To solve this problem, we need to find the probability that there is at least one person from each rank (professors, associate professors, assistant professors, and instructors) on the randomly formed committee of size 7. The ranks available are:

• 7 Professors
• 7 Associate Professors
• 9 Assistant Professors
• 12 Instructors

Step 1: Total Number of Ways to Form a Committee

The total number of ways to form a committee of 7 people from 35 individuals (7 professors + 7 associate professors + 9 assistant professors + 12 instructors) is given by the combination formula: $\binom{35}{7} = \frac{35!}{7!(35-7)!}$ Calculating this gives the total number of possible committees.

Step 2: Number of Ways to Form a Committee with at Least One Person from Each Rank

For the committee to have at least one person from each rank, we need to distribute the 7 committee members across the four ranks (professors, associate professors, assistant professors, and instructors) such that no rank is left out. We can break this into cases based on how many members from each rank are selected, ensuring that each rank has at least one member.

Let’s consider the distribution of the 7 committee members. We need to find valid combinations of selecting members from each rank that sum up to 7. The possible distributions are:

• 1 Professor, 1 Associate Professor, 1 Assistant Professor, 4 Instructors
• 1 Professor, 1 Associate Professor, 2 Assistant Professors, 3 Instructors
• 1 Professor, 2 Associate Professors, 2 Assistant Professors, 2 Instructors
• 2 Professors, 1 Associate Professor, 2 Assistant Professors, 2 Instructors

For each case, we calculate the number of ways to choose members from each rank using the combination formula.

Case 1: 1 Professor, 1 Associate Professor, 1 Assistant Professor, 4 Instructors

The number of ways to select the committee in this case is: $\binom{7}{1} \times \binom{7}{1} \times \binom{9}{1} \times \binom{12}{4}$

Case 2: 1 Professor, 1 Associate Professor, 2 Assistant Professors, 3 Instructors

The number of ways to select the committee in this case is: $\binom{7}{1} \times \binom{7}{1} \times \binom{9}{2} \times \binom{12}{3}$

Case 3: 1 Professor, 2 Associate Professors, 2 Assistant Professors, 2 Instructors

The number of ways to select the committee in this case is: $\binom{7}{1} \times \binom{7}{2} \times \binom{9}{2} \times \binom{12}{2}$

Case 4: 2 Professors, 1 Associate Professor, 2 Assistant Professors, 2 Instructors

The number of ways to select the committee in this case is: $\binom{7}{2} \times \binom{7}{1} \times \binom{9}{2} \times \binom{12}{2}$

Step 3: Total Number of Favorable Outcomes

We sum the number of favorable outcomes from each case:

$\text{Total favorable outcomes} = \binom{7}{1} \times \binom{7}{1} \times \binom{9}{1} \times \binom{12}{4} + \binom{7}{1} \times \binom{7}{1} \times \binom{9}{2} \times \binom{12}{3} + \binom{7}{1} \times \binom{7}{2} \times \binom{9}{2} \times \binom{12}{2} + \binom{7}{2} \times \binom{7}{1} \times \binom{9}{2} \times \binom{12}{2}$

Step 4: Calculate the Probability

The probability of having at least one person from each rank is the ratio of the number of favorable outcomes to the total number of possible outcomes: $P = \frac{\text{Total favorable outcomes}}{\binom{35}{7}}$

Would you like me to calculate the actual values for these combinations and the probability, or do you want to proceed with the general method?