A bag has 4 green and 6 yellow marbles. Three marbles are taken out at random and put aside without anyone seeing their colours. One more marble is then drawn from the bag. What is the probability that it is green?

Nobody saw the three marbles that were put aside, so nothing has been learned about the bag: the marble drawn next is equally likely to be any of the original 10, and P(green) = 4/10 = 2/5. The long way agrees: the set-aside marbles hold 0, 1, 2 or 3 greens with probabilities 20/120, 60/120, 36/120 and 4/120, leaving 4, 3, 2 or 1 green among 7, so P = (20·4 + 60·3 + 36·2 + 4·1)/(120·7) = 336/840 = 2/5. The other options assume a particular set was removed: 4/7 (all yellow), 3/7 (one green), 2/7 (two green).