From a group of 12 people, you randomly select 3 of them. Find the probability that at least one of them is among the 3 oldest people in the group

Answer 1

There are 12C3 = 220 ways, group of 3 people from 12
There are 9C3 = 84 ways, none of the three oldest.

220-84 = 136 groups where at least one member is 3 oldest.

136/220 = 0..618

Answer 2

Alternatively: It's the complement of the probability that none of them is among that group.

Let E[n] be the indexed event that the [n]th person selected is not one of the three oldest.

1 - P(E1) P(E2|E1) P(E3|E1 and E2)
= 1 - (9/12) (8/11) (7/10)
= 0.61818....