simple conditional probability problem using total probability A theft is committed by one of two suspects, Aaron and Becca.
Initially, there is equal evidence against both of them. In further
investigation at the crime scene, it is found that the guilty
individual had a blood type found in 10% of the population.
Suspect Aaron does match this blood type, whereas the blood type
of Suspect Becca is unknown.
a) Given this new information, what is the probability that
Aaron is the guilty party?
b) Given this new information, what is the probability that
Becca’s blood type matches that found at the crime scene?
for part one I have arrived at the following solution
 P(A) = Aaron is guilty = 0.5
P(B)= $P(A^c)$ = Becca is guilty  = 0.5
P(G) = guilty individual has a blood type  = 0.1
other information 
Aaron matched the blood type found in guilty individuals
part a 
using bayes rule
$ P(A/G) = \frac{P(G/A).P(A)}{P(G)}$
applying law of total probability
$ P(A/G) = \frac{P(G/A).P(A)}{P(G/A).P(A) + P(G/A^c).P(A^c)}$
$P(G/A) = 1$ since A's blood matches with those found in guilty population
$P(G/A^c) = P(G/B)= 0.1$  Since the suspect has 10% chanes of having the blood found in guilty.
$ P(A/G) = \frac{1*0.5)}{1*0.5 + 0.1*0.5} = \frac{10}{11} = 0.90$
any hints about the second part which asks for the probability that beccas blood matches that found at the crime scene given that the guilty have 10 percent chances of having that blood . thanks

If there was a population of A and B pairs with independent blood types, all  being suspected in the manner described 
1) 81% of the pairs, neither is type X
2) 9% of the pairs A only is type X
3) 9% of the pairs B only is type X
4) 1% of the pairs both type X.
suppose A has type X
I'm left with the pairs from 2 and 4   (10% of pairs - that makes sense)
now for each pair, and with equal probability, A & B committed a crime, and left blood - I have then got all of the 1% from 4) come up with left blood X and A has blood X, and I have half of the pairs in 2) Where it was A that left blood X (and A has blood X, if B left blood, blood X wasn't found)
so I've got 1% + 4.5% of pairs are flagged up in the way described, but only 1/2% of them has B guilty (half of the 1%)  so then 0.5 / 5.5 = 1/11 is prob B guilty, so 10/11 A guilty
ok so i agreed with that part
B is type X in 1 of the 5.5% groups (and guilty in half of those), so chance of blood group X is 1 / 5.5 = 2/11
This isn't really the method you wanted though, so sorry it's posted as answer.
