The word ANDROMEDA consists of the letters A, N, D, R, O, M, E, D, A. There are 9 letters in total.

We need to find the probability of selecting 4 letters such that there is at least one D and exactly one A.

Method 1:

Consider the scenarios:

1. Scenario 1 (AD^): Select 1 A, 1 D, and 2 other letters from the remaining 7 letters (excluding the second D). The number of ways to do this is \(\binom{2}{1} \times \binom{2}{1} \times \binom{5}{2} = 40\).
2. Scenario 2 (ADD^): Select 1 A, 2 Ds, and 1 other letter from the remaining 6 letters. The number of ways to do this is \(\binom{2}{1} \times \binom{1}{1} \times \binom{6}{1} = 10\).

\(Total number of favorable selections = 40 + 10 = 50.\)

Total number of ways to select 4 letters from 9 = \(\binom{9}{4} = 126\).

Thus, the probability is \(\frac{50}{126} = \frac{25}{63}\).