A shop's card machine breaks down on any given day with probability 0.1, independently of what happens on other days. What is the probability that it breaks down on at least one of three consecutive days?

Go through the complement. On a single day the machine survives with probability 0.9, and the days are independent, so it survives all three with probability $0.9^3=0.729$. Hence P(at least one breakdown) = 1 − 0.729 = 0.271. Trap 0.729 is that complement left unsubtracted. Trap 0.3 adds the three daily risks, 0.1 + 0.1 + 0.1, which double-counts the days on which more than one breakdown occurs; probabilities may be added only for outcomes that cannot happen together. Trap $0.001=0.1^3$ is 'breaks down on all three days'.