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In some city, 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. We know that the eyewitness tell the truth in 75% of the cases (and lies on the other 25%). What is the probability that the cab was indeed yellow?

What I have tried:

P(yellow) / [P(yellow) + P(white, but reported yellow)]

0.8 / [0.8 + 0.2(1 - 0.75)] = 0.94

I am unsure if I used the correct thought process for this question.

brownjamba30
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1 Answers1

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Let's assume there are $100$ accidents. In $20$ of those accidents, the cab was white. In $5$ of those $20$ "white" accidents, the cab will be reported yellow.

In the other $80$ accidents, the cab was yellow. But it will only be reported yellow in $60$ of those $80$ accidents.

We know only that the cab was reported yellow. As discussed above, in $60$ of those $65$ reports, the cab was indeed yellow. Thus, the correct answer is $\frac{60}{65}=\frac{12}{13}$. Your answer is not correct because not all of the yellow accidents will actually be reported as yellow. In other words, instead of the $0.8$ a priori probability that the cab was actually yellow, you should be using the smaller $0.6$ probability that the cab was correctly reported to be yellow.

Robert Shore
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  • Correct and well explained - but why answer a duplicate question? – Ethan Bolker Jun 10 '22 at 00:16
  • @EthanBolker Because it wasn't flagged as a duplicate when I answered it, and because I believe the concrete nature of my explanation is absent from prior discussions and a useful contribution to the discussion. – Robert Shore Jun 10 '22 at 01:42