1. **Problem 1:** A bucket contains 3 red, 4 yellow, and 5 purple balls. One ball is taken without replacement. Find the probability that the first ball is red and the second is purple. 2. Total balls initially: $3 + 4 + 5 = 12$. 3. Probability first ball is red: $\frac{3}{12} = \frac{1}{4}$. 4. After removing one red ball, remaining balls: $12 - 1 = 11$. 5. Probability second ball is purple: $\frac{5}{11}$. 6. Combined probability (both events): $$\frac{1}{4} \times \frac{5}{11} = \frac{5}{44}$$. --- 7. **Problem 2:** A die is thrown twice. Find the probability of getting 4 or 5 on the first throw and 2 or 3 on the second throw. 8. Probability first throw is 4 or 5: $\frac{2}{6} = \frac{1}{3}$. 9. Probability second throw is 2 or 3: $\frac{2}{6} = \frac{1}{3}$. 10. Combined probability: $$\frac{1}{3} \times \frac{1}{3} = \frac{1}{9}$$. --- 11. **Problem 3:** A bucket contains 2 red, 4 yellow, and 5 purple balls. One ball is taken and replaced. Another ball is taken. Find the probability first ball is red and second is purple. 12. Total balls: $2 + 4 + 5 = 11$. 13. Since the first ball is replaced, total remains 11 for second draw. 14. Probability first ball red: $\frac{2}{11}$. 15. Probability second ball purple: $\frac{5}{11}$. 16. Combined probability: $$\frac{2}{11} \times \frac{5}{11} = \frac{10}{121}$$. **Final answers:** - Problem 1: $\frac{5}{44}$ - Problem 2: $\frac{1}{9}$ - Problem 3: $\frac{10}{121}$