If the two selected balls have different colors, what is the probability that the selected ball from the first box was green?

Question

There are two boxes. The first box contains $2$ red balls and $5$ green balls; the second box contains $8$ red balls and $10$ green balls. One ball is selected from each box. Each ball is equally likely to be selected. If the two selected balls have different colors, what is the probability that the selected ball from the first box was green?
(a) $0.47619$
(b) $0.66667$
(c) $0.33333$
(d) $0.41667$

I can't really find an answer. I tried using Bayes' theorem. However, I wasn't met with an choice, $P(B_1 \mid G)$. I can't have any idea to find it. I got about $0.425$ for answer. However, nothing of those options.

Answer 1

Using the definition of conditional probability $(P(A \mid B) = P(A \cap B)/P(B)$), we have \begin{align} & P(\text{first ball is green} \mid \text{two balls have different colors}) \\ &= \frac{P(\text{two balls have different colors}, \text{first ball is green})}{P(\text{two balls have different colors})} \\ &= \frac{P(\text{first green, second red})}{P(\text{first green, second red}) + P(\text{first red, second green})} \\ &= \frac{(5/7)(8/18)}{(5/7)(8/18) + (2/7)(10/18)}. \end{align}

Answer 2

Knowing that the colors are different, there were 20 ways to draw red/green and 40 ways to draw green red, so 40 out of 60 ways of 66.6666… % the number from the first box is green.