# Probability for Sampling Schemes

I'm currently doing my homework for my stats class and I have a question for one of my exercises:

Consider a population P of size N. We define a new sampling scheme in the following way: we first select a sample SA using a simple random sampling without replacement of size n1. Then, we sample SB in P outside of SA according to a simple random sampling without replacement of size n2. We obtain in a such a way the final sample S as S = SA ∪ SB.

Obtain the probability P(S = s) for a realization s (probability mass function). Is it equivalent to a simple random sampling with sample size n = n1 + n2?

My questions is:

I don't really understand the question and don't know what to answer. What kind of PDF am I supposed to describe? What does he mean by P(S=s) for some realization s?

In my understanding the question is asking what the probability is that the sample will contain a certain random variable. Can someone maybe give me a more concrete example so I can work it out?

Unless I am interpreting wrong, here is my translation of the problem

In a population of size $$P$$, take a simple random sample of size $$n_1$$ without replacement. Next, without putting anything from this first sample back, take a second simple random sample of size $$n_2$$. Let our sample $$S$$ be the union of these two smaller samples. Is this the same as taking one large simple random sample of size $$n_1 + n_2$$ from the original population of size $$P$$ ?

A realization is, put simply, an observed value

For $$\mathbb{P}(S=s)$$, remember what $$S$$ (big s) represents. It is defined as the union of two disjoint samples, so it would be a set of objects. $$s$$ (little s) could be any particular set of objects from the population.

• thanks for your answer. I see how the sampling scheme works, but how do I put this formally into a probability mass function since i dont know what s is exactly. My attempt so far was to show that either way you can make the same amount of distinct samples, which implies that the probability is the same. But how can i exactly show that with a probability mass function? – Nicola Zaugg Nov 25 '18 at 12:04
• We have $n_1+n_2$ objects being chosen from $P$ total objects as our sample. So there are ${P}\choose{n_1+n_2}$ different possible samples we could have. And if each sample is equally likely, the probability for any particular sample being selected would be $1 / {P}\choose{n_1+n_2}$ – WaveX Nov 25 '18 at 16:47