The first digit of (1) is not zero, so the first digit can only be selected from 1 to 9, and the probability of getting 8 is1/9; The second digit can be any one of 10 numbers from 0 to 9, and the probability of getting 8 is110; The first digit to 8 and the second digit to 8 are independent of each other, so the probability that the first two digits are 8 is1/9×110 =1/90.
(2) At least one of the first two digits does not exceed 8, including the following situations: only the first digit does not exceed 8; Only the second digit does not exceed 8; The first two digits are no more than 8. This event happens to be the opposite event with the first two digits exceeding 8. Since the probability that both the first two digits exceed 8 (that is, both the first two digits are 9) is1/9×10 =1/90, the probability that at least one of the first two digits does not exceed 8 is1-kloc-0//.
(3) The two events "the first two are different" and "the first two are the same" are independent events, and there are nine possibilities for "the first two are 1, 2, 3, 4, 5, 6, 7, 8 or 9, and the probability of each case is 1/90, the first two.
I hope the above explanation will help you understand this problem.