A certain hereditary disease can be passed from a mother to her children. Given that the mother has the disease, her children independently will have it with probability 1=2. Given that she doesn't have the disease, her children won't have it either. A certain mother, who has probability 1=3 of having the disease, has two children.

(a) Find the probability that neither child has the disease.

(b) Is whether the elder child has the disease independent of whether the younger child has the disease? Explain.

(c) The elder child is found not to have the disease. A week later, the younger child is also found not to have the disease. Given this information, nd the probability that the mother has the disease.

(a) P=0.694

(b) It is independent, beacuse the probability of having the disease for the children depends only on her mother condition (if she has the disease or not), not the condition of his brothers and sisters.

(c) P=0.25

Step-by-step explanation:

(a) If the mother has 0.33 probabilities of having the disease, the probability of the children having the disease is equal to the product of the probability of the mother having the disease (0.33) and the probability of inherit it (0.50).

So the probability of one child of having the disease is 0.33*0.5=0.167. The probability of not having the disease is then (1-0.167) = 0.833

The probability of both children to not have the disease is 0.833^2=0.694.

(b) It is independent, beacuse the probability of having the disease for the children depends only on her mother condition (if she has the disease or not), not the ones of his brothers and sisters.

(c) If the mother has the disease, the child have a probability of 0.5 of having the disease.

The probability, given that the mother has the disease, of both child not having the disease is 0.5^2=0.25.