1. **State the problem:** We want to find the probability of choosing three letters from the word FACETIOUS such that the letters are A, E, and I in any order. 2. **Identify total letters and favorable letters:** The word FACETIOUS has 9 letters: F, A, C, E, T, I, O, U, S. 3. **Total number of ways to choose 3 letters:** Since order does not matter, the total number of ways to choose 3 letters from 9 is given by the combination formula: $$\binom{9}{3} = \frac{9!}{3!(9-3)!} = \frac{9 \times 8 \times 7}{3 \times 2 \times 1} = 84$$ 4. **Number of favorable outcomes:** We want the letters A, E, and I specifically. Since these three letters are distinct and all present in the word, there is exactly 1 way to choose these three letters. 5. **Calculate the probability:** Probability is the number of favorable outcomes divided by the total number of outcomes: $$P = \frac{1}{84}$$ 6. **Final answer:** The probability of choosing A, E, and I in any order when selecting 3 letters from FACETIOUS is: $$\boxed{\frac{1}{84}}$$