Hardy-Weinberg Principle... help with a question?

A population this size is really too small to use Hardy-Weinberg equations accurately, but with these numbers, it comes out exact.  Count how many B alleles there are in the population- there are 40 (20 from the 10 BBs and 10 from the 20 Bbs).  Since there are 80 total alleles and 40 of them are B then the frequency of B is 40/80 = 0.5.  Therefore, the frequency of b must be the rest, or 0.5 also.  In Hardy-Weinberg, p = frequency of dominant allele and q = frequency of recessive allele.

So, p?2 + 2pq  q?2 = 0.25 + 0.5 + 0.25 = 1, which is exactly what you have 0.25 BB, 0.5 Bb, and 0.25 bb.