A prize spinner is split into three unequal sectors: orange measuring 144°, teal measuring 126° and pink measuring 90°. The pointer is spun twice and on each spin is equally likely to stop anywhere. What is the probability that at least one of the two spins stops on pink?

Pink takes 90° of the full 360°, so one spin stops on pink with probability 90/360 = 1/4 and misses pink with probability 3/4. The two spins are independent, so P(no pink at all) = $(3/4{)}^2=9/16$ and therefore P(at least one pink) = 1 − 9/16 = 7/16. Trap 9/16 is that complement left unsubtracted — the commonest slip in 'at least one' questions. Trap $1/16=(1/4{)}^2$ is 'pink on both spins', a far stricter demand. Trap 1/2 doubles the single-spin chance of 1/4, which counts the both-pink case twice; probabilities may be added only for outcomes that cannot occur together.