In a city with one hundred taxis, 1 is blue and 99 are green. A witness observes
ID: 3004294 • Letter: I
Question
In a city with one hundred taxis, 1 is blue and 99 are green. A witness observes a hit-and-run by a taxi at night and recalls that the taxi was blue, so the police arrest the blue taxi driver who was on duty that night. The driver proclaims his innocence and hires you to defend him in court. You hire a scientist to test the witness' ability to distinguish blue and green taxis under conditions similar to the night of accident. The data suggests that the witness sees blue cars as blue 99% of the time and green cars as blue 2% of the time. Write a speech for the jury to give them reasonable doubt about your clients guilt. Your speech need not be longer than the statement of this question. Keep in mind that most jurors have not taken this course, so an illustrative table may be easier for them to understand than fancy formulas.Explanation / Answer
From scientist's observations:
Probability that blue car is seen as blue under test conditions=0.99
Thus, probability that blue car is seen as green under test conditions=1-0.99=0.01
Probability that green car is seen as blue under test conditions=0.02
Thus, probability that blue car is seen as blue under test conditions=1-0.02=0.98
Out of the hundred cars, probability the a car is blue = 0.01
Out of the hundred cars, probability the a car is green = 0.99
Thus probabililty that the witness identified a blue car as blue = probability that the car is blue*probability that the witness identifies a blue car as blue under test conditions = 0.01*0.99=0.0099
Probabililty that the witness identified a green car as blue = probability that the car is green*probability that the witness identifies a green car as blue under test conditions = 0.99*0.02=0.0198
Thus probability that the car identified by the witness as blue was actually a blue car= 0.0099/(0.0099+.0198) = 1/3
So in 1 out of 3 cases the car identified as blue would actually be a blue car. In 2 out of 3 cases, the car identified as a blue car would actually be a green car! So the chances that the witness's identification is wrong is twice than what it is right.