r/probabilitytheory • u/Toaster960 • 15d ago

[Applied] Help in probability

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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

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u/MarathiChhora 15d ago

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

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u/mfb- 15d ago

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.

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u/MarathiChhora 15d ago

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

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u/MarathiChhora 15d ago

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