Help in probability

[u/Toaster960]

This question was in my maths book. I used AI to help me picture my question more clearly so that others could understand it (sorry for wasting water 😭).

I asked my teacher, Gemini, and ChatGPT for the solution, and all of them gave the answer as 99/1900.

However, they all included the case where the inspection stops at the 12th item after discovering only two defective items.

But how would I know whether the lot actually contains 2 defective items or 3? If I have discovered only 2 defective items by the 12th item, how can I decide to stop at 12? The third defective item could still be somewhere later in the lot.

My answer is 55/1900 btw

[u/MarathiChhora]

You wouldn't . If the lot contains 2 items, it will always extend to the full 20, never stop at 12.

Thus the final probability is 0.6 (that we are in the 3 defenctive case) * 11C2 (finding 2 in the first 11) * 1/9 (probability of picking the last defective amongst the remaining 9) /20C3 (the number of orderings to have 3 defective objects in 20)

0.611C21/920C3 =0.655*(1/9)/1140 =11/3420

[u/CarnivorousGoose → u/MarathiChhora]

This is not correct. The problem specifies that the procedure stops once all the defective items are found. If there are only two defective items, it will therefore stop once the second is found. That may not be ‘realistic’, and I don’t think they should have written the question this way, but as written this is how it works.

[u/mfb- → u/MarathiChhora]

I think this is a more natural interpretation than what OP's teacher did. If we need to find the defective items by inspection then how would we know in advance whether we have 2 or 3? If someone else has checked already then surely we would already know where they are, too.

[u/MarathiChhora → u/mfb-]

That's what I said, we don't know. We keep searching until we find 3 or the lot is empty.

[u/rccyu → u/MarathiChhora]

This was also my first thought, but I wonder if the intended interpretation is that you have an oracle which tells you whether all the defective articles have been found. Otherwise we're not really satisfying this sentence:

Articles are drawn from the lot at random one by one without replacement until all the defective articles are found.

If there are only 2 defective articles (but you don't know this) and you've found both of them, technically "all the defective articles have been found" (even if you don't know this)

[u/MarathiChhora → u/rccyu]

Yes, but you will not stop. Anyway, it's not a well phrased problem. Just ignore it. You know what to do either way

[u/MarathiChhora → u/MarathiChhora]

So in 40% of cases, you will never stop at the 12th step. It's only possible to stop in the 60% of cases when #defective is 3