1. **State the problem:** We have three bags A, B, and C with red and white counters. Ratios and counts are given, and counters are moved between bags. We want the probability that the number of white counters in bag C is even after these moves. 2. **Given data:** - Bag A: red:white ratio = 7:3 (total counters unknown, but ratio suffices for probability) - Bag B: 2 red, 9 white counters - Bag C: 4 red, 11 white counters 3. **Step 1: Probability of drawing a red or white counter from bag A** - Total counters in A = 7 + 3 = 10 - Probability red from A = $\frac{7}{10}$ - Probability white from A = $\frac{3}{10}$ 4. **Step 2: After moving one counter from A to B, update counts in B** - If red moved: B has 3 red, 9 white - If white moved: B has 2 red, 10 white 5. **Step 3: Probability of drawing red or white from updated B** - Case 1 (red moved from A): total in B = 3 + 9 = 12 - $P(\text{red from B}) = \frac{3}{12} = \frac{1}{4}$ - $P(\text{white from B}) = \frac{9}{12} = \frac{3}{4}$ - Case 2 (white moved from A): total in B = 2 + 10 = 12 - $P(\text{red from B}) = \frac{2}{12} = \frac{1}{6}$ - $P(\text{white from B}) = \frac{10}{12} = \frac{5}{6}$ 6. **Step 4: Update counts in C after moving one counter from B to C** - Original C: 4 red, 11 white - We want the probability that white counters in C are even after adding one counter from B. 7. **Step 5: Analyze parity of white counters in C** - Initial white counters in C = 11 (odd) - Adding a red counter does not change white count (remains 11, odd) - Adding a white counter increases white count by 1, making it 12 (even) 8. **Step 6: Calculate total probability that white counters in C are even** - Probability white counters in C even = sum over all paths where a white counter is added to C 9. **Step 7: Calculate combined probabilities** - Path 1: A red, then B white - $P = P(\text{red from A}) \times P(\text{white from B} | \text{red from A}) = \frac{7}{10} \times \frac{3}{4} = \frac{21}{40}$ - Path 2: A white, then B white - $P = P(\text{white from A}) \times P(\text{white from B} | \text{white from A}) = \frac{3}{10} \times \frac{5}{6} = \frac{15}{60} = \frac{1}{4}$ 10. **Step 8: Sum probabilities for white counter added to C** $$\frac{21}{40} + \frac{1}{4} = \frac{21}{40} + \frac{10}{40} = \frac{31}{40}$$ 11. **Final answer:** The probability that the number of white counters in bag C is even after the moves is $\boxed{\frac{31}{40}}$ or 0.775.