B)$. In order to solve the problem, we have to set $P(A) \geq 0,90$, so after a little algebra we find $P(B) \leq 0,10$. Since we have $n=1000$ items and $k=10$ defectives, I thought that $P(B) = \prod_{a=0}^{m} \big( \frac{990 - a}{1000 - a} \big) $. So find $m$, we have to solve $P(B) \leq 0,10 $. Putting this in wolframalpha, we find $m \geq 204 $.

I think this is the correct answer, but I thought that I am perhaps missing something, as it seemed that the solution is too easy. Am I doing this right? Or am I missing something?

Thanks in advance.

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You are doing this right (except that the product for $P(B)$ should stop at $a=m-1$ instead of $a=m$). – Did Feb 12 '12 at 13:33
Thank you! (more text) – Max Muller Feb 12 '12 at 14:11
@Max: You can fill the space with empty math code (e.g. ${}{}{}$). – joriki Feb 13 '12 at 5:46
@joriki thank you for the tip. – Max Muller Feb 15 '12 at 16:00

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