simple conditional probability problem using total probability

Question

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

Answer 1

Calculations are correct, but your logic has gone a bit awry.

$P(G)$ = guilty individual has a blood type $= 0.1$

No.   The guilty individual certainly has the blood type (it is evidence).   An innocent person has that type with probability $0.10$.

That is, the probability the criminal's blood type matches Aaron when given guilt is $\mathsf P(M_A\mid A)=1$, but when given innocence is $\mathsf P(M_A\mid A^\complement)=0.1$ (assuming blood type is independent of all other incriminating evidence).

Now find $\mathsf P(A\mid M_A)$ by the usual methods

$$\mathsf P(A\mid M_A) = \dfrac{\mathsf P(A)\mathsf P(M_A\mid A)}{\mathsf P(A)\mathsf P(M_A\mid A)+\mathsf P(A^\complement)\mathsf P(M_A\mid A^\complement)} = \dfrac{1}{1+0.1}=0.\dot{\overline{90}}$$

Then $\mathsf P(B\mid M_A)=0.\dot{\overline{09}}$ and, by total probability, the probability that the blood type will match Becca when given that it matches Aaron is::

$$\begin{align}\mathsf P(M_B\mid M_A) ~=~& \mathsf P(M_B\mid A)\mathsf P(A\mid M_A)+\mathsf P(M_B\mid B)\mathsf P(B\mid M_A) \\[1ex] \vdots~& \\[1ex]~=~& \bbox[white]{\color{white}{\frac 2{11}}}\end{align}$$

Answer 2

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.