Bayesian analysis 
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.


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 $P(G|w) \approx 0.585$.
