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!

Assume that one is hiring from a large pool of equally qualified applicants of whom half are women and half are men. The number of women hired out of $n$ is $X.$ Suppose that $p$ is the probability that any one hire will be a women.
Perhaps you want to test
The null hypothesis $H_0: p = 1/2$ against $H_a: p < 1/2.$
For $n = 16, x = 2$ the P-value of this test is given in R as follows:
binom.test(2, 16, .5, alt="less")

        Exact binomial test

data:  2 and 16
number of successes = 2, number of trials = 16, p-value = 0.00209
alternative hypothesis: 
 true probability of success is less than 0.5
95 percent confidence interval:
 0.0000000 0.3438252
sample estimates:
probability of success 
                 0.125 

The P-value can also be found as follows:
pbinom(2, 16, .5)
[1] 0.002090454

Perhaps a more even-handed approach would
be to test $H_0: p= 1/2$ against $H_a: p\ne 1/2,$
perhaps allowing for the possibility that the
hiring process might be biased in either direction.
It turns out that the P-value of this 2-sided test
is twice the P-value of the 1-sided test.
binom.test(2, 16, .5)

        Exact binomial test

data:  2 and 16
number of successes = 2, number of trials = 16, p-value = 0.004181
alternative hypothesis: 
 true probability of success is not equal to 0.5
95 percent confidence interval:
 0.0155136 0.3834762
sample estimates:
probability of success 
                 0.125 

A major difficulty in interpreting results of these
tests arises in justifying the assumption that "a large pool of equally qualified applicants of whom half are women and half are men" was used--or is available. Without addressing such much more
difficult issues, counts of women hired may lead
to 'statistical significance', but not likely to persuasion. @whuber's Comment provides useful advice.
For data with 350 women hired out of 1150, the P-values of one- and two-sided tests are shown below.
Both of the P-values are very nearly $0,$
indicating that the observed $\hat p =X/n = 350/1150 = 0.304$ is inconsistent with $p = 1/2.$
binom.test(350, 1150, .5, alt="less")$p.val
[1] 2.388788e-41
binom.test(350, 1150, .5)$p.val
[1] 4.777577e-41

