Bayesian analysis

Question

A New York City cab was involved in a hit-and-run accident last night. Five witnesses reported the incident, four of whom said that the cab was green and one of whom said that the cab was yellow. Assume each witness correctly identifies the colour of a cab with probability 2/3. It is known that 85% of registered cabs in New York City are yellow and 15% are green. Based on this information, what is the probability that the cab was green?

Answer choices:

• 41.5%
• 15.0%
• 58.5%
• 88.9%

My Progress

• I don't know what the prior probability is and have particular difficulty calculating the likelihood.

• I attempted to use binomial expansion to calculate the likelihood this way:

$$\frac{5!}{4!(5-4)!} \times 0.15^4 \times 0.85^1$$

• But when I calculate the likelihood and subsequently the posterior probability, it does not match any of the above choices.

Answer 1

We want to compute what is the probability that the car was green (G), given the witnesses, $P(G|w)$.

From Bayes theorem, we need to compute

$$P(G|w) = \frac{P(w|G)P(G)}{P(w)}$$

We know $P(G)=0.15$ and $P(Y)=0.85$ (Y stands for yellow). Lets compute $P(w)$.

By definition, $$P(w) = \sum_{c\in \{G,Y\}} P(w,c) = \sum_{c\in \{G,Y\}} P(w|c)P(c) = P(w|G)P(G) + P(w|Y)P(Y).$$

Lets compute $P(w|c)$. This is the probability of the particular witnesses outcome given the color of the car. The probability of 4 out of 5 observing G given $c=G$ is $P(w|G) = 5\times (\frac{2}{3})^4\frac{1}{3}$. Likewise, $P(w|Y) = 5\times (\frac{1}{3})^4\frac{2}{3}$.

Plugging in in the equation of $P(w)$, and in the equation for $P(G|w)$ in the numerator, and doing the math, you should get to one of the answers.

Checking @lulu answer, indeed this gives $P(G|w) \approx 0.585$.

Answer 2

Your prior is $.15$ since, at the start, you have no reason to reject the hypothesis that cab color is independent of accident proclivity. That is, knowing that a cab was involved in an accident should tell us nothing about the color.

Now, we have witness testimony and we must incorporate that new information. There are two ways we could have obtained this particular witness testimony:

Path I: The cab really was Green. Given our prior, we think this is a $.15$ probability event. Given that the cab was Green the probability that exactly $4$ out of $5$ witnesses would report "Green" is $5\times \left( \frac 23 \right)^4\times \frac 13$. Thus this path contributes $$.15\times 5\times \left( \frac 23 \right)^4\times \frac 13=0.049382716$$

To the total probability.

Path II: The cab really was Yellow. Given our prior, we think this is a $.85$ probability event. Given that the cab was Yellow the probability that exactly $4$ out of $5$ witnesses would report "Green" is $5\times \left( \frac 13 \right)^4\times \frac 23$. Thus this path contributes $$.85\times 5\times \left( \frac 13 \right)^4\times \frac 23=0.034979424$$

To the total probability.

Finally, the total probability is $0.034979424+0.049382716=0.08436214$ and the answer you want is then $$\frac {0.049382716} {0.08436214}=0.585365854\sim \fbox {.585}$$