Question

I'm looking for feedback on the problem below from Jeffreys' probability primer. I think (a) is 0.0009 and (b) is 1 in 1000. Is this correct?

(a) In an urn with 1000 balls, one is green and the rest are red. A ball is drawn at random and seen by no one but a slightly colorblind witness, who reports that the ball was green. What is your probability that the witness was right on this occasion, if his reliability in distinguishing red from green is .9, i.e., if pr(He says it is X|It is X) = .9 when X = Red and when X = Green?

(b) Suppose that the balls were all numbered, from 1 to 1000, and the witness knows this fact. A ball is drawn, and he tells me that it was numbered 25, what is the probability that he is right?" In answering you are to "assume that, there being no apparent reason why he should choose one number rather than another, he will be likely to announce all the wrong ones equally often."

Thanks!

Answer 1

Part (a) involves applying Bayes' theorem to combine the witness' reliability at identifying colors with the base rate probability of drawing the green ball. Part (b) is straightforward since the probability of the witness correctly identifying ball number 25 is simply the probability of that ball being drawn, which is 1 in 1000.

Step-by-step explanation:

For part (a), let's consider the witness' reliability and the actual chance of drawing a green ball. The probability the witness is correct, Pr(Correct), given they say the ball is green, combines the chance of them being correct when they say it's green (Pr(He says Green | It is Green) = 0.9) and the initial chance of drawing the green ball (Pr(Green) = 1/1000). Using Bayes' theorem:

Pr(Green | He says Green) = [Pr(He says Green | It is Green) * Pr(Green)] / [Pr(He says Green)].

Where Pr(He says Green) combines the true positives and false positives—witness claims green when it is green, and when it's actually red but he misidentifies it. Therefore, Pr(He says Green) = 0.9 * 1/1000 (true positive rate) + 0.1 * 999/1000 (false positive rate), as he has a 10% chance of misidentifying any of the red balls as green.

After calculating, you'll see that (a) is not 0.0009 as suggested, but rather a different, higher value once you account for both the witness's accuracy and the base rates.

For part (b), if the witness claims the ball is number 25, since he's equally likely to announce all the wrong numbers and there are 999 wrong numbers, the probability of him being right is just the probability of drawing ball number 25, which is 1 in 1000. So, the answer to (b) is indeed 1 in 1000.