A committee of 3 is being formed randomly from the employees at a school: 6 administrators, 36 teachers, and 4 staff. What is the probability that all 3 members are administrators? Express your answer as a fraction in lowest terms or a decimal rounded to the nearest millionth.
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Step 1 ：There are a total of 46 people in the school, which includes 6 administrators, 36 teachers, and 4 staff.

Step 2 ：A committee of 3 people is to be formed randomly from these 46 people.

Step 3 ：The total number of ways to form a committee of 3 people from 46 is given by the combination formula \(C(n, k) = \frac{n!}{k!(n-k)!}\), where \(n\) is the total number of people and \(k\) is the number of people in the committee. Substituting \(n = 46\) and \(k = 3\), we get \(C(46, 3) = 15180\) ways.

Step 4 ：The number of ways to choose 3 administrators out of 6 is also given by the combination formula. Substituting \(n = 6\) and \(k = 3\), we get \(C(6, 3) = 20\) ways.

Step 5 ：The probability that all 3 members are administrators is then the number of ways to choose 3 administrators divided by the total number of ways to form a committee. So, the probability is \(\frac{20}{15180} = 0.0013175230566534915\).

Step 6 ：Rounding this to the nearest millionth, we get \(0.001317\).

Step 7 ：Final Answer: The probability that all 3 members are administrators is approximately \(\boxed{0.001317}\).