Two fair dice are thrown together. What is the probability that the product of the two numbers shown is a multiple of 6?

Work through the first die. If it shows 1 the other must be 6 (1 way); a 2 needs a multiple of 3, that is 3 or 6 (2 ways); a 3 needs an even number, 2, 4 or 6 (3 ways); a 4 again needs a multiple of 3 (2 ways); a 5 needs a 6 (1 way); and a 6 allows anything (6 ways). That is 1 + 2 + 3 + 2 + 1 + 6 = 15 of the 36 outcomes, so the probability is 15/36 = 5/12. Counting only the 11 outcomes that actually contain a six gives 11/36 and misses (2,3), (3,2), (3,4) and (4,3) — the product needs a factor 2 AND a factor 3, and the two dice may supply them separately. Recovering just one of those four missing pairs gives 12/36 = 1/3, and 1/6 keeps only the row in which the first die shows 6.