1. **State the problem:** We want to find the probability that Kiaan takes two counters that are not white when he draws one counter, replaces it, and then draws another. 2. **Identify total counters and probabilities:** Total counters = 16 white + 3 green + 1 black = 20 counters. Probability of not white = $ \frac{4}{20} $ because 3 green + 1 black = 4 counters. 3. **Use the multiplication rule for independent events:** Since Kiaan replaces the counter, the draws are independent. The probability of two not white counters is: $$ P(\text{not white, then not white}) = P(\text{not white}) \times P(\text{not white}) $$ 4. **Calculate the probability:** $$ P = \frac{4}{20} \times \frac{4}{20} = \frac{4 \times 4}{20 \times 20} = \frac{16}{400} $$ 5. **Simplify the fraction:** $$ \frac{16}{400} = \frac{\cancel{16}}{\cancel{400}} = \frac{1}{25} $$ 6. **Final answer:** The probability that Kiaan takes two counters that are not white is $ \frac{1}{25} $.