1. **Problem statement:** A bag contains 6 red, 4 blue, 2 green, and 3 yellow marbles. Three marbles are picked at random. Find the probability that 2 are blue and 1 is yellow. 2. **Formula and rules:** The probability of an event is given by $$\text{Probability} = \frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}}$$ We use combinations to count outcomes because order does not matter: $$\binom{n}{k} = \frac{n!}{k!(n-k)!}$$ 3. **Calculate total number of ways to pick 3 marbles:** $$\binom{6+4+2+3}{3} = \binom{15}{3} = \frac{15 \times 14 \times 13}{3 \times 2 \times 1} = 455$$ 4. **Calculate favorable outcomes (2 blue, 1 yellow):** Number of ways to pick 2 blue marbles: $$\binom{4}{2} = \frac{4 \times 3}{2 \times 1} = 6$$ Number of ways to pick 1 yellow marble: $$\binom{3}{1} = 3$$ Total favorable outcomes: $$6 \times 3 = 18$$ 5. **Calculate probability:** $$\frac{18}{455}$$ 6. **Final answer:** The probability that 2 marbles are blue and 1 is yellow is $$\boxed{\frac{18}{455}}$$ This corresponds to option (c).