Solution: - Total number of 2-digit numbers \(=99-10+1=90\)
- Number of 2-digit numbers with 0 in them = 10, 20, 30...90 \(= 9\)
- Therefore, total number of 2-digit numbers that do not have 0 in them \(=90-9=81\)
An alternate way to count this would be by multiplying \(9 \times 9\)
- From 11 to 19
- Number of digit with unit digit greater than the tens digit \(=8\) (12, 13, 14...19)
- From 21 to 29
- Number of digit with unit digit greater than the tens digit \(=7\) (23, 24, 25...29)
- .
- .
- .
- From 81 to 89
- Number of digit with unit digit greater than the tens digit \(=1\) (89)
- From 91 to 99
- Number of digit with unit digit greater than the tens digit \(=0\)
- Therefore, total number of 2-digit numbers with unit digit greater than the tens digit \(=8+7+6+5+4+3+2+1=36\)
Thus, the required probability \(=\frac{36}{81}=\frac{4}{9}\)
Hence the right answer is
Option B