A hospital keeps three standby generators that start independently of one another when the mains fail, with probabilities 1/2, 2/3 and 3/4 respectively. On a night when the mains fail, what is the probability that exactly two of the three generators start?

The three start with probabilities 1/2, 2/3, 3/4 and therefore fail to start with probabilities 1/2, 1/3, 1/4. 'Exactly two start' splits into three cases, one for whichever generator stays silent: first and second start, third does not — (1/2) × (2/3) × (1/4) = 2/24; first and third start, second does not — (1/2) × (1/3) × (3/4) = 3/24; second and third start, first does not — (1/2) × (2/3) × (3/4) = 6/24. Adding: 2/24 + 3/24 + 6/24 = 11/24. Trap 1/4 = 6/24 is the probability that ALL THREE start, which 'exactly two' rules out. Trap 17/24 = 11/24 + 6/24 answers 'at least two start'. Trap 1/24 = (1/2) × (1/3) × (1/4) is the probability that none of them starts.