Drawing red and blue balls without knowing how many balls are in the bag This is a lot like the blue and red balls problem, except we don't know how many balls are in the bag, only the relation of red to blue balls, is this problem possible to solve?
Let's say we have a population $x$ of an unknown size $n$ we know that a fraction $f$, let's say $25\%$ or $\frac{1}{4}$ of that population share a trait (like being a blue ball).
We now randomly select a sample $y$ of a known size $k$, the expected result is that $k\times f$ balls in our sample are blue, but what is the probability that a number of balls, $p$, in our sample are blue.
So, lets say that I have a bag with $\frac{1}{4}$ blue balls and $\frac{3}{4}$ red balls, from that bag I take 40 balls. What are the odds that 15 of those balls are blue?

For any given $x$ you can compute it.  You can compute the chance of drawing specifically $15$ blue balls then $25$ red balls, then multiply by $40 \choose 15$ for the number of orders.  The answer will be tiny.  If you are drawing with replacement $x$ doesn't matter.  The chance of getting a blue ball stays $2\%$, so the chance of $15$ blue balls out of $40$ is $${40 \choose 15}\frac {49^{25}}{50^{40}}\approx 8\cdot 10^{-16}$$
