Binomial distribution for gender discrimination? I could be on the wrong path, but I'm trying to use binomial distribution to identify gender discrimination and am having issues getting consistent results based on trial size(?)... I think. Small numbers work as expected, large numbers do not.
An example that works: 16 people are hired. 2 are women. What is the chance this happened randomly? (x)=2, (n)=16, (p)=.5. Example below [from R] show what I expected.
pbinom(2, size=16, prob=0.5) 
#0.0021 -> .2% chance that there is no gender discrimination

An example that doesn't work: 1150 people are hired. 350 are women. What is the chance this happened randomly? (x)=350, (n)=1150, (p)=.5. Example below [from R] does not show what I expected.
pbinom(350, size=1150, prob=0.5) 
# 2.388788e-41 -> ?!

Thanks in advance. Any suggestions better approached please!

In basic hypothesis testing, we articulate a null hypothesis, choose an ordering of results, collect data, and then find the probability, given the null hypothesis, of getting results that are, according to the ordering we chose, as extreme or more extreme as the results that we found.
That's it. We find the probability of the results given the hypothesis. Nothing else.
We do not find the probability of the hypothesis given results. We do not find the probability of some hypothesis other than the null hypothesis.
In your case, the null hypothesis is that each hiring decision is a Bernoulli trial, each one is independent of the others, and the probability for each is 0.5. Your ordering is "more men is more extreme". The number 0.0021 you found is the probability of getting as many men or more, given those premises. It is not the probability of there being no gender discrimination. Not only are you confusing $P(E|H)$ and $P(H|E)$, there are many alternative hypotheses other than "there in gender discrimination". For instance, there could be fewer qualified female candidates, or the hirings could be not independent.
As for the effect of larger sample size, the number of standard deviations scales with the square root of the sample size. That is, in the examples you gave, the sample size was multiplied by 71.875, and square root of that is 8.48, so deviations from the expected mean of 0.50 represent 8.48 as many standard deviations.
