1. **State the problem:** Amit, Tan, and Raju shared 36 rambutans. Amit took 12 rambutans, and Tan and Raju took the rest. The number of rambutans taken by Tan is twice the number taken by Raju. Among Tan's rambutans, 3 are rotten. We need to find the probability that if Tan chooses one rambutan to eat, it is not rotten. 2. **Identify known values:** - Total rambutans = 36 - Amit's rambutans = 12 - Tan's rambutans = $2 \times$ Raju's rambutans - Rotten rambutans with Tan = 3 3. **Calculate rambutans taken by Tan and Raju:** Since Amit took 12, the remaining rambutans for Tan and Raju are: $$36 - 12 = 24$$ Let Raju's rambutans be $x$. Then Tan's rambutans are $2x$. So, $$x + 2x = 24$$ $$3x = 24$$ $$x = 8$$ Therefore, Raju took 8 rambutans and Tan took: $$2 \times 8 = 16$$ 4. **Calculate the number of good (not rotten) rambutans Tan has:** Total rambutans Tan has = 16 Rotten rambutans = 3 Good rambutans = $$16 - 3 = 13$$ 5. **Calculate the probability that Tan picks a good rambutan:** Probability = \frac{\text{Number of good rambutans}}{\text{Total rambutans Tan has}} = $$\frac{13}{16}$$ **Final answer:** The probability that Tan picks a rambutan that is not rotten is $$\frac{13}{16}$$.