1. **Problem 21:** A die is rolled twice. Find the probability that the difference of the two rolls is 2 or the sum is 7, expressed as a fraction \(\frac{a}{b}\) in lowest terms. Then find the product \(a \times b\). 2. **Step 1:** Total outcomes when rolling a die twice is \(6 \times 6 = 36\). 3. **Step 2:** Find outcomes where difference is 2: - Possible pairs: (1,3), (3,1), (2,4), (4,2), (3,5), (5,3), (4,6), (6,4) = 8 outcomes. 4. **Step 3:** Find outcomes where sum is 7: - Possible pairs: (1,6), (6,1), (2,5), (5,2), (3,4), (4,3) = 6 outcomes. 5. **Step 4:** Check overlap (difference 2 and sum 7): - Pairs with sum 7 and difference 2: (3,4) and (4,3) are difference 1, so no overlap. 6. **Step 5:** Total favorable outcomes = 8 + 6 = 14. 7. **Step 6:** Probability \(= \frac{14}{36} = \frac{7}{18}\) (lowest terms). 8. **Step 7:** Product \(a \times b = 7 \times 18 = 126\). --- 9. **Problem 22:** A box contains papers numbered 0 to 9. Two papers are drawn without replacement. Let \(n\) be the probability that the two-digit number formed is prime (leading zero allowed, e.g., 01 = 1). Find \(\left( \frac{30n}{4} \right)^2\). 10. **Step 1:** Total ways to pick 2 papers in order: \(10 \times 9 = 90\). 11. **Step 2:** List all two-digit primes from 01 to 99 with digits 0-9: - Primes with two digits (including leading zero): 2,3,5,7,11,13,17,19,23,29,31,37,41,43,47,53,59,61,67,71,73,79,83,89,97. 12. **Step 3:** Count how many pairs correspond to these primes: - For example, 01=1(not prime), 02=2(prime), 03=3(prime), 05=5(prime), 07=7(prime), 11=11(prime), etc. 13. **Step 4:** Valid pairs (first digit, second digit) that form prime: - (0,2),(0,3),(0,5),(0,7),(1,1),(1,3),(1,7),(1,9),(2,3),(2,9),(3,1),(3,7),(4,1),(4,3),(4,7),(5,3),(5,9),(6,1),(6,7),(7,1),(7,3),(7,9),(8,3),(8,9),(9,7). 14. **Step 5:** Count = 25. 15. **Step 6:** Probability \(n = \frac{25}{90} = \frac{5}{18}\). 16. **Step 7:** Calculate \(\left( \frac{30n}{4} \right)^2 = \left( \frac{30 \times \frac{5}{18}}{4} \right)^2 = \left( \frac{150/18}{4} \right)^2 = \left( \frac{25/3}{4} \right)^2 = \left( \frac{25}{12} \right)^2 = \frac{625}{144}\). --- 17. **Problem 23:** Litchi buys 6 chocolate chip cookies and 5 butter cookies for 138. Boy buys 3 chocolate chip and 7 butter cookies for 123. Find how much One pays for 5 chocolate chip and 4 butter cookies and how much less than the total of Litchi and Boy. 18. **Step 1:** Let \(x\) = price of chocolate chip cookie, \(y\) = price of butter cookie. 19. **Step 2:** Equations: \(6x + 5y = 138\) \(3x + 7y = 123\) 20. **Step 3:** Multiply second by 2: \(6x + 14y = 246\) 21. **Step 4:** Subtract first from this: \((6x + 14y) - (6x + 5y) = 246 - 138\) \(9y = 108 \Rightarrow y = 12\) 22. **Step 5:** Substitute \(y=12\) into first: \(6x + 5(12) = 138 \Rightarrow 6x + 60 = 138 \Rightarrow 6x = 78 \Rightarrow x = 13\) 23. **Step 6:** One buys 5 chocolate chip and 4 butter cookies: \(5x + 4y = 5(13) + 4(12) = 65 + 48 = 113\) 24. **Step 7:** Total Litchi and Boy pay: \(138 + 123 = 261\) 25. **Step 8:** Difference: \(261 - 113 = 148\) --- 26. **Problem 24:** Arrow trajectory given by \(y = -\frac{x^2}{2} + 2x + \frac{3}{2}\), find height at \(x=3\). 27. **Step 1:** Substitute \(x=3\): \[ y = -\frac{3^2}{2} + 2(3) + \frac{3}{2} = -\frac{9}{2} + 6 + \frac{3}{2} \] 28. **Step 2:** Calculate: \( -\frac{9}{2} + \frac{3}{2} = -\frac{6}{2} = -3 \) 29. **Step 3:** So, \( y = -3 + 6 = 3 \) 30. **Answer:** The arrow is 3 meters above the ground at 3 meters from the shooting point.