1. **State the problem:** There are 15 buttons: 3 red, 2 pink, and 10 blue. Pete takes 3 buttons at random. We want the probability that after taking 3 buttons, there is still at least one pink button left in the box. 2. **Understand the problem:** Initially, there are 2 pink buttons. To have at least one pink button left, Pete must take at most one pink button (0 or 1 pink buttons taken). 3. **Total number of ways to choose 3 buttons from 15:** $$\binom{15}{3} = \frac{15 \times 14 \times 13}{3 \times 2 \times 1} = 455$$ 4. **Calculate the number of ways Pete takes 0 pink buttons:** - Choose all 3 buttons from the 13 non-pink buttons (3 red + 10 blue = 13): $$\binom{13}{3} = \frac{13 \times 12 \times 11}{3 \times 2 \times 1} = 286$$ 5. **Calculate the number of ways Pete takes exactly 1 pink button:** - Choose 1 pink button from 2: $$\binom{2}{1} = 2$$ - Choose 2 non-pink buttons from 13: $$\binom{13}{2} = \frac{13 \times 12}{2} = 78$$ - Total ways for exactly 1 pink button: $$2 \times 78 = 156$$ 6. **Total favorable outcomes (0 or 1 pink button taken):** $$286 + 156 = 442$$ 7. **Calculate the probability:** $$P = \frac{442}{455}$$ 8. **Simplify the fraction if possible:** 442 and 455 share a common factor 1 only, so fraction is already in simplest form. **Final answer:** $$\boxed{\frac{442}{455}}$$ This is the probability that there is still at least one pink button left in the box after Pete takes 3 buttons.