This question looks interesting.

Note that from the specification we have , since we have and of computers of the two different versions, and once we sampled there are computers of 1 version, we will stop. So we must have (pigeonhole principle) and hence the bound.

In order to make sure which version of computer is more, you will need to check that there are more than computer of that version. So we have and similarly, by pigeonhole principle, . i.e. The support of is

If you are able to determine the version on the -th check, it means that you have checked computers of a certain version in the previous checks, and have that version again on the -th check. So the probability mass function of is given by

where as stated earlier.