A box holds 5 green tokens and 3 yellow tokens. Tokens are taken out one at a time at random and are NOT put back, and the drawing stops as soon as a yellow token appears. What is the probability that at most two draws are needed?

'At most two' covers two separate cases. The first token is yellow: 3/8. Or the first is green and the second yellow: (5/8) × (3/7) = 15/56, the denominator dropping to 7 because the green token is not put back. Adding gives 21/56 + 15/56 = 36/56 = 9/14. Each option is one piece of the work: 3/8 keeps only the first case, 15/56 only the second, and 5/14 is the complement — the probability that more than two draws are needed, which is just (5/8)(4/7), the chance that the first two tokens are both green. As a check, 9/14 + 5/14 = 1.