Let's assume the following:

Population size: N
Individuals with a particular feature: x% of N
Sample size: y% of N

What's the chance that we get at least one individual with that particular feature in our sample based on N, x, y? Sampling is done without replacement.

(N - population size, x - percent of individuals with the particular feature, y - sampling percent)

Here's an example:

Population size: 10
Individuals with a particular feature: 30% of 10 = 3
Sample size: 30% of 10 = 3

The chance of picking at least one individual with that particular feature is 70.83%.

  • $\begingroup$ Is sampling done with replacement or without? $\endgroup$ – paw88789 May 23 at 16:08
  • $\begingroup$ Without replacement. Updated the question. $\endgroup$ – Razvan May 23 at 16:18

If you can convert the percentages of the population to integers (as in your example):

Say there are $c=N\cdot\frac{x}{100}$ individuals in the population with the characteristic; and the sample size is $k=N\cdot \frac{y}{100}$.

Then the probability of at least one person having the characteristic would be $$1-\frac{\binom{N-c}{k}}{\binom{N}{k}}$$

The numerator of the fraction in the preceding expression comes from choosing $k$ people without the characteristic.

If you can't get exact numbers, but if the sample is quite small compared to the population (i.e., $y$ is a very small percentage), then sampling without replacement may be a reasonable approximation to sampling with replacement. Then the probability would be approximately equal to $$1-\left(1-\frac{x}{100}\right)^{\frac{yN}{100}}$$


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