There are two fairly simple approaches I can see that one could take to solve this problem. They both give the same answer of course. But let's use the first approach I'm thinking of: Direct probability calculations.

The question we want to answer takes the form: P(Both people chosen are divorced | At least one of them is divorced). Using the definition of conditional probability we can rewrite this as:

P(Both people chosen are divorced AND at least one of them is divorced) / P(At least one of them is divorced)

The numerator clearly reduces to P(Both people chosen are divorced). We can then look at what the probability of both being divorced is

P(Both people chosen are divorced)

= (# of ways we can choose two divorced people) / (# of ways we can choose two people from this population)

= Choose(612, 2) / Choose(1669, 2)

Where Choose(n, k) = n!/(k!(n-k)!)

The denominator is P(At least one of them is divorced) - we'll take a less direct approach to calculating this by recalling that P(A) = 1 - P(A') where A' is the complement of A.

P(At least one of them is divorced) = 1 - P(It is not true that at least one of them is divorced) = 1 - P(Neither are divorced)

If we're looking to calculate P(Neither are divorced) we can use a similar calculation as before:

P(Neither are divorced)

= (# of ways to choose two people that haven't been divorced) / (# of ways to choose two people from this population)

= Choose(1057, 2) / Choose(1669, 2)

So the probability of interest is:

[Choose(612,2) / Choose(1669,2)] / [1 - Choose(1057,2)/Choose(1669,2)]