Bayes theorem question 
A car accident occurs and of the 5 witnesses, 4 of them saw a green
car, and 1 of them saw a yellow car. In the universe, $85\%$ of cars are
yellow and $15\%$ of cars are green. Witnesses report correctly $2/3$ time.
What is the probability that the car is green?

So I'm trying to figure out $P(G \mid D)$ where $G$ is green and $D$ is the data.
So... the way I think of it is that I have to weight the probability of the data given that the car is green against the total probability of the data (true positives + false positives)
In my mind, that's something like
$$\frac{\text{true positive}}{\text{true positive} + \text{false positive}}$$
The likelihood of a true positive is: $\left(\frac{2}{3}\right)^4 \cdot \frac{1}{3} \cdot 0.15 = 0.0098765432 $ right?
Is the false positive: $\left(\frac{1}{3}\right)^4 \cdot \frac{2}{3} = 0.00699588477$
Am I on track? Am I using the word likelihood correctly?
 A: I calculate a 58.5 % chance that the car is green.   See that attached image
Note that this is the same result that you got, except you didn't normalize your .009876 and .0069958.   I normalized every step, but you don't have to every step.  You do need to normalize at the end though

I like to think of Bayes theorem as a "Normalized Weighted Average", I have a blog post about that here  http://www.fairlynerdy.com/an-intuitive-guide-to-bayes-theorem/
Edit to describe normalizing a bit
here is the bayes theorem equation with some labels.  You did the prior and the likelihood, but not the normalizing constant.  That makes the total probability of all the outcomes sum to 1.0 at the end

In this case the normalizing constant is the sum of all the probabilities given your results, which were 4 green witnesses, 1 yellow.  That sum is (.009876 + .006995)  so the probability the car is green is  .009876 / (.009876 + .006995) = 58.5%
A: We are given $P(G)=15\%$ prior probability and are looking for $P(G|data)$ posterior probability.
By Bayes' Theorem we have
$P(G|data) = \frac{P(data|G)P(G)}{P(data)}$
By law of total probability we have
$P(G|data) = \frac{P(data|G)P(G)}{P(data|G)P(G) + P(data|Y)P(Y)}$
Observe that the data is that there are 4 correct and 1 incorrect if it was green, and vice-versa (usage?) otherwise. Hopefully, one can see the relevance of the binomial distribution here.
Thus,
$P(data|G) = \binom{5}{4}(2/3)^4(1/3)^1 \tag{**}$

 
 
 
 
Appendix
 
 
 
 
Actually $$data := \{\text{W1 reports Y} \cap \text{W2 reports G} \cap \text{W3 reports G} \cap \text{W4 reports G} \cap \text{W5 reports G}\}$$ 
$$\cup \{\text{W1 reports G} \cap \text{W2 reports Y} \cap \text{W3 reports G} \cap \text{W4 reports G} \cap \text{W5 reports G}\}$$
$$\cup \dots \cup \{\text{W1 reports G} \cap \text{W2 reports G} \cap \text{W3 reports G} \cap \text{W4 reports G} \cap \text{W5 reports Y}\}$$
Thus, if the car is green then 
$$data = data_a \tag{*}$$
where
$$data_a := \{\text{W1 is not right} \cap \text{W2 is right} \cap \text{W3 is right} \cap \text{W4 is right} \cap \text{W5 is right}\}$$ 
$$\cup \{\text{W1 is right} \cap \text{W2 is not right} \cap \text{W3 is right} \cap \text{W4 is right} \cap \text{W5 is right}\}$$
$$\cup \dots \cup \{\text{W1 reports G} \cap \text{W2 reports G} \cap \text{W3 is right} \cap \text{W4 is right} \cap \text{W5 is not right}\}$$
and $(*)$ is showing the equivalence of the events which is
$\forall \omega$, sample points $\in \Omega$ the sample space, we have
$$\omega \in data \iff \omega \in data_a$$
if $\omega \in G$
where
$G, Y, data, data_a, \{\text{Wi reports G}\}, \{\text{Wi reports Y}\}$ are events, that is, subsets of the sample space $\Omega$.
So really, the justification for $(**)$ is that
$P(data|G) = P(data_a|G)$
because
$P(data \cap G) = P(data_a \cap G) \tag{***}$
because of (*).
Observe that if $\omega \in Y$, then $(***)$ is simply $0=0$
