Think of a two-digit number. What is the probability that it has different digits?

9/10

Step-by-step explanation:

The first two digit number is 10 and the last is 99. That's a total of 99-10+1 numbers in all. That simplifies to 90. (Just like if we wanted to see how many numbers was 3,4,5, we would do 5-3+1=3 to get the total number.

Anyways, let's consider first how many 2 digjt numbers whose digits are equal. You have 11 22,33,44 55,66,77,88,99 which is 9 numbers total.

So the amount of 2 digits number whose digits differ is 90-9=81.

The probability that a 2 digit number have different digits is 81/90.

This can reduce. Divide top and bottom by 9 giving 9/10.

i don't even know man. edge inuity be messing with us. i had to ask a question too.

no

step-by-step explanation: