An integer greater than 9 and less than 100 is randomly chosen. What is the probability that its digits are different?

From the \(10\) to \(99\), there are \(90\) numbers. Out of these numbers, \(9\) have the same digits(\(11, 22, 33, 44, 55, 66, 77, 88, 99\)). So your probability would be \(\frac{81}{90} = \frac{9}{10}\). 

 

- Daisy