1. **State the problem:** We have a bag with 5 red, 6 blue, and 4 green marbles, totaling $5 + 6 + 4 = 15$ marbles. 2. **What is asked:** Find the probability that all three marbles drawn without replacement are blue. 3. **Formula for probability without replacement:** $$P(\text{all blue}) = \frac{\text{number of ways to choose 3 blue marbles}}{\text{number of ways to choose any 3 marbles}} = \frac{\binom{6}{3}}{\binom{15}{3}}$$ 4. **Calculate combinations:** $$\binom{6}{3} = \frac{6!}{3!\times(6-3)!} = \frac{6\times5\times4}{3\times2\times1} = 20$$ $$\binom{15}{3} = \frac{15!}{3!\times(15-3)!} = \frac{15\times14\times13}{3\times2\times1} = 455$$ 5. **Calculate probability:** $$P = \frac{20}{455}$$ 6. **Simplify fraction:** $$\frac{20}{455} = \frac{\cancel{5} \times 4}{\cancel{5} \times 91} = \frac{4}{91}$$ 7. **Convert to decimal:** $$\frac{4}{91} \approx 0.043956$$ 8. **Round to nearest 1000th:** $$0.044$$ **Final answer:** The probability that all three marbles drawn are blue is approximately $0.044$.