1. **State the problem:** We want to find the probability that the first letter in a random arrangement of the letters in "ARIZONA" is A. 2. **Count total letters and repetitions:** The word "ARIZONA" has 7 letters: A, R, I, Z, O, N, A. Note that A appears twice. 3. **Total number of distinct arrangements:** The total number of ways to arrange these letters is given by the formula for permutations of multiset: $$\frac{7!}{2!}$$ where 7! is the factorial of 7 and 2! accounts for the two identical A's. 4. **Number of favorable arrangements (first letter is A):** Fix the first letter as A. Now arrange the remaining 6 letters: A, R, I, Z, O, N. The number of ways to arrange these is: $$6! / 1! = 6!$$ since there is only one A left. 5. **Calculate the probability:** $$\text{Probability} = \frac{\text{Number of favorable arrangements}}{\text{Total number of arrangements}} = \frac{6!}{\frac{7!}{2!}}$$ 6. **Simplify the expression:** $$\frac{6!}{\frac{7!}{2!}} = \frac{6!}{\frac{7 \times 6!}{2}} = \frac{6!}{7 \times 6! / 2} = \frac{6! \times 2}{7 \times 6!} = \frac{2}{7}$$ 7. **Final answer:** $$\boxed{0.286}$$ The probability that the first letter is A is approximately 0.286.