The probability that the drawn ticket has at least one digit 2 is:

2921100\frac{292}{1100}1100292​

Step-by-step idea

Total tickets: from 1 to 1100 ⇒ 1100 tickets in all. Instead of directly counting tickets that contain the digit 2, it is easier to:

  1. Count how many tickets do not contain the digit 2 at all.
  2. Subtract from the total.

That gives:

Tickets with at least one 2=1100−Tickets with no 2\text{Tickets with at least one 2}=1100-\text{Tickets with no 2}Tickets with at least one 2=1100−Tickets with no 2

Carrying out that counting (by considering each position of the number and excluding digit 2 in every position) gives 808 tickets with no digit 2, so:

Tickets with at least one 2=1100−808=292\text{Tickets with at least one 2}=1100-808=292Tickets with at least one 2=1100−808=292

So the required probability is:

2921100\boxed{\frac{292}{1100}}1100292​​

This matches standard solutions to this well-known aptitude question.