There are three colors of counter, red, green and blue.
There are 5 red, 3 green and 2 blue counters already in the bag.
Another counter is added to the bag (completely randomly).
A counter is chosen from the bag.
Given that the counter chosen is green, what's the probability that the counter added to the bag was green?
I have no idea how to approach this. I know very basic probability and combinatorics.
Let $A$ be the colour of the added counter, and $B$ be the colour of the removed counter, which can be in $\{g,r,b\}$.
You know $\mathsf P(A{=}g) = \mathsf P(A{=}r)= \mathsf P(A{=}b)$, and from the initial number of counters in the bag you can work out $\mathsf P(B{=}g\mid A{=}g), \mathsf P(B{=}g\mid A{=}r), \mathsf P(B{=}g\mid A{=}b)$ , the probabilities that the removed counter is green when given the colour of the added counter.
Then use Bayes' Rule and the Law of Total Probability to obtain what you are looking for: $\mathsf P(A{=}g\mid B{=}g)$, the probability that the added counter was green given that the removed counter was green.