Three distinct integers are chosen uniformly at random from the set
$$\{2021, 2022, 2023, 2024, 2025, 2026, 2027, 2028, 2029, 2030\}.$$
Compute the probability that their arithmetic mean is an integer.
To solve this problem, we need to determine the probability that the arithmetic mean of three distinct integers chosen from the set \(\{2021, 2022, 2023, 2024, 2025, 2026, 2027, 2028, 2029, 2030\}\) is an integer. 

1. **Identify the remainders when divided by 3:**
   - The set \(\{2021, 2022, 2023, 2024, 2025, 2026, 2027, 2028, 2029, 2030\}\) can be divided into three groups based on their remainders when divided by 3:
     - Numbers with remainder 0: \(\{2022, 2025, 2028\}\)
     - Numbers with remainder 1: \(\{2021, 2024, 2027, 2030\}\)
     - Numbers with remainder 2: \(\{2023, 2026, 2029\}\)

2. **Calculate the number of ways to choose three numbers such that their arithmetic mean is an integer:**
   - The arithmetic mean of three numbers \(a, b, c\) is \(\frac{a+b+c}{3}\). For this to be an integer, \(a + b + c\) must be divisible by 3.
   - There are four cases to consider:
     1. All three numbers have remainder 0 when divided by 3.
     2. All three numbers have remainder 1 when divided by 3.
     3. All three numbers have remainder 2 when divided by 3.
     4. The three numbers have distinct remainders (0, 1, and 2).

3. **Count the number of ways for each case:**
   - **Case 1:** All three numbers have remainder 0.
     - There are \(\binom{3}{3} = 1\) way to choose 3 numbers from the 3 numbers with remainder 0.
   - **Case 2:** All three numbers have remainder 1.
     - There are \(\binom{4}{3} = 4\) ways to choose 3 numbers from the 4 numbers with remainder 1.
   - **Case 3:** All three numbers have remainder 2.
     - There are \(\binom{3}{3} = 1\) way to choose 3 numbers from the 3 numbers with remainder 2.
   - **Case 4:** The three numbers have distinct remainders.
     - There are \(3\) choices for the number with remainder 0, \(4\) choices for the number with remainder 1, and \(3\) choices for the number with remainder 2.
     - Thus, there are \(3 \times 4 \times 3 = 36\) ways to choose one number from each group.

4. **Calculate the total number of favorable outcomes:**
   - The total number of favorable outcomes is \(1 + 4 + 1 + 36 = 42\).

5. **Calculate the total number of possible outcomes:**
   - The total number of ways to choose 3 numbers from 10 is \(\binom{10}{3} = \frac{10 \times 9 \times 8}{3 \times 2 \times 1} = 120\).

6. **Calculate the probability:**
   - The probability is \(\frac{42}{120} = \frac{7}{20}\).

The final answer is \(\boxed{\frac{7}{20}}\).