A university is putting together a hiring committee of 6 members for the next university provost. The eligible pool of members consists of 4 administrative personnel, 9 faculty, and 10 students. Suppose the committee is chosen at random and with no restrictions. What is the probability that each subgroup has 2 members on the committee? Express your answer as a decimal rounded to the nearest millionth.

To solve this problem, we use combinations and the basic principles of probability.

### Step 1: Understand the problem
We are forming a committee of \(6\) members. The eligible pool contains:
- \(4\) administrative personnel,
- \(9\) faculty,
- \(10\) students.

We are tasked with finding the probability that the committee contains exactly \(2\) administrative personnel, \(2\) faculty, and \(2\) students.

### Step 2: Total ways to choose the committee
The total pool of eligible members is:
\[
4 + 9 + 10 = 23.
\]
The total number of ways to choose \(6\) members from \(23\) is:
\[
\binom{23}{6} = \frac{23 \times 22 \times 21 \times 20 \times 19 \times 18}{6 \times 5 \times 4 \times 3 \times 2 \times 1} = 100947.
\]

### Step 3: Ways to select the specific composition
We want exactly \(2\) administrative personnel, \(2\) faculty, and \(2\) students. For each subgroup:
- The number of ways to choose \(2\) administrative personnel from \(4\) is:
  \[
  \binom{4}{2} = \frac{4 \times 3}{2 \times 1} = 6.
  \]
- The number of ways to choose \(2\) faculty from \(9\) is:
  \[
  \binom{9}{2} = \frac{9 \times 8}{2 \times 1} = 36.
  \]
- The number of ways to choose \(2\) students from \(10\) is:
  \[
  \binom{10}{2} = \frac{10 \times 9}{2 \times 1} = 45.
  \]

The total number of ways to select \(2\) administrative personnel, \(2\) faculty, and \(2\) students is:
\[
6 \times 36 \times 45 = 9720.
\]

### Step 4: Calculate the probability
The probability is the number of favorable outcomes divided by the total outcomes:
\[
P = \frac{9720}{100947}.
\]

### Step 5: Simplify and round
Performing the division:
\[
P \approx 0.09634.
\]

### Final Answer:
The probability that the committee contains \(2\) members from each subgroup is approximately:
\[
\boxed{0.096340}.
\]

Would you like further explanation or to explore related questions?

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#### Related Questions:
1. What if the committee required \(3\) faculty, \(2\) students, and \(1\) administrative personnel?
2. What is the probability that there are no administrative personnel on the committee?
3. How many total committees can be formed with at least one member from each group?
4. What is the expected number of students in a randomly chosen committee of \(6\)?
5. If the committee required a majority of students, what is the probability of forming such a committee?

#### Tip:
When calculating probabilities involving combinations, ensure that all restrictions (like specific subgroup counts)

A university is putting together a hiring committee of 6 members for the next university provost. The eligible pool of members consists of 4 administrative personnel, 9 faculty, and 10 students. Suppose the committee is chosen at random and with no restrictions. What is the probability that each subgroup has 2 members on the committee? Express your answer as a decimal rounded to the nearest millionth.

Learn how to calculate the probability of forming a 6-member committee with exactly 2 administrative personnel, 2 faculty, and 2 students, chosen from pools of 4, 9, and 10 members respectively. This detailed solution includes combinatorics, probability principles, and step-by-step calculations. Suitable for Grades 10-12.

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