p = dominant allele frequency
q = recessive allele frequency
p^2 = frequency of the homozygous dominant individuals
2pq = frequency of the heterozygous individuals
q^2 = frequency of the homozygous recessive individuals
p^2 + 2pq + q^2 = 1
Since 1 in 1000 are born with Tay-Sachs that is the frequency of aa, the homozygous recessive.
We also need to find q the recessive allele frequency,
q^2 = 1/1000
q = 0.0316
The probability is equal to
since we know q, we can find p,
p=1-q = 0.968
0.0612 is the proportion of the population that is heterozygous. Therefore, to find the probability that two people both have it we just square this number.