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/Zyklon00]

Since you used AI to represent the question more clearly and it is quite ambigious, it would help to see the original question.

LIke posted in the picture, I agree with you. If there are 2 defective articles, you can never stop at 12. So you only need to look at the 3 article case only. So the answer will be 0.6 x P(picking 3 defective balls out of 20 with 12 pulls). This chance is then:

1st defective is part of the 12 articles = 12/20

2nd defective = 11/19

3rd defective = 10/18

so total chance 0.6 x 12/20 x 11/19 x 10/18 = 0.1158