Conditional Probability Summation Rule Problem

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 gulity", $M$ for "A matching the blood type" and $N$ for "B matching the blood type". Then $$P(A|M) = P(A|M,N)P(N|M) + P(A|M,N^{C})P(N^{C}|M)$$ $$P(A|M) = \frac{1}{2}\frac{1}{10} + 1\frac{9}{10} = \frac{19}{20}$$ Here it is inferred that the blood type of B is independant of that of A and that if both have the matching blood type they are equally likely to be guilty.

Where is the flaw in the application of the summation rule above? The correct answer is $\frac{10}{11}$.

Answer 1

Let the blood type found at crime seen be X.

Probability that A is gulity prior to the new evidence $P(A)=1-P(B)=0.5$

Probability that blood type of X is found given A is gulty $=$ Probability that A has the blood type $P(X|A)=1$

Probability that blood type of X is found given B is guilty $=$ Probability that B has the blood type $P(X|B)=0.1$

Now it is known that blood type is X, and given exactly one of A or B is guilty,

$$P(A|X) = \frac{P(X|A)P(A)}{P(X)} = \frac{P(X|A)P(A)}{P(X|A)P(A)+P(X|B)P(B)} = \frac{10}{11}$$

Answer 2

Using Bayes' theorem: $$ \Pr(A\mid M) = \frac{\Pr(M|A) \Pr(A)}{\Pr(M\mid A) \Pr(A) + \Pr(M \mid B) \Pr(B)} $$ and $\Pr(A) = \Pr(B) = \tfrac{1}{2}$ since both have equal evidence against them, $\Pr(M\mid A) = 1$ since match is certain if $A$ is indeed guilty, and $\Pr(M \mid B) = \Pr(M) = \tfrac{1}{10}$ since the match of $A$ is not dependent on the guilt of $B$, you get $\Pr(A \mid M) = \tfrac{10}{11}$.