24. Two natural numbers are given at random; determine the probability that their product is divisible by 3 without a remainder; that when divided by 3, the remainder is 1; the remainder is 2.
24. A number can be of one of three types: $3 m, 3 m+1$, $3 m+2$, and another can be of one of the types: $3 n, 3 n-1-1, 3 n+2$; from the nine equally possible cases (each type of the first number can be combined with each type of the second), in five cases the product gives a remainder of 0, in two cases a remainder of 1, and in two cases a remainder of 2. Therefore, the corresponding probabilities will be: $\frac{5}{9} ; \frac{2}{9} ; \frac{2}{9}$.