Witness Reliability and Bayes' Theorem
A murder has been committed. A witness claims the murderer drove a red car. In this city, all cars are either red or blue, with a fraction $r$ of all cars being red. The witness correctly identifies the color of a car with probability $a$ (and is wrong with probability $1-a$), regardless of the car's actual color.
Given the witness's statement that the car was red, what is the probability the car was actually red?
Evaluate for $r = 13/20$ and $a = 3/5$.
Hints
- Condition on the witness saying "red" and apply Bayes' theorem. You need the prior probability of a red car and the likelihood that the witness says red given the actual color.
- The witness says red either because the car IS red and the witness is correct (probability $ar$), or because the car is blue and the witness is wrong (probability $(1-a)(1-r)$).
- The posterior is $\frac{ar}{ar + (1-a)(1-r)}$. For the numerical answer, compute the numerator $\frac{3}{5} \cdot \frac{13}{20} = \frac{39}{100}$ and denominator terms.
Worked Solution
How to Think About It: This is a classic Bayes' theorem problem where you need to update a prior belief (the base rate of red cars) using noisy evidence (the witness statement). The witness says "red," but they are only right with probability $a$. The question is whether this testimony moves the needle, and by how much. Your gut should say: if most cars are already red ($r > 0.5$) and the witness says red, the posterior should be higher than the prior -- the testimony reinforces the base rate. But the witness is not very reliable ($a = 0.6$), so the update should be moderate.
Quick Estimate: Prior probability of red is $r = 0.65$. The witness says red and is right 60% of the time. Rough Bayesian reasoning: the "evidence for red" is $a \cdot r = 0.6 \times 0.65 = 0.39$. The "evidence for blue but witness says red" is $(1-a)(1-r) = 0.4 \times 0.35 = 0.14$. Posterior $\approx 0.39 / (0.39 + 0.14) = 0.39/0.53 \approx 0.736$. So the witness testimony bumps the probability from 65% to about 74%.
Approach: Apply Bayes' theorem directly, conditioning on the witness saying "red."
Formal Solution:
Let $R$ = the car is red, and $W$ = the witness says the car is red.
- Prior: $P(R) = r$, $P(R^c) = 1 - r$
- Likelihood: $P(W | R) = a$ (witness is correct when car is red)
- False positive: $P(W | R^c) = 1 - a$ (witness is wrong when car is blue)
By Bayes' theorem:
$$P(R | W) = \frac{P(W|R) \cdot P(R)}{P(W|R) \cdot P(R) + P(W|R^c) \cdot P(R^c)}$$
$$= \frac{ar}{ar + (1-a)(1-r)}$$
Plugging in $r = 13/20$ and $a = 3/5$:
$$P(R|W) = \frac{\frac{3}{5} \cdot \frac{13}{20}}{\frac{3}{5} \cdot \frac{13}{20} + \frac{2}{5} \cdot \frac{7}{20}}$$
$$= \frac{\frac{39}{100}}{\frac{39}{100} + \frac{14}{100}} = \frac{39}{53}$$
Answer: $P(R|W) = \dfrac{ar}{ar + (1-a)(1-r)}$. For $r = 13/20$, $a = 3/5$: $P(R|W) = \dfrac{39}{53} \approx 0.736$.
Intuition
This is one of the most classic examples of Bayesian updating and it illustrates a point that trips up many people: the reliability of evidence depends on the base rate. Here the witness is only 60% accurate, which is barely better than a coin flip. But because red cars already make up 65% of the population, the testimony pushes the posterior from 65% to about 74% -- a modest update.
The really interesting case is when the base rate and the evidence conflict. If red cars were rare (say $r = 0.1$) but the witness says red with $a = 0.6$ accuracy, the posterior would be $0.06/(0.06 + 0.36) \approx 14\%$ -- the witness says red but the car is still probably blue. This is the same phenomenon as the false positive paradox in medical testing: unreliable evidence about rare events is dominated by false positives. In a trading context, this is exactly how you should think about analyst recommendations or news signals -- always weight the signal reliability against the base rate.