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A witness testifying in a hit-and-run case indicates that a green cab hit the plaintiff in the case. You are aware that 85% of the cabs in the city are blue and the other 15% are green. However, the defense attorney conducts an experiment recreating the conditions at the scene, and the witness, when shown green and blue cabs is only able to correctly differentiate them 80% of the time. Assume that prior to the witness' testimony, you would believe there was a 15% probability that a green cab was involved. Thus, what is the probability that in fact a green cab was involved in the accident?

I know that this is the equation I need to use P(G|W) = P(W|G)P(G)/P(W) To get P(W|G) tried multiplying .80 (how many times the wirness can get them right) by .15 (the number of green cars in the city) to get .12 I then multipled that by .15 which is the probability the car was green then divide by .80 the probability the witness was correct. I got .0225 which was not correct.

StubbornAtom
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mad
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    Welcome to MSE. You'll get a lot more help, and fewer votes to close, if you show that you have made a real effort to solve the problem yourself. What are your thoughts? What have you tried? How far did you get? Where are you stuck? This question is likely to be closed if you don't add more context. Please respond by editing the question body. Many people browsing questions will vote to close without reading the comments. – saulspatz Feb 14 '20 at 23:50
  • Titles are also important. There are a lot of questions about probability and Bayes' Rules. There are somewhat less questions about blue and green cabs or hit-and-run. It might help searchers to focus on those keywords. – Graham Kemp Feb 14 '20 at 23:59
  • Ohhh thank you. – mad Feb 15 '20 at 00:44

1 Answers1

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Think about it this way. Suppose (somehow) the same thing happens $100$ times. Initially we think that a blue cab will be involved $85$ times and a green cab $15$ times. In the first case, the witness will correctly say the cab is blue $68$ times ($80\%$ of $85$) and incorrectly say it was green $17$ times. In the cases where the cab is green, he correctly says it's green $12$ times and says it's blue $3$ times.

So he says the cab is green $29$ times, and he's right $12$ times, and our new estimate that the cab is green is $$\frac{12}{29}\approx41.38 \%$$

You should try to fit this into your Bayes' rule formula.

saulspatz
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