# Which probability result is greater? Using hypergeometric distribution or binomial distribution?

Suppose I have $$N$$ samples, and 1/4 of them are bad. I draw $$n$$ samples $$(n, and I want to know what's the probability that less than 1/3 are bad. I know it should use hypergeometric distribution to calculate, that is:

using hypergeometric distribution

And I understand that if N approaches to infinity, then we can use binomial distribution to calculate the value, that is:

using binomial distribution

My question is: when $$N$$ is not infinity, then which probability is greater? Using hypergeometric distribution or binomial distribution? Is there any result in math?

The only simple relationship is that the hypergeometric is less variable than the binomial (because it comes from sampling without replacement rather than with replacement). The mean is the same and the variance is smaller by a factor $$(N-n)/(N-1)$$
So, in your example, the hypergeometric gives you more chance of getting a answer near $$n/4$$, and less chance of getting an answer far from $$n/4$$. The question now is whether $$n/3$$ is 'near' or 'far from' $$n/4$$ in this sense. For large enough $$n$$ and $$N$$, $$n/4$$ will be far from $$n/3$$, so the probability of getting more than $$n/3$$ bad out of $$n$$ will be smaller with the hypergeometric than the binomial.
I don't see any easy way to decide how large $$n$$ has to be, apart from direct calculation or simulation.