High school mathematics probability problem, who can tell me how to calculate several basic events for similar problems, what is the answer to this question and how to do it?

This topic only considers the first two bits, not the other bits.

1)

All 8, only 1 case.

p 1 =( 1 * 1)/(9 * 10)

= 1/90

2)

You can choose 1- (the probability that the first two digits are greater than 8).

P2 = 1-( 1 * 1/9 * 10)

=89/90

Or discuss it in the following three categories.

(1) First place > 8, second place ≤8.

(2) First place ≤8, second place >; eight

(3) The first position is ≤8, and the second position is ≤8.

P2 =(8 * 9+ 1 * 9+ 1 * 8)/(9 * 10)= 89/90

3)

You can choose 1- (the probability that the first two digits are equal).

P3 = 1-(9/9 * 10)= 9/ 10

or

Select the first number and randomly select a different number as the second number.

P3=9*9/(9* 10)