Confusing Bayes Theorem Example I'm trying to find out the probability that I have a disease given that it is in my family history so $P(disease|history)$.
If the rate of having a disease is $P(disease)=\frac{1}{1000}$ and the rate that those who have the disease have a family history of the disease is $P(history|disease)=1/10$. Then using Bayes' rule:
$P(disease|history) = \frac{P(history|disease)P(disease)}{P(history)}$
Now, I know that I have a family history so $P(history)=1$, right? So the right-hand side of the equation becomes $\frac{1/10000}{1}$, however this means that the probability of having the disease has gone down given that I have a family history even though the priors suggest otherwise. Have I misinterpreted the family history prior $P(history)$?
 A: Any statement of probability depends on some background information, call it $I$. Now we have two scenarios, we can consider background information $I_1$ which is your state of knowledge not knowing whether you have a family history of this disease, or background information $I_2=I_1\land \text{history}$.
It's certainly the case that $P(\text{history}\mid I_2)=1$ assuming $I_2$ isn't self-contradictory, but presumably the data that you refer to as $P(\text{disease})$ and $P(\text{history}\mid\text{disease})$ is $P(\text{disease}\mid I_1)$ and $P(\text{history}\mid\text{disease}\land I_1)$ respectively. Bayes law is then: $$P(\text{disease}\mid\text{history}\land I_1)=\frac{P(\text{history}\mid\text{disease}\land I_1)P(\text{disease}\mid I_1)}{P(\text{history}\mid I_1)}$$
You can't use $P(\text{history}\mid I_2)$ in place of $P(\text{history}\mid I_1)$. If you decide to use $I_2$ everywhere, then $P(\text{history}\mid\text{disease}\land I_2)= 1$ and $P(\text{disease}\mid I_2)$ is already what you want to know by definition!
