1. **Problem statement:** Find the probability that there are no green leaves in box A finally after the described process. 2. **Setup:** Box A initially has 1 green and 3 red leaves (total 4). Box B initially has 4 green and 1 red leaves (total 5). 3. **Step 1: Transfer a leaf from A to B.** - Probability of transferring green leaf from A to B: $\frac{1}{4}$ - Probability of transferring red leaf from A to B: $\frac{3}{4}$ 4. **Step 2: Transfer a leaf from B to A after step 1.** We consider two cases: **Case 1: Green leaf transferred from A to B** - Box B now has 5 green and 1 red (total 6). - Probability of transferring green leaf back from B to A: $\frac{5}{6}$ - Probability of transferring red leaf back from B to A: $\frac{1}{6}$ **Case 2: Red leaf transferred from A to B** - Box B now has 4 green and 2 red (total 6). - Probability of transferring green leaf back from B to A: $\frac{4}{6} = \frac{2}{3}$ - Probability of transferring red leaf back from B to A: $\frac{2}{6} = \frac{1}{3}$ 5. **Final composition of box A:** We want the probability that box A has no green leaves finally. - Initially, box A has 1 green leaf. - To have no green leaves finally, the green leaf must be transferred out and not returned. 6. **Calculate probability:** - Probability green leaf transferred out from A to B: $\frac{1}{4}$ - Then, green leaf not returned from B to A: $1 - \frac{5}{6} = \frac{1}{6}$ So, probability for this scenario: $\frac{1}{4} \times \frac{1}{6} = \frac{1}{24}$ - If a red leaf is transferred out first, box A still has the green leaf, so no chance of zero green leaves finally. 7. **Final answer:** $$\boxed{\frac{1}{24}}$$