I have a sample of children that are aged 5 years up to 7 years. In total there are 2000 children from 160 centres. The purpose of this exercise was to give each child an assessment and record the result.
Two sub-groups had a higher chance of selection than the rest of the children:
children age 5 years and 11 months to 6 years and 1 month
children who were within three weeks of starting at the centre when the assessment took place
I need to work out how many children would have been selected into the above groups, if they hadn't had a higher chance of selection (or, how much to weight the results of the children in these groups by to account for the higher chance of selection).
So more specifically, if I sampled 10 children aged 5 to 7 years from each of 160 centres, how many would be aged 5 years and 11 months to 6 years and 1 month, by chance? And how many would be within their first three weeks of starting at the centre?
I'm assuming that the distribution of ages in the population of 5 to 7 year olds is uniform.
Let the entire population have size $N$. Let $N_*$ be the population satisfying some criteria (eg. between 5 years and 11 months to 6 years and 1 month).
Suppose you sample $n$ from your population of $N$, and of your sample a total of $n_*$ come from the subpopulation $N_*$. Then this problem can be reduced to a simple binomial distribution problem.
You have $n$ trials, and each trial has probability of success $\frac{N_*}{N}$. Hence the probability of getting $n_*$ children from your sample is simply
$$P = {n \choose n_*} \frac{N_*}{N}^{n_*} \left(1-\frac{N_*}{N} \right)^{n-n_*}$$
Similarly, the probability of getting at least $n_*$ is
$$P = \sum_{k=n_*}^n {n \choose k} \frac{N_*}{N}^{k} \left(1-\frac{N_*}{N} \right)^{n-k}$$
Note that the binomial distribution assumes independent trials which is not strictly true here, since you're sampling without replacement. However, if the population as a hole is large, this dependence is negligible.
Hence to work out the probability you only need to estimate $N_*$. If you assume age distribution is uniform, then this is fairly straightforward, though non-uniform birthdates complicates it.