Conditional probability using Bayes's theorem

Question

From Blitzstein, Introduction to Probability (2019 2 edn), Chapter 2, Exercise 25, p 87.

25. A crime is committed by one of two suspects, A and B. Initially, there is equal evidence against both of them. In further investigation at the crime scene, it is found that the guilty party had a blood type found in 10% of the population. Suspect A does match this blood type, whereas the blood type of Suspect B is unknown.

(a) Given this new information, what is the probability that A is the guilty party?

So here is my approach. Let $A$ stands for "A guilty", $G$ for "Individual having the rare matching blood type" and $B$ for "B is guilty". Then using Baye's theorem $$P(A|G)P(G) = P(G|A)P(A)$$ $$P(A|G) = \frac{1 * \frac{1}{2}}{\frac{1}{10}} = \frac{10}{2}$$ Clearly I did something wrong. The solution replaced P(G) with $(\frac{1}{2} + \frac{1}{10})$.

Why am I wrong in saying P(G) is simply $\frac{1}{10}$ because it is the unconditional probability of anybody in the general population having guilty blood type? What did I miss here?

Answer 1

For sake of simplicity, call the blood type $O$.

We should first compute the probability that blood type $O$ is observed at the scene. Given that $A$ is guilty, the probability that blood type $O$ is observed is $1$. Otherwise, given that $B$ is guilty, because their blood type is unknown, there is a $\frac{1}{10}$ probability that blood type $O$ is observed. Since our prior assigns $\frac{1}{2}$ to both suspects $A, B$, then the probability of the evidence is $\frac{1}{2} \left( 1 + \frac{1}{10} \right) = \frac{11}{20}$.

Therefore, the probability that $A$ is guilty increases to $\frac{\frac{1}{2} (1)}{\frac{11}{20}} = \boxed{\frac{10}{11}}$. This makes sense because the blood type is not common.

Answer 2

$P(G_A|B_p) = \frac{P(G_A) \times P(B_p|G_A)}{P(G_A) \times P(B_p|G_A) + P(\neg G_A) \times P(B_p|\neg G_A)}$

$P(G_A|B_p) = \frac{0.5 \times 1}{0.5 \times P(B_p|G_A) + 0.5 \times P(B_p|\neg G_A)}$⁰⁰

$G_A$ = $A$ is guilty
$P(B_p)$ = Blood test is positive

We have what's known as prevalence = $10 \% = \frac{1}{10} = 0.1$

Then there's the blood test itself. Tests have properties, pertinent to the answer, and they are sensitivity (the fraction of those with the blood group in question it reports as positive) and specificity (the fraction of those that don't have the blood group ($0.9$) it reports as negative). Vide supra the denominator.

In short, the probability of A having the incriminating blood group isn't $10 \%$. It's $P(G_A) \times P(B_p|G_A) + P(\neg G_A) \times P(B_p|\neg G_A) = \text{denominator}$.
That is to say we need info on how, sensu latissimo, reliable the blood test is.

In this case $\text{denominator} > 0.5 \iff P(B_p|G_A) = 1$