One number is picked at random from 1 to 60, every number being equally likely. What is the probability that it is divisible by 3 or by 5, but NOT by both?

Among 1 to 60 there are 60/3 = 20 multiples of 3, 60/5 = 12 multiples of 5, and 60/15 = 4 numbers divisible by both. 'One or the other but not both' therefore counts (20 − 4) + (12 − 4) = 16 + 8 = 24 numbers, so the probability is 24/60 = 2/5. Trap 7/15 = 28/60 is the plain union 20 + 12 − 4, which still admits the 4 multiples of 15 that the word 'not both' bars. Trap 1/15 = 4/60 reports those barred multiples of 15 instead. Trap 1/3 = 20/60 counts the multiples of 3 alone.