STATISTICS
posted by MIA .
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

There are 12C3 = 220 ways, group of 3 people from 12
There are 9C3 = 84 ways, none of the three oldest.
22084 = 136 groups where at least one member is 3 oldest.
136/220 = 0..618 
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(E2E1) P(E3E1 and E2)
= 1  (9/12) (8/11) (7/10)
= 0.61818....