# Bayes' Ratio: Dramatic Taxicab Example

We already observed that, for certain problems, it may be more convenient to use the Bayes' ratio to evaluate comparative odds of two events than Bayes' theorem itself. For three events $A$, $B$, $C$

$\frac{p(A|C)}{p(B|C)} = \frac{p(C|A)}{p(C|B)}\cdot \frac{p(A)}{p(B)}.$

A. Tversky and D. Kahneman give a dramatic example that makes use of the latter formula:

A cab was involved in a hit and run accident at night. Two cab companies, the Green and the Blue, operate in the city. You are given the following data:

1. 85% of the cabs in the city are Green and 15% are Blue.
2. a witness identified the cab 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?

Solution

### References

1. A. Tversky, D. Kahneman, Evidential impact of base rates, in Judgement under uncertainty: Heuristics and biases, D. Kahneman, P. Slovic, A. Tversky (editors), Cambridge University Press, 1982  A cab was involved in a hit and run accident at night. Two cab companies, the Green and the Blue, operate in the city. You are given the following data:

1. 85% of the cabs in the city are Green and 15% are Blue.
2. a witness identified the cab 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?

### Solution

Let $G$ be the event of the delinquent being Green. Let $B$ be the event of the delinquent being Blue. Finally, let $W$ be the witness' report. Clearly,

\begin{align} \frac{p(B|W)}{p(G|W)} & = \frac{p(W|B)}{p(W|G)}\cdot \frac{p(B)}{p(G)} \\ & = \frac{0.8}{0.2}\cdot \frac{0.15}{0.85} \\ & = \frac{12}{17}. \end{align}

Since $p(G|W) + p(B|W) = 1$, it follows that

$\displaystyle p(B|W) = \frac{12}{12 + 17} \approx 0.41.$

meaning that, in spite of the witness testimony, the hit-and-run cab is more likely to be Green than Blue. 