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20% of the cabs are white and the other 80% are yellow. A cab was involved in an accident and ran away. An eyewitness to the accident claims that the cab was yellow. Knowing that eyewitness tell the truth in only 75% of the cases, the police officer decides to ask a second independent witness about the accident. The second witness also reports that the cab was yellow.

  1. Compute a new probability that the cab was yellow given two independent witnesses claiming that the cab was yellow.

What I have tried so far is that the P(2 witnesses telling the truth) = (3/4)^2 / [(3/4)^2 + (1/4)^2] = 9/10. Then using Bayes Theorem again, I would account for the color of the cabs by: (0.9 * 0.8) / [(0.9 * 0.8) + (0.1 * 0.2)] = 0.97.

brownjamba30
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    Hint: When two witnesses agree, either they are both telling the truth, or they are both lying. Given that the witnesses are independent, the probability that they are both telling the truth is $(3/4)^2 = (9/16)$ and the probability that they are both lying is $(1/4)^2 = (1/16)$. Therefore, given that the possibility that exactly one of the witnesses is telling the truth has been eliminated, the probability that they are both telling the truth has changed to $$\frac{(9/16)}{(9/16) + (1/16)} = \frac{9}{10}.$$ ...see next comment – user2661923 Jun 12 '22 at 18:25
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    However, this is not the answer to question 1. You still have to use the data that $80%$ of the cabs are yellow and $20%$ of the cabs are not yellow. You need Bayes Theorem, which is actually the approach that I took in the previous comment. Further, you now should have a reasonable tool that will allow you to edit your posting to show work, as indicated in this article on MathSE protocol. – user2661923 Jun 12 '22 at 18:29
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    As a technical note: you also need to assume that lying (or not) is independent of the true color of the cab. Nothing in the problem suggests that, though it seems clear that one is meant to assume it. It makes a big difference if, say, witnesses just prefer to say yellow and are far more inclined to lie when the true color is white. – lulu Jun 12 '22 at 18:48
  • Thank you for the tip. I used the Bayes Theorem accounting for the color of the cabs (as seen in my edit) and got 0.97 as the probability that two independent witnesses claimed that the cab was yellow. – brownjamba30 Jun 12 '22 at 19:56
  • There are numerous cab/witness questions already on the site such as https://math.stackexchange.com/questions/66263/use-bayes-methods-to-find-the-probability-that-the-cab-involved-in-the-accident – Henry Jun 13 '22 at 01:03
  • This is different than any other "similar" question as it considers two independent witnesses rather than one. – brownjamba30 Jun 13 '22 at 02:34

1 Answers1

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What I have tried so far is that the P(2 witnesses telling the truth) = (3/4)^2 / [(3/4)^2 + (1/4)^2] = 9/10.

Well, to be clear, that would be the probability for both witnesses telling the truth given that they make the same claim.

Then using Bayes Theorem again, I would account for the colour of the cabs by: (0.9 * 0.8) / [(0.9 * 0.8) + (0.1 * 0.2)] = 0.97.

Sure, that is correct (to two decimal places), but of course:

Let $A$ be the event that the cab is actually yellow, and $B$ the event that both claim it was yellow.

$$\begin{align}\mathsf P(A\mid B)&=\dfrac{\mathsf P(A)\,\mathsf P(B\mid A)}{\mathsf P(A)\,\mathsf P(B\mid A)+\mathsf P(A^\complement)\,\mathsf P(B\mid A^\complement)}\\[1ex]&=\dfrac{0.80\cdot 0.75^2}{0.80\cdot 0.75^2+0.20\cdot 0.25^2}\\[3ex]&=\dfrac{0.80\cdot 0.75^2/(0.75^2+0.25^2)}{0.80\cdot 0.75^2/(0.75^2+0.25^2)+0.20\cdot 0.25^2/(0.75^2+0.25^2)}\\[1ex]&=\dfrac{0.80\cdot 0.90}{0.80\cdot 0.90+0.20\cdot 0.10}\\[1ex]&=\dfrac{36}{37}\end{align}$$

So we could have skipped "the truth" and focussed just on "the claim" and "the actual".

Still you were not wrong; you just got there with extra steps.

Graham Kemp
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