Inclusion Probability in Simple Random Sampling (SRS) Without Replacement

Question

Imagine that we want to choose a sample of size $ n $ from a population of size $ N $. Let $ j $ be a unit contained within the population of size $ N $.

What is the inclusion probability of unit $ j $ for simple random sampling without replacement? In other words, what is the probability that our sample of size $ n $ will contain $ j $?

According to Thompson M.E. (1997):

"Since there are $ {N-1}\choose{n-1} $ samples of size $ n $ which contain $ j $ and each has probability $ 1/{{N}\choose{n}} $, then $ \pi_j = {{N-1}\choose{n-1}} \big/{{N}\choose{n}} = \frac{n}{N} $."

I understand that the probability of obtaining any given sample is $ 1/{{N}\choose{n}} $. However, I don't understand the other claim, which states that there are $ {N-1}\choose{n-1} $ samples of size $ n $ that contain unit $ j $... Any help or guidance would be appreciated. Thanks!

Answer 1

Side note: there is a far easier way to compute $\pi_j$. A simple random sample can be obtained by randomly permuting the $N$ units and choosing the first $n$. It should be clear that the probability that unit $j$ ends up in the first $n$ after permuting is $n/N$.

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The number of samples of size $n$ which contain $j$ is simply counting the number of ways you can choose the remaining $n-1$ elements of your sample. There are $N-1$ other units to choose from, so there are $\binom{N-1}{n-1}$ ways.