1. **Problem statement:** A box contains 5 red, 3 blue, and 2 green marbles. Two marbles are drawn without replacement. Find the probability that exactly one red and one blue marble are drawn. 2. **Total marbles:** There are $5 + 3 + 2 = 10$ marbles in total. 3. **Total ways to draw 2 marbles:** The number of ways to choose any 2 marbles from 10 is given by the combination formula: $$\binom{10}{2} = \frac{10 \times 9}{2} = 45$$ 4. **Favorable outcomes:** We want exactly one red and one blue marble. - Number of ways to choose 1 red marble from 5: $\binom{5}{1} = 5$ - Number of ways to choose 1 blue marble from 3: $\binom{3}{1} = 3$ 5. **Total favorable pairs:** Multiply the two: $$5 \times 3 = 15$$ 6. **Probability:** The probability is the ratio of favorable outcomes to total outcomes: $$P = \frac{15}{45} = \frac{1}{3} \approx 0.33$$ **Final answer:** The probability that exactly one red and one blue marble are drawn is $0.33$. This corresponds to option C.