Selecting probabilities

Question

The grade appeal process at a university requires that a jury be structured by selecting four individuals randomly from a pool of ten students and thirteen faculty.​ (a) What is the probability of selecting a jury of all​ students? (b) What is the probability of selecting a jury of all​ faculty? (c) What is the probability of selecting a jury of two students and two faculty?

Answer 1

This problem fundamentally concerns "how many ways can you pick X number of objects from a total of Y objects". For example, if I had 13 red balls and I wanted to pick 4 of them, there are 13C4 ways (13! / (4! * 9!)). the "!" represents factorial symbol

In that example, we implicitly assumed that the "order" of picking the balls doesn't matter. If the order did matter, we use permutation, so a combination is essentially a permutation, but accounting for "double counting" because order doesn't matter.

In this problem, I would argue that picking out a jury of students doesn't really depend on the order. For example, a jury of "A, B, C" is the same as a jury of "C, B, A" except you "changed the order"

Now onto the parts

A) Choose 4 students from a total of 10 students. That's 10C4 ("10 choose 4") as a combination

B) Same idea as part A, but we are selecting 4 faculty from a total of 13 faculty

C) We apply the idea from parts A and B, but with the restriction of 2 student, 2 faculty as your jury. Well the condition says the jury must comprise of 2 student AND 2 faculty. Generally when you see an "AND", we have to multiply!

You know how to pick a jury of 2 students and jury of 2 faculty individually. Just multiply those pieces together, so we have

10C2 * 13C2 ways to have jury of 2 students, 2 faculty