The question is: 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 ${1\over2}$. Given that she doesn't have the disease, her children won't have it either. A certain mother, who has probability ${1\over 3}$ of having the disease, has two children.
$(a)$ Find the probability that neither child has the disease.
$(b)$ Is whether the first child has the disease independent of whether the second child has the disease? Explain.
$(c)$ The first child is found not to have the disease. A week later, the second child is also found not to have the disease. Given this information, find the probability that the mother has the disease.
I got ${25\over 36}$ for an (a) by calculating $P(\text{a child has no disease})$ and squaring it because the problem said mother to child is independent.
For $(b)$, is it evident that they are independent because the question suggest that mother to child is independent?
And I have a problem with $(c)$.
How can I solve these questions?
Best Answer
Let $$P(Mother\;has\;disease) = P(M);$$ $$P(elder\;child\;has) = P(C1);$$ $$P(younger\;child\;has) = P (C2)$$
(a) $$P(C1^c, C2^c)$$ $$=P(C1^c, C2^c|M) . P(M) + P(C1^c, C2^c|M) x P(M^c)$$ $$={1\over 2} × {1\over 2} × {1\over 3} + 1 × {2\over 3} = {3\over 4}$$
(b) No, they are not independent. If the elder child has the disease, it means the mother must have the disease as it can be passed only through mother, which implies that the probability of younger child getting the disease goes up.
(c) $$P(M^c|C1^c, C2^c)={P(C1^c, C2^c) . P(M^c)\over P(C1^c, C2^c)}= {\frac{1}{4} . {1\over 3}\over {3\over 4}} = {1\over 9}$$