Bayesian Probability - two criminals, blood, and various things

[Astrofiend]

Homework Statement

I am trying to work out what I believe to be a fairly simple 'odds ratio' problem using Bayesian probability for my physics class. Here is the statement of the problem:

"Two criminals have left traces of their own blood at the scene of a crime. A suspect,
Oliver, is tested and found to have type `O' blood. The blood groups of the two traces
are found to be of type `O' (a common type in the local population, having frequency
60%) and of type `AB' (a rare type, with frequency 1%). Do these data (type `O' and
`AB' blood were found at scene) give evidence in favour of the proposition that Oliver
was one of the two people present at the crime?
(b) What are the odds that Oliver was one of the two people at the scene of the
crime? Is the evidence in favor of this hypothesis?
(c) Suppose Oliver had blood type `AB'. What is the odds ratio now?"

I have tried to calculate part b as below:

Homework Equations

The odds ratio comparing two hypotheses is given by:

<br /> <br /> O_1_2 = \frac{p(M_1 \mid D,I)}{p(M_2 \mid D,I)} = \frac{p(M_1 \mid I) p(D \mid M_1,I) }{p(M_2 \mid I) p(D \mid M_2,I} = \frac{p(D \mid M_1,I)}{p(D\mid M_2,I)}<br /> <br />

Where I have applied this by making M1 the hypothesis that Oliver was present at the crime scene, M2 is the counter hypothesis, D is the data that 'o' type blood was present at the scene and I - the prior information, I'm not sure about.

The Attempt at a Solution

So, I equate p(D \mid M_1,I) to 1, reasoning that, assuming o type blood is left by the criminals, if oliver is there he will leave o type blood.

I equate p(D \mid M_2,I) to 0.6, reasoning that if oliver isn't there, then there is a 60% chance that someone from the general population will leave o type blood.

But I'm not sure what to do next. There is no information in the question that relates to the 'prior' I, which surely comes into this analysis as p(M_1 \mid I) and p(M_2 \mid I) in the equation. This would relate to the likelihood of Oliver being at the scene essentially by random chance as a member of the general population I think, but there is no such info in the question.

Am I completely off track? Am I trying to analyse with the wrong equation? Have I made errors of reasoning? Any help would be much appreciated!

 

[D H]

You can't arbitrarily say p(M_1|I)=p(M_2|I), which is exactly what you did in here:

O_1_2 = \frac{p(M_1 \mid D,I)}{p(M_2 \mid D,I)} = \frac{p(M_1 \mid I) p(D \mid M_1,I) }{p(M_2 \mid I) p(D \mid M_2,I} = \frac{p(D \mid M_1,I)}{p(D\mid M_2,I)}

Consider the case of minimal knowledge: Somebody alive somewhere in the world committed the crime, meaning there are 7 billion equiprobable suspects. In this case, P(M_1|I) = 1/(7e9) and P(M_2|I)=1-P(M_1|I)\approx 1. You wiped out a factor of 1.4×10^-10. Ignoring this factor is a form of the prosecutor's fallacy.

 

[Astrofiend]

Thanks for that - yep, I had come around to that conclusion. What about the other conditional probability assignments that I made? Are they valid?