Prior Probability for schizophrenia The chance of a person having schizophrenia in a country is 10%. A family has 3 people. 
Suppose that at least one person has schizophrenia in the family, whats the probability that everyone has schizophrenia?
I used Bayes theorem 
p(D) = having schizophrenia = 0.1
p(A) = Probability of all three having schizophrenia = 0.1^3
p(B) = at least one person having schizophrenia = 1 - complement B = 1-0.9^3 = 0.271
So I got the following:
P(A|B) = P(A)P(B|A)/P(B)= 0.00369
PART 2
Suppose that one is selected from the family and has schizophrenia, whats the probability that everyone has schizophrenia?
P(C) = probability the one selected from family has schizophrenia = 0.1
P(A|C) = P(A)P(C|A)/P(C) = 0.01
I actually got a higher probability for this question so I am wondering whether I have done this right? because given one has schizophrenia shouldn't the probability for the second Q be lower than first Q?
Any help is appreciated.Thanks
 A: The answer to part 2 should be expected to be higher than the answer to part 1. Why? Because in part 1, we weight all of the cases in which at least one person has schizophrenia equally, while in part 2 we weight the cases in which multiple people have it higher, proportionally to the number of people who have it. If $p_k$ is the probability of exactly $k$ people having schizophrenia, the answer to part 1 is $\frac{p_3}{p_1+p_2+p_3}$, while the answer to part 2 is $\frac{3p_3}{p_1+2p_2+3p_3}$. If two people have it, that's two chances to pick someone that does. If three people have it, it's three chances to pick someone that does.
Under the assumptions here, with $p_1$ substantially larger than the other $p_k$, the second probability will be significantly larger.
One note: the assumption of independence used here is blatantly false in the real world. Schizophrenia is strongly inheritable; as a family is assumed to contain people related to each other, the real-world probabilities of multiple people having it will be much higher than the numbers here.
