A committee of 3 is being formed randomly from the employees at a school: 6 administrators, 36 teachers, and 5 staff. what is the probability that all 3 members are administrators? express your answer as a fraction
step 1
Find out the total possible results
Calculate the combination 47C3
![47C3=(47!)/(3!(47-3)!)=16,215](https://img.qammunity.org/2023/formulas/mathematics/college/urwcrmnagfoinof05sd21vjqb53ase5e6g.png)
step 2
Calculate the combination 6C3 (administrators)
![6C3=(6!)/(3!(6-3)!)=20](https://img.qammunity.org/2023/formulas/mathematics/college/7rskfx1omr0cvgsvanqxt9lwi2rzqv0cqq.png)
therefore
the probability is
P=20/16,215
simplify
P=4/3,243