October 21, 2013 6 Comments
“In a certain town blue and green cabs operate in a ratio of 85 to 15, respectively. A witness identifies a cab in a crash as green, and the court is told [based on a test] that in the relevant light conditions he can distinguish blue cabs from green ones in 80% of cases. [What] is the probability (expressed as a percentage) that the cab involved in the accident was blue?” (See my notes on Cohen for a discussion of alternatives.)
For bonus points …. if you were involved , what questions might you reasonably ask before estimating the required percentage? Does your first answer imply some assumptions about the answers, and are they reasonable?
My thoughts below:
If, following Good, we use
P(A|B:C) to denote the odds of A, conditional on B in the context C,
Odds(A1/A2|B:C) to denote the odds P(A1|B:C)/P(A2|B:C), and
LR(B|A1/A2:C) to denote the likelihood ratio, P(B|A1:C)/P(B|A2:C).
Then we want P(blue| witness: accident), which can be derived by normalisation from Odds(blue/green| witness : accident).
We have Odds(blue/green: city) and the statement that the witness ”can distinguish blue cabs from green ones in 80% of cases”.
Let us suppose (as I think is the intention) that this means that we know Odds(witness| blue/green: test) under the test conditions. This looks like a job for Bayes’ rule! In Odds form this is
Odds(A1/A2|B:C) = LR(B|A1/A2:C).Odds(A1/A2:C),
as can be verified from the identity P(A|B:C) = P(A&B:C)/P(B:C) whenever P(B:C)≠0.
If we ignore the contexts, this would yield:
Odds(blue/green| witness) = LR(witness| blue/green).Odds(blue/green),
as required. But this would only be valid if the context made no difference. For example, suppose that:
Green cabs have many more accidents than blue ones.
The accident was in an area where green cabs were more common.
The witness knew that blue cabs were much more common than green and yet was still confident that it was a green cab.
In each case, one would wish to re-assess the required odds. Would it be reasonable to assume that none of the above applied, if one didn’t ask?
My notes on probability.