To solve this problem, we need to compute the probability that at least one person from each rank is selected on a random committee of 7 members from the faculty.

Step 1: Total Number of Ways to Form the Committee

We first calculate the total number of ways to select 7 members from the entire faculty, which consists of:

• 7 professors
• 7 associate professors
• 9 assistant professors
• 12 instructors

The total number of faculty members is:

$7 + 7 + 9 + 12 = 35 \text{ members}.$

The total number of ways to form a committee of 7 members from these 35 people is given by the combination formula:

$\binom{35}{7}.$

Step 2: Counting Favorable Outcomes (At Least One from Each Rank)

We now need to count the number of ways to form a committee with at least one member from each rank. Since the committee must include at least one professor, one associate professor, one assistant professor, and one instructor, we will start by assigning one member from each rank and then choose the remaining 3 members freely from any rank.

Step 2.1: Choose One Person from Each Rank

• 1 professor from 7 professors: $\binom{7}{1}$,
• 1 associate professor from 7 associate professors: $\binom{7}{1}$,
• 1 assistant professor from 9 assistant professors: $\binom{9}{1}$,
• 1 instructor from 12 instructors: $\binom{12}{1}$.

The number of ways to select one person from each rank is the product of these combinations:

$\binom{7}{1} \times \binom{7}{1} \times \binom{9}{1} \times \binom{12}{1} = 7 \times 7 \times 9 \times 12.$

Step 2.2: Choose the Remaining 3 Members

After selecting one person from each rank, we have 4 members chosen and need to select 3 more members from the remaining faculty members, which consist of:

• 6 remaining professors (since 1 professor has already been selected),
• 6 remaining associate professors (since 1 associate professor has already been selected),
• 8 remaining assistant professors (since 1 assistant professor has already been selected),
• 11 remaining instructors (since 1 instructor has already been selected).

The total number of remaining members is:

$6 + 6 + 8 + 11 = 31.$

The number of ways to select the remaining 3 members from these 31 faculty members is:

$\binom{31}{3}.$

Step 2.3: Total Number of Favorable Outcomes

The total number of favorable outcomes is the product of the ways to choose 1 person from each rank and the ways to select the remaining 3 members:

$\left(7 \times 7 \times 9 \times 12\right) \times \binom{31}{3}.$

Step 3: Probability Calculation

The probability that there is at least one person from each rank on the committee is the ratio of favorable outcomes to total outcomes:

$P = \frac{(7 \times 7 \times 9 \times 12) \times \binom{31}{3}}{\binom{35}{7}}.$

Now we can compute the values of the combinations:

$\binom{35}{7} = \frac{35 \times 34 \times 33 \times 32 \times 31 \times 30 \times 29}{7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1},$ $\binom{31}{3} = \frac{31 \times 30 \times 29}{3 \times 2 \times 1}.$

Finally, substituting these into the probability formula gives the exact value.

Would you like me to compute the final value for you?