§1. The a priori probability that a fair coin produces five heads and no tails is 1/2^5=1/32≈3%. So, it is important to note that what we observe is a priori unlikely, if there is no sexual discrimination in Starfleet.
§2. Given h=5 and t=0 (respectively 5 heads and no tails), the posterior probability density function of r (where r is the uniform probability that the hypothetical coin yields heads) conditional on the observed values of h and t is 6*r^5 [
https://en.wikipedia.org/wiki/Checking_whether_a_coin_is_fair]. The expected value of r is therefore E(r)=int(6*r^6,r=0..1)=6/7≈85%.
The expected number of female
Constitution-class captains is therefore 12/7≈2 [
edit - the value 12/7 is actually the expected number of females under the expected distribution, and it should have been designated as an approximation of the actual expected number of females; see §4 for the full calculation showing that the expected number of female starship captains is in fact exactly one]. The corresponding biased coin obeys a Bernoulli distribution [
https://en.wikipedia.org/wiki/Bernoulli_distribution], under which the expected deviation from the mean of 6/7≈.85 is sqrt(6)/7≈.35. Under this expected distribution, the number of male captains would be expected to be in the range 6 to 12 around the expected value of about 10. While 6 female captains are possible under this distribution, having none at all is at least as likely.
§3. The best estimator of r is p=h/(h+t). For confidence interval Z, the error of p as the true value of r is no more than E=Z/(2*sqrt(h+t)), which is in fact a conservative bound when p is not close to .5, as is the case for us. Since h=5 and t=0, we have p=1. Therefore, 90% of the time, conservatively r>63% (with Z=1.6449). So, 90% of the time, no more than 4 female captains are expected [
edit - see §4].