Simple Bayesian probability question Assume that I evaluate the probability that a murder suspect is guilty to 10%. Then I find a blood trace at the scene and only 1% of the population has this blood type. The suspect has this blood type and the test is accurate (say 100% accuracy for simplicity). What is now the probability that the suspect is guilty?
posterior = prior x likelihood / normalization = .1 x 1 / .01 = 10
What am I missing here?
 A: Short answer: The denominator is missing a term. It's missing the "innocent and evidence" bit which is .009
Long answer:
You want to calculate the probability of guilty given that we have the evidence of the rare blood type. Assuming that if someone is guilty, we are always going to find his blood type as evidence then:
$P(Guilty|Evidence) = \frac{P(Evidence|Guilty)  P(Guilty)}{P(Evidence)}$
$P(Guilty|Evidence) = \frac{P(Evidence|Guilty)  P(Guilty)}{P(Guilty)P(Evidence|Guilty) + P(NotGuilty)P(Evidence|NotGuilty)}$ 
$P(Guilty|Evidence) = \frac{1 \times 0.10}{0.10 \times 1 + 0.90 \times  0.01}$
$P(Guilty|Evidence) = 0.917$
A: You're calculating the posterior probability of a particular suspect being guilty, not a random person from the general population being guilty. Therefore, the normalisation term is not the probability of someone in the general population having the blood type, it's the probability of the suspect having that blood type given the prior probability that they're guilty. Since they must have the blood type if they're guilty, this is equal to
$$\begin{align*}
\mathbb{P}(\textrm{suspect has blood type}) =& \mathbb{P}(\textrm{suspect guilty})\\
&+ \mathbb{P}(\textrm{suspect innocent}) \mathbb{P}(\textrm{suspect has blood type if innocent})\\
=& 0.1 + 0.9 \times 0.01\\
=& 0.109,
\end{align*}$$
which leads to a valid posterior probability.
