1. **State the problem:** We have a bag with 5 red, 3 blue, and 6 green marbles, totaling $5 + 3 + 6 = 14$ marbles. We draw 3 marbles without replacement and want the probability that all 3 are green. 2. **Formula for probability without replacement:** $$P(\text{all green}) = \frac{\text{number of ways to choose 3 green}}{\text{number of ways to choose any 3}} = \frac{\binom{6}{3}}{\binom{14}{3}}$$ 3. **Calculate combinations:** $$\binom{6}{3} = \frac{6!}{3!\times(6-3)!} = \frac{6\times5\times4}{3\times2\times1} = 20$$ $$\binom{14}{3} = \frac{14!}{3!\times(14-3)!} = \frac{14\times13\times12}{3\times2\times1} = 364$$ 4. **Calculate probability:** $$P = \frac{20}{364}$$ 5. **Simplify the fraction:** $$\frac{20}{364} = \frac{\cancel{20}}{\cancel{364}} = \frac{5}{91}$$ 6. **Final answer:** The exact probability that all three marbles drawn are green is $$\boxed{\frac{5}{91}}$$ This means if you randomly draw 3 marbles without replacement, there is a $\frac{5}{91}$ chance all are green.