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I have a problem where a shipment contains $K$ good and $N-K$ defective components. And we pick at random $ n\leq K$ components and test them. I am trying to show that the probability $p$ that $k$ of the tested components are good is equal to: $$p = \frac{\binom{K}{k}\binom{N-K}{n-k}}{\binom{N}{n}}$$

So I know the following:

  • $\binom{N}{n}$ is the number of ways of picking $n$ element distinct subsets from the $N$ total components.
  • $\binom{K}{k}$ is the number of ways of picking $k$ element distinct subsets from the good components.
  • $\binom{N-K}{n-k}$ is the number of ways of picking $n-k$ element distinct subsets from the $N-k$ defective components.

Now, here is where I get confused.

We are trying to show that the probability $p$ that $k$ of the tested components are good is equal to the equation above. And I also know that the probability of an event consisted of $r$ elements equals $\frac{r}{N}$ (classical probability). I understand why we divided by $\binom{N}{n}$, but I do not understand why we multiplied by $\binom{N-K}{n-k}$ if we are trying to find the probability that $k$ are good.

Doesn't it defeat the purpose of multiplying by the probability of the defective components? And similarly, what if we were trying to find the probability $p$ that $k$ of the tested components were defective?

1 Answers 1

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You need to multiply by $\binom{N-K}{n-k}$ to specify which defective components you are picking. The reason you need to multiply by this factor is because $n$ might be less than $N$. If they were equal, then once you have chosen the $k$ good components, this is equivalent to choosing $N-k$ defective ones and so there wouldn't be a need to multiply both.