Have a probability question I'm not sure how to solve. Hopefully this makes sense:

1) Lets say we have a population of 1000 unique books.

2) First, we take a random sample (without replacement) of 500 of those 1000 books. Lets call this S1.

3) Second, we take another random sample (without replacement) of 750 of those same 1000 books. Lets call this S2.

What is the probability that >=100 books from S2 exist in S1?

What is the probability that <=100 books from S2 exist in S1?

What is the probability that exactly 100 books from S2 exist in S1?


1 Answer 1


The distribution of the number of items in both samples is the hypergeometric distribution. Suppose we let $\mathcal{S}_1$ and $\mathcal{S}_2$ denote the two samples from your initial population of $N=1000$ items, and let the operation $| \cdot |$ denote the size of a set. Since the samples were formed independently via simple random sampling without replacement with $|\mathcal{S}_1| = 500$ and $|\mathcal{S}_2| = 750$ you have:

$$\mathbb{P}(|\mathcal{S}_1 \cap \mathcal{S}_2| = s) = \frac{{750 \choose s}{250 \choose 500-s}}{{1000 \choose 500}} \quad \quad \quad \text{for all } 250 \leqslant s \leqslant 500.$$

(Note that there must be at least $250$ items in both samples.) So, you have:

$$\begin{equation} \begin{aligned} \mathbb{P}(|\mathcal{S}_1 \cap \mathcal{S}_2| \geqslant 100) &= 1, \\[6pt]\mathbb{P}(|\mathcal{S}_1 \cap \mathcal{S}_2| \leqslant 100) &= 0, \\[6pt] \mathbb{P}(|\mathcal{S}_1 \cap \mathcal{S}_2| = 100) &= 0. \\[6pt] \end{aligned} \end{equation}$$

  • $\begingroup$ Thank you for the response! I'll try and process this in my head :) $\endgroup$
    – JD1
    Oct 30, 2018 at 21:07
  • $\begingroup$ I think when I tried to scale my real data down to a test example (like the one here), I didn't think through it enough. I was expecting a probability different than simply 1 or 0. If S1 and S2 stayed the same, but the original N was changed to 20,000 vs 1000, I think it would give an answer more in the realm of what I was looking for. Will try and use the formula here and plug in the new numbers. Thanks again! $\endgroup$
    – JD1
    Oct 30, 2018 at 21:15

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.