1. **Problem statement:** We have 4 red, 3 blue, and 5 green identical balls in a bag. We pick 3 balls at random without replacement. We want to find the probability that all three balls are of different colors. 2. **Total number of balls:** $$4 + 3 + 5 = 12$$ 3. **Total ways to pick 3 balls from 12:** Using combinations, the total number of ways is $$\binom{12}{3} = \frac{12!}{3! \times 9!} = 220$$ 4. **Number of favorable outcomes (3 balls all different colors):** We want one red, one blue, and one green ball. Number of ways to pick one red ball: $$\binom{4}{1} = 4$$ Number of ways to pick one blue ball: $$\binom{3}{1} = 3$$ Number of ways to pick one green ball: $$\binom{5}{1} = 5$$ Total favorable ways: $$4 \times 3 \times 5 = 60$$ 5. **Probability calculation:** $$P(\text{3 different colors}) = \frac{\text{favorable outcomes}}{\text{total outcomes}} = \frac{60}{220} = \frac{3}{11}$$ **Final answer:** $$\boxed{\frac{3}{11}}$$