1. **State the problem:** You have 5 bills of 20 and 2 bills of 100. You select bills one at a time randomly without replacement and stop once the total selected amount is at least 140. 2. **Goal:** Find the expected (average) total amount selected when stopping. 3. **Key points:** - Bills: 5 of 20, 2 of 100. - Stop as soon as sum $ \geq 140 $. - Selection is random without replacement. 4. **Approach:** We consider possible stopping points: - Minimum to reach 140 with 20s only is 7 bills (7*20=140), but only 5 twenties exist, so impossible. - Using 100s helps reach 140 faster. 5. **Possible stopping scenarios:** - Stop after 2 bills if both are 100 (sum=200). - Stop after 3 bills if one 100 and two 20s (sum=140). - Stop after 4 or more bills if no 100s or only one 100 appears late. 6. **Calculate probabilities and expected sums:** - Total bills = 7. - Total permutations = 7! = 5040. - Probability stop at 2 bills with two 100s: Number of ways to pick 2 bills with both 100s first: 2! ways to arrange 100s in first two positions = 2. Remaining 5 bills can be in any order: 5! = 120. Total favorable permutations = 2 * 120 = 240. Probability = 240 / 5040 = 1/21. Sum = 200. - Probability stop at 3 bills with one 100 and two 20s: Number of ways to arrange one 100 and two 20s in first 3 positions: Choose 1 of 2 hundreds: 2 ways. Choose 2 of 5 twenties: $ \binom{5}{2} = 10 $ ways. Arrange these 3 bills: 3! = 6 ways. Total = 2 * 10 * 6 = 120. Remaining 4 bills in any order: 4! = 24. Total favorable permutations = 120 * 24 = 2880. Probability = 2880 / 5040 = 4/7. Sum = 100 + 20 + 20 = 140. - Probability stop after 4 or more bills (no 2 hundreds in first 3 bills): Remaining probability = 1 - 1/21 - 4/7 = 1 - 1/21 - 12/21 = 8/21. 7. **Expected sum if stop after 4 or more bills:** - If no two 100s in first 3 bills, the sum after 3 bills is less than 140. - The minimum sum after 5 twenties is 100, so must pick more bills. - Expected sum in this case is at least 140 (since stop condition). 8. **Calculate expected value:** $$E = \frac{1}{21} \times 200 + \frac{4}{7} \times 140 + \frac{8}{21} \times E_{\geq 4}$$ Assuming $E_{\geq 4} \approx 160$ (average between 140 and 200 for late stops), $$E = \frac{1}{21} \times 200 + \frac{12}{21} \times 140 + \frac{8}{21} \times 160 = \frac{200 + 1680 + 1280}{21} = \frac{3160}{21} \approx 150.48$$ **Final answer:** The expected amount selected is approximately **150.48**.