# A marginal probability distribution on the number of ways to draw coloured balls from a box

The question has the following setup:

A box contains 5 yellow balls, 3 red, from which 4 are drawn one at a time, at random, without replacement. Let $X$ be the number of yellow balls on the first two draws, and $Y$ the number of yellow balls on all 4 draws. Find the marginal probability for $X$ and then $Y$.

Note this is not a duplicate as far as I can tell.

The suggested answer says:

Since $X$ is the number of yellow balls in the first 2 draws without replacement, the marginal distribution of $X$ is Hypergeometric with marginal probability function $$f(x) = \frac{{5\choose x}{3\choose 2-x}}{{8\choose 2}}$$

Now, the interpretation of Hypergeometric I learned is that you have $N$ trials, either a success or a failure, with $r$ successes (and $N - r$ failures). If you pick $n$ of these at random, the probability that $x$ are successes is $$\frac{{r\choose x}{N - r\choose n-x}}{{N\choose n}}$$ So clearly, $n = 2$, $N = 8$, and $r=5$, implying the interpretation for this case is that the trials are simply the picking of the balls, and success is "if the ball is yellow".

Now, I don't know why but I can't seem to equate these two cases in my head. Like, in the question, when we're choosing balls, we're doing so in order, we're not just choosing a $2$-subset of balls and then another one, we're choosing an "ordered" $4$-tuple of balls. If we just chose 2 $2$-subsets of balls, there would be multiple different orders that they correspond to.

So, I can't see how we can simply call this a case of hypergeometric distribution? In hypergeometric distribution, don't we also count the cases where the $2$ balls drawn are the last two? This doesn't make any sense to me.