How Many Parts to Sample for a Complete Set with 95% Confidence?

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To achieve a 95% confidence level in obtaining at least one of each part numbered 1 to 8 from a bin, a sample size greater than 40 parts may be necessary, as initial attempts with 40 parts resulted in missing numbers. The probability of obtaining all eight parts in the first sample is calculated to be approximately 0.0024, indicating a low likelihood of success. The discussion raises concerns about the inefficiency of the sampling method, questioning why parts cannot be visually selected from the bin. It is suggested that this scenario resembles the coupon collector's problem, which involves determining the number of samples needed to collect all unique items. Understanding the probability dynamics with each additional sample is crucial for improving the sampling strategy.
Bandit127
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I could do with some help here about a problem I had at work today.

I have a process that dumps 8 parts in a bin every cycle. Each part is numbered, 1 to 8. (It is a big bin and it will contain an equal quantity of each number).

I need to measure one part of each number, so I need to grab a sample of parts from the bin.

How many parts do I need to pull out of the bin before I have a 95% chance of getting at least one of each number? (My guess was 40 parts, but in two samples of 40 parts I had two numbers missing from each sample. Loads of 5s but no 1s for example).

So, I worked out that the probability of getting 1 to 8 in the first 8 moulds is 8!/88 or about 0.0024.

But there I got stuck on how that probability changes with the next set of 8 parts I pull from the bin. And so on until I have a ~95% chance of getting them all.
 
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Bandit127 said:
I could do with some help here about a problem I had at work today.

I have a process that dumps 8 parts in a bin every cycle. Each part is numbered, 1 to 8. (It is a big bin and it will contain an equal quantity of each number).

I need to measure one part of each number, so I need to grab a sample of parts from the bin.

How many parts do I need to pull out of the bin before I have a 95% chance of getting at least one of each number? (My guess was 40 parts, but in two samples of 40 parts I had two numbers missing from each sample. Loads of 5s but no 1s for example).

So, I worked out that the probability of getting 1 to 8 in the first 8 moulds is 8!/88 or about 0.0024.

But there I got stuck on how that probability changes with the next set of 8 parts I pull from the bin. And so on until I have a ~95% chance of getting them all.

So I have to ask -- why can't you look into the bin and select 8 parts that have different numbers? Why do you have to pull a part out of the bin before looking at the number? Seems like a very inefficient way to design a process, IMO.
 
I was reading a Bachelor thesis on Peano Arithmetic (PA). PA has the following axioms (not including the induction schema): $$\begin{align} & (A1) ~~~~ \forall x \neg (x + 1 = 0) \nonumber \\ & (A2) ~~~~ \forall xy (x + 1 =y + 1 \to x = y) \nonumber \\ & (A3) ~~~~ \forall x (x + 0 = x) \nonumber \\ & (A4) ~~~~ \forall xy (x + (y +1) = (x + y ) + 1) \nonumber \\ & (A5) ~~~~ \forall x (x \cdot 0 = 0) \nonumber \\ & (A6) ~~~~ \forall xy (x \cdot (y + 1) = (x \cdot y) + x) \nonumber...

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