There are 12 people in a room, where each person holds a card with a 2-digit number. What is the probability that there are at least two people holding cards with identical numbers

First, note there are 90 2-digit numbers. (There are 99 numbers from 1 - 99; now we subtract 9 to get rid of 1-9 because those are single-digit numbers)

Let A be the event that at least two people in the room have the same 2-digit number.

It is easier to calculate P(A) = 1 - P(A') where A' is A-complement. A-complement is the event such that at least 2 people in the room don't have the same number (i.e. everyone has distinct numbers.)

P(A') = (90/90) * (89/90) *(88/90) * (87/90) *(86/90) * (85/90) *(84/90) * (83/90) *(82/90) * (81/90) *(80/90) * (79/90) 

This is because the first person can have any of the 90 possible numbers. The second person only has 89 possible options since one of the 2-digit numbers was taken by person one, and so on so forth.

P(A') = (1/90)^{12} * (90*89*...*79)

So the desired probability is P(A) = 1 - P(A') which is approximately 53.55%