Given the parents do not have the disease we know that for a parent we we have , , and we have the following cases:
parent 1 BB and parent 2 BB with probability
parent 1 BB and parent 1 bB with probability
parent 1 bB and parent 2 BB with probability
parent 1 bB and parent 2 bB with probability
We know that the probability of any children having the disease in the first three of these cases is 0, which leaves the third case
(when the number of children who have the disease is ).