There are a total of $8+4+2+4+2+4=24$ balls in the three containers.
The probability of selecting container I is $1/3$, and the probability of selecting a green ball from container I is $4/12=1/3$.
The probability of selecting container II or III is also $1/3$, and the probability of selecting a green ball from either container is $4/6=2/3$.
Therefore, the probability of selecting a green ball is $(1/3)(1/3)+(1/3)(2/3)+(1/3)(2/3)=1/9+2/9+2/9=5/9$.
The answer is: \frac{5}{9}
Below is a math problem. Provide a detailed, step-by-step solution.

### Problem:
Container I holds 8 red balls and 4 green balls; containers II and III each hold 2 red balls and 4 green balls. A container is selected at random and then a ball is randomly selected from that container. What is the probability that the ball selected is green? Express your answer as a common fraction.

### Answer:
