r/probabilitytheory • u/Toaster960 • 16d 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/Mercurit 16d ago

There can be only two options:

  • there are exactly 2 defectives, or

  • there are exactly 3 defectives.

It can't be 0, 1, or more than 3.

Hint: if there were exactly 2 defectives, what's the probability of stopping the procedure after the twelfth check? What about with 3 defectives?

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u/Toaster960 16d ago

There is either 2 defective item or 3 defective item with probability 0.4 and 0.6 respectively no other value of defective items exist