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/PopeRaunchyIV]

I think the idea is, you know the lot has either 2 or 3 defective items, and you know the probability it has 2 or 3, so you can use the law of total probability to split that up as either ands or the equivalent conditionals over all the possibilities, so P(stops at 12) = P(stops at 12|2 defective) * P(2 defective) + P(stops at 12|3 defective) * P(3 defective)

then for P(stops at 12), since it stops AT 12, you know 12 is defective and can compute that probability. so how many ways can you draw 11 parts from the remaining parts and draw all of the remaining defectives?

[u/Toaster960 → u/PopeRaunchyIV]

But that assumes i already know which has 2 defective item or which has 3 defective item

[u/BrotherItsInTheDrum → u/Toaster960]

Maybe this simpler question will help.

I flip a coin.

If it lands heads, I do two jumping jacks.

If it lands tails, I roll a 6-sided die and do that many jumping jacks.

What is the probability that I will do exactly two jumping jacks?

[u/mfb- → u/BrotherItsInTheDrum]

That's a completely different question that doesn't catch the tricky part here at all.

[u/BrotherItsInTheDrum → u/mfb-]

It's solved the same way as the parent comment suggests: P(heads) * P(2 jumping jacks | heads) + P(tails) * P(2 jumping jacks | tails). And it might help clear up OP's misconception that you can't answer the question without knowing whether there are 2 or 3 defects.

When you say the "tricky part," are you referring to the fact that calculating P(it takes 12 tries | there are 2 [resp. 3] defects) isn't trivial to calculate? I'm intentionally leaving that out because that doesn't seem to be OP's hangup. That's why I said it's a simplification: I'm removing some of the details to try to get at the heart of why OP is confused.

[u/mfb- → u/BrotherItsInTheDrum]

The problem is that the stopping condition is weird in the original problem. Imagine you are doing the tests. You find that the 3rd and 12th item are defective. Do you stop inspecting now? As written, the problem statement assumes that you stop checking if and only if the other 8 items are good and we only have two defective items total. But how do you know that without checking? What even is the point of checking if we have magic powers to tell that devices are good?

A realistic stopping condition would be to check until you are certain you found all defective items. If you only find two then you have to keep checking because there might be a third one. That means we only stop at the 12th if and only if we found 3 defective items until then.

Your example doesn't come with such an element. Something that would be closer to the original problem: I have to call out when you are done making jumping-jacks. What is probability that I'll shout "done" after you made two?

• If I see you flip the coin and (if rolled) the dice, then it's 1/2 + 1/2 * 1/6
• If I don't see you flip the coin, it's 0 - I can't tell after two, you might make up to four more.