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Suppose the marble population has $n$ marbles. And based on prior knowledge I know that 30% of them are red, 50% of them are green, and 20% of them are blue. If I want to sample 2 marbles from my population, and I wished to calculate the probability of getting at least 1 red marble, would I assume that I'm drawing with, or without replacement? i.e whether the first draw is independent from the second one.

So in general $P($at least 1 red marble$) = 1 - P($no red marble)

In the drawing WITH replacement case:

$P($at least 1 red marble$) = 1 - (0.70)^2$

In the drawing WITHOUT replacement case:

$P($at least 1 red marble$) = 1 - \frac{\binom{n - 0.3*n}{2}}{\binom{n}{2}}$

So I am confused by whether I should consider the first drawing event to be dependent or independent of the second one. I am leaning towards the former because I have prior knowledge of the frequencies at which the marbles are present in the population. So say if my first draw is a red marble, then I know my chances of getting a second red marble is lower, as there are only a fixed a mount of red marbles (in this case 30%) present in the population. Is my line of thinking valid?

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If you are taking two marbles out at the same time, so they cannot physically be the same marble, then this is sampling without replacement.

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Your population is finite - having $n$ elements - so supposing that sampling is without replacement is the reasonable way. Hence, your second solution is correct.

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