I'm not sure if this is a tractable question or not. And I am unsure where to start. I've attempted the chained combinations formula. but at even these sample sizes excel breaks. So I'm not sure my approach is correct.
- Population size: 1000 units
- Sample 1 size: 300 units
- Sample 2 size: 450 units
Order does not matter. Each sample is taken independently but from the same population.
Assuming normal distribution, is it possible to determine:
- A. Probability that At least 1 unit from sample 1 is contained in sample 2?
- B. Probability that Sample 1 is entirely contained in sample 2?
So each sample is drawn without replacements (i.e. no repeats within each sample), but of course after sample $1$ everything is replaced before drawing sample $2$.
The number of ways to draw both samples is ${1000 \choose 300} \times {1000 \choose 450}$
Part A: Drawing both samples in a way they don't overlap, means the second sample must come from the remaining $700$ not in the first sample. The number of ways is ${1000 \choose 300} \times {700 \choose 450}$
These calculations might break excel, but they don't break e.g. wolfram alpha which says:
$$P(\text{no overlap}) = {{1000 \choose 300} \times {700 \choose 450} \over {1000 \choose 300} \times {1000 \choose 450}} \approx 2.4 \times 10^{-101}$$
$$P(\text{at least one overlap}) = 1 - P(\text{no overlap}) \approx 1 - 2.4 \times 10^{-101} \approx 1$$
Part B: Drawing both samples in a way that sample $1$ is a subset of sample $2$, means sample $2$ only needs to draw the $450-300 = 150$ extra units from the $700$ not in sample $1$. The number of ways is ${1000 \choose 300} \times {700 \choose 150}$ and again wolfram alpha says:
$$P(\text{sample $1 \subset$ sample $2$}) = {{1000 \choose 300} \times {700 \choose 150} \over {1000 \choose 300} \times {1000 \choose 450}} \approx 1.8 \times 10^{-141} \approx 0$$
Hope this answers your question...?