I saw the Wikipedia article which says that the probability of drawing a green marble in the $i^{th}$ draw is $$ {\displaystyle P(G_{i})={\frac {K}{N}},} $$ given that $N$ describes the number of all marbles in the urn and $K$ describes the number of green marbles, then $N − K$ corresponds to the number of red marbles.
But I cannot understand why the probability of $i$th draw can be determined by population parameters alone.
If it is done "without replacement", is the $i$th probability influenced by the prior state of drawing? It seems like the hypergeometric distribution has a somewhat memoryless property although I know the geometric distribution is the only one with that property.
Somewhat informally, the green marble positions are uniformly distributed, there is nothing special about the $i^{th}$ position, and at the end of the process all $K$ green marbles have been distributed, thus this probability is $K/N.$
Alternatively count the number of favourable configurations, so assume the $i^{th}$ position has a green marble. The number of ways of distributing the remaining $K-1$ green and $N-K$ red marbles in the remaining positions is $$ \binom{N-1}{K-1}. $$ Dividing this by the total number of configurations $\binom{N}{K}$ gives the same expression, $K/N.$