A cab was involved in a hit and run accident at night.
probability/cab Added material
Problem
A cab was involved in a hit and run accident at night. Two cab companies,
the Green and the Blue, operate in the city. Here is some data:
a) Although the two companies are equal in size, 85% of cab
accidents in the city involve Green cabs and 15% involve Blue cabs.
b) A witness identified the cab in this particular accident as Blue.
The court tested the reliability of the witness under the same circumstances
that existed on the night of the accident and concluded that the witness
correctly identified each one of the two colors 80% of the time and failed
20% of the time.
What is the probability that the cab involved in the accident was
Blue rather than Green?
If it looks like an obvious problem in statistics, then consider the
following argument:
The probability that the color of the cab was Blue is 80%! After all,
the witness is correct 80% of the time, and this time he said it was Blue!
What else need be considered? Nothing, right?
If we look at Bayes theorem (pretty basic statistical theorem) we
should get a much lower probability. But why should we consider statistical
theorems when the problem appears so clear cut? Should we just accept the
80% figure as correct?
Show solution
The police tests don't apply directly, because according to the wording, the witness, given any mix of cabs, would get the right answer 80% of the time. Thus given a mix of 85% green and 15% blue cabs, he will say 20% of the green cabs and 80% of the blue cabs are blue. That's 20% of 85% plus 80% of 15%, or 17%+12% = 29% of all the cabs that the witness will say are blue. Of those, only 12/29 are actually blue. Thus P(cab is blue | witness claims blue) = 12/29. That's just a little over 40%. Think of it this way... suppose you had a robot watching parts on a conveyor belt to spot defective parts, and suppose the robot made a correct determination only 50% of the time (I know, you should probably get rid of the robot...). If one out of a billion parts are defective, then to a very good approximation you'd expect half your parts to be rejected by the robot. That's 500 million per billion. But you wouldn't expect more than one of those to be genuinely defective. So given the mix of parts, a lot more than 50% of the REJECTED parts will be rejected by mistake (even though 50% of ALL the parts are correctly identified, and in particular, 50% of the defective parts are rejected). When the biases get so enormous, things starts getting quite a bit more in line with intuition. For a related real-life example of probability in the courtroom see People v. Collins, 68 Cal 2d319 (1968).
Additional material
In article <173611398FS86.NELSON@msu.edu> NELSON@msu.edu (Doug Nelson) writes: >Assume that the testing was done by presenting an equal number each of >green and blue cards, and that all answers were accurately recorded (no >hard of hearing officer doing the testing, etc.). What is the >assumption being made in presenting/using the 80% correct figure, and >what is the real range of probabilities, given that you have a better >understanding of the test results? The standard solution assumes that the witness is equally likely to be wrong about the color of a green cab as to be wrong about the color of a blue cab. Another interpretation is that the witness was tested by being shown a large number of blue or green cards and being asked to identify them, and the witness was correct in 80% of the cases. For example, we may have shown the witness 100 cards and gotten 80 correct and 20 incorrect answers. But this doesn't fully specify what the results of the test were. For example, if we showed 50 blue cards and 50 green, and the witness identified 10 of the blue cards as green and 10 of the green cards as blue, then we get the answer in the archive. But suppose *all* the witnesses' errors occurred when he was looking at a card that was blue and mistakenly said it was green. (This can occur only if at least 20% of the test cards were blue, of course.) In this case the witness never said blue unless the card actually was blue. So our best estimate for the likelihood that the cab was blue, given that the witness said it was, is 100%. (The full story is more complicated, of course, because the test results might have been due to luck rather than the witness's accuracy. But that's a question of confidence levels, not estimated likelihood.) On the other hand, suppose all the witnesses' errors occurred when he was looking at a card that was green and mistakenly said it was blue. (This can occur only if at least 20% of the test cards were green, of course.) Assuming equal numbers of cards of each color, that means the witness is mistakenly identifying 40% of green cards as blue. The results if such a witness were to view 100 random-selected accidents in the city would be: 15 accidents caused by blue cabs, witness says "blue" 34 accidents (.4*85) caused by green cabs, witness says "blue" 51 accidents (.6*85) caused by green cabs, witness says "green" So 49 times out of 100, the witness will say "blue," but he'll be right only 15/49 (about 30.6%) of the time. -- David A. Karr (karr@cs.cornell.edu)