Hardy Weinberg principle?

In the Hardy-Weinberg formula, p + q = 1, where p is the frequency of one allele and q is the frequency of the other allele (often p is used as the frequency of the dominant allele and q is used as the frequency of the recessive allele).  Since there are only two, then the total has to add to 1.

If you square both sides of the above equation, you get:  
p^2 + 2pq + q^2 = 1, where p^2 is the frequency of homozygotes for one allele, 2pq is the frequency of heterozygotes and q^2 is the frequency of homozygotes for the other allele.

So, if q^2 = 1/17000 = 0.0000588, then q = 0.00767
so p = 1 - 0.00767 = 0.99233 (the frequency of the dominant allele).  As an aside, note that the frequency of homozygous dominant individuals would be p^2 = 0.99233 * 0.99233 = 0.9847

and finally, the frequency of heterozygotes (carriers of Albinism) is:  2pq = 2 * 0.99233 * 0.00767 = 0.0152 (or 1.52%)