Three machines work independently. On a given day they break down with probabilities 1/5, 1/4 and 1/3 respectively. What is the probability that at least one of them breaks down on that day?

Each machine keeps working with probability 4/5, 3/4 and 2/3, so P(none breaks down) = (4/5)(3/4)(2/3) = 24/60 = 2/5 and therefore P(at least one breaks down) = 1 − 2/5 = 3/5. Adding the three probabilities gives 1/5 + 1/4 + 1/3 = 47/60, which double-counts the days on which two or three machines fail, and 1/60 is the chance that ALL three break down.