1
$\begingroup$

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

$\endgroup$
  • $\begingroup$ what's ten divided by eleven equal to? $\endgroup$ – Cato Sep 21 '16 at 11:27
  • $\begingroup$ From where we are. A has the rare blood group, then in 10% of cases, B also has the same blood group - in that case they are equal again, and there is therefore an overall 5% chance that B is guilty - leaving A at 95%, born out by the 90% chance only he has the rare blood group, and pus the 5% chance B has the blood group, but A did it - making A apparently 95% $\endgroup$ – Cato Sep 21 '16 at 11:32
  • $\begingroup$ B) is 10% isn't it? B's blood type is independent of A's blood type - so it is still 10% $\endgroup$ – Cato Sep 21 '16 at 11:35
  • $\begingroup$ I'm not sure about this, i'm thinking $\endgroup$ – Cato Sep 21 '16 at 11:36
  • $\begingroup$ @Andrew Deighton any more thoughts on this thanks $\endgroup$ – alex Sep 21 '16 at 12:11
2
$\begingroup$

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::

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

$\endgroup$
  • $\begingroup$ can you give the complete answer @Graham Kemp $\endgroup$ – alex Sep 21 '16 at 12:43
  • $\begingroup$ @alex Yes, but can you try it yourself first? Next step, apply the Law of Total Probability partitioning on who's guilty. $\endgroup$ – Graham Kemp Sep 21 '16 at 12:49
  • $\begingroup$ I m not getting it @Graham Kemp $\endgroup$ – alex Sep 21 '16 at 13:37
  • $\begingroup$ @Graham Kemp - thanks for posting this, very informative. $\endgroup$ – Cato Sep 21 '16 at 16:49
0
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.