1. **Problem statement:** We have a bowl with 8 balls: 4 red, 3 blue, and 1 yellow. We draw one ball, note its color, replace it, and then draw a second ball. We want to find the probability of drawing exactly one red ball in these two draws. 2. **Formula and rules:** Since the draws are with replacement, the draws are independent. The probability of an event happening exactly once in two independent trials is given by: $$P(\text{exactly one red}) = P(\text{red on first, not red on second}) + P(\text{not red on first, red on second})$$ 3. **Calculate individual probabilities:** - Probability of red ball in one draw: $$P(R) = \frac{4}{8} = \frac{1}{2}$$ - Probability of not red ball in one draw: $$P(\neg R) = 1 - P(R) = 1 - \frac{1}{2} = \frac{1}{2}$$ 4. **Calculate combined probabilities:** $$P(R \text{ then } \neg R) = P(R) \times P(\neg R) = \frac{1}{2} \times \frac{1}{2} = \frac{1}{4}$$ $$P(\neg R \text{ then } R) = P(\neg R) \times P(R) = \frac{1}{2} \times \frac{1}{2} = \frac{1}{4}$$ 5. **Sum the probabilities:** $$P(\text{exactly one red}) = \frac{1}{4} + \frac{1}{4} = \frac{2}{4} = \frac{1}{2} = 0.5$$ 6. **Convert to percentage:** $$0.5 \times 100 = 50\%$$ **Final answer:** The probability of drawing exactly one red ball in two draws with replacement is **50%**.