Suppose the city has a total of 100 cabs. This means 85 are Green and 15 are Blue. The witness correctly identifies the colour of a cab 80% of the time and wrongly identifies the colour of a cab 20% of the time. In other words, of the 85 Green cabs, the witness would, on average, identify 68 of them (80% of 85) as Green and 17 of them (20% of 85) as "Blue". Likewise, of the 15 Blue cabs, the witness would, on average, identify 12 of them (80% of 15) as Blue and 3 of them (20% of 85) as "Green". So overall, out of these 100 cabs, the witness "sees" 29 of them as Blue (17 "false" Blues plus 12 true Blues). Therefore, the probability that the cab involved in the accident was Blue rather than Green is 12/29, or approximately 41%.
Wednesday, November 24, 2010
A Base-ic Solution
The "obvious" and wrong answer to yesterday's question is 80%. It's wrong because it ignores the base rate information that we have been given, namely that 85% of the city's cabs are Green and 15% are Blue. Again, Bayes' Theorem is the standard mathematical technique to unravel this problem. But here's a simple, common-sense solution: