Very wordy. I like my answer better
SOLUTION: Let's try some smaller numbers and see what happens. Obviously if the house is all men or all women, the sex of the last person beamed up will be determined. If there are equal numbers of men and women, then by symmetry, the probability that the last person beamed up is a woman would be 1/2. So the simplest interesting case is, say, one man and two women.
In that case, if the man is beamed up first (probability: 1/3), the last person beamed up will be a woman. Suppose a woman is beamed up first; if she is followed by a man (who is then beamed back down), we are down to the symmetric case where the probability of ending with a woman is 1/2. Finally, if a second woman follows the first (probability 2/3 x 1/2 = 1/3), the man will be last to be beamed up. Putting the cases together, we get probability 1/2 that the last person beamed up is a woman. Is it possible that 1/2 is the answer no matter how many men and women are present, as long as there's at least one of each?
Looking more closely at the above analysis, it seems that the sex of the last person beamed up is determined by the next-to-last saucer — the one that reduces the house to one sex. To see why this is so, it is useful to imagine that the Xylofonian acquisition process operates the following way: Each time a flying saucer arrives, the current inhabitants of the house arrange themselves in a uniformly random permutation, from which they are beamed up left to right.
For example, if the inhabitants at one saucer's arrival consist of males Amit and Boris and females Carol, Dina and Esme, and they arrange themselves "Dina, Esme, Boris, Carol, Amit," then the saucer will beam up Dina, Esme, and Boris, then will beam Boris back down again, and take off with just the females Dina and Esme. The remaining folks, Boris, Amit, and Carol, will now re-permute themselves in anticipation of the next saucer's arrival.
We see that a saucer will be the next to last just when the permutation it encounters consists of all men followed by all women, or all women followed by all men. But no matter how many of each sex are in the house at this point, these two events are equally likely! Why? Because if we simply reverse the order of a such a permutation, we go from all-men-then-all-women to all-women-then-all-men, and vice versa.
There's just one more observation to make: If both men and women are present initially, then one saucer will never do, thus there always will be a next-to-last saucer. When that comes — even though we do not know in advance which saucer it will be — it is equally likely to depart with the rest of the men, or the rest of the women.