1. **State the problem:** We have a bag with 6 red, 7 blue, and 3 green marbles, totaling $6 + 7 + 3 = 16$ marbles. We want to find the probability that when drawing 3 marbles without replacement, all 3 are red. 2. **Formula for probability without replacement:** $$P(\text{all red}) = \frac{\text{number of ways to choose 3 red marbles}}{\text{number of ways to choose any 3 marbles}} = \frac{\binom{6}{3}}{\binom{16}{3}}$$ 3. **Calculate combinations:** $$\binom{6}{3} = \frac{6!}{3!\times(6-3)!} = \frac{6 \times 5 \times 4}{3 \times 2 \times 1} = 20$$ $$\binom{16}{3} = \frac{16!}{3!\times(16-3)!} = \frac{16 \times 15 \times 14}{3 \times 2 \times 1} = 560$$ 4. **Calculate probability:** $$P = \frac{20}{560}$$ 5. **Simplify the fraction:** $$P = \frac{\cancel{20}}{\cancel{560}} = \frac{1}{28}$$ 6. **Final answer:** The exact probability that all three marbles drawn are red is $$\boxed{\frac{1}{28}}$$