A caretaker carries 5 similar-looking keys, exactly one of which opens the store room. He tries them one at a time at random and never tries the same key twice. What is the probability that the store room opens on his third try?

For the third try to be the winning one, the first two must both fail and the third must succeed: $(4/5)\times (3/4)\times (1/3)=1/5$. There is a shorter way to see it — the right key is equally likely to sit in any of the 5 positions of the trying order, so each position carries probability 1/5. Trap 3/5 answers 'within the first three tries', which is 3 of those 5 positions. Trap 1/3 treats the third try as a fresh 1-in-3 pick and forgets that the first two must have failed for the third try to happen at all. Trap $16/125=(4/5{)}^2\times (1/5)$ is the answer only if a failed key were dropped back into the bunch and could be tried again.