A squad has 10 players, among them a captain and a vice-captain. Four of the ten are picked at random to represent the squad at a ceremony. What is the probability that both the captain and the vice-captain are among the four picked?

Total selections = C(10,4) = 210. A selection holding both named players fills its other 2 places from the remaining 8: C(8,2) = 28. Probability = 28/210 = 2/15. A quicker route: the captain is in the four with probability 4/10, and given that, the vice-captain fills one of the 3 remaining places out of the 9 remaining players, 3/9 — and (4/10) × (3/9) = 2/15. Trap 1/3 = C(8,4)/210 = 70/210 is the chance that NEITHER of them is picked. Trap 2/3 is the complement of that, 'at least one of the two is picked'. Trap 1/15 = 14/210 halves the correct count as though the pair carried an order that had to be divided out.