I recently studied sampling from one of online courses and got to know about various sampling methods but i'm confused regarding the probability of an element being selected in a sample. In that course the lecturer just wrote down that the probability of an element being selected in a sample of size $n$ from the population of size $N$ for both with replacement(order matter) and without replacement(order does not matter) is $n/N$. I'm not able to derive it and also not able to get the results for the remaining cases. Assume that the set A = {$a_1,a_2,a_3,....,a_N$} be the population set and you sample $n$ elements from this set using one of the below:
- Simple Random Sampling with Replacement(order matters)
- Simple Random Sampling with Replacement(order does not matter)
- Simple Random Sampling without Replacement(order matters)
- Simple Random Sampling without Replacement(order does not matter)
So what is the probability that an element, say $a_i$ is one of the $n$ elements being sampled using above methods.
Edit: (what i mean by order matters)
take population set to be A = {1,2,3} now select a sample of size 2 these can be {1,1}{1,2}{1,3}{2,1}{2,2}{2,3}{3,1}{3,2}{3,3}(depending on the method used) now in the case of order matters {1,2} and {2,1} are counted twice but in case where order does not matter these are counted only once so the samples become {1,1}{1,2}{1,3}{2,2}{2,3}{3,3}
Edit 2: Further clarification
Suppose set A={1,2,3} and then simple random sampling is performed selecting 2 elements using one of the above methods, i want to know the probability of selection of each element. In short what is probability of selection of "1" when sampling is performed. For case (4) it is easy to derive:
number of samples = 3 ({1,2},{1,3},{2,3})
number of samples in which "1" is present = 2 ({1,2},{1,3})
so the probability of selection = 2/3 = n/N (n = 2, N = 3)
but how to derive the probability of including a specified unit in a sample(in general) for this case and also for other cases.