A pen stand holds 3 blue and 4 red pens. Two pens are taken out together at random and a friend reports, truthfully, that at least one of them is blue. Given this, what is the probability that both are blue?

Of the C(7,2) = 21 possible pairs, the all-red ones number C(4,2) = 6, so the report 'at least one blue' rules those out and leaves 21 − 6 = 15 equally likely pairs. Both blue accounts for C(3,2) = 3 of them, so the probability is 3/15 = 1/5. Trap 1/7 = 3/21 answers the unconditional question and wastes the information given. Trap 5/7 = 15/21 is the probability of the CONDITION itself, not of the event asked about. Trap 4/7 = 12/21 counts the pairs holding exactly one blue pen.