What is the probability that a random 2 digit number (00-99) does not end in 5

11/99 will have a 5 then 88/99 will not have a 5

Sample space = {00,10,11,...99} = 100 Total outcomes

from 00 to 99 you can form 100 numbers with 2 digits (including 00).
So any 2 digit number could be written T (tens) and U (unit)  → TU

T can take any value from 0 to 9, then there are 10 ways to choose T
U can take any value from 0 to 9 EXCEPT 5, then there are 9 ways
to choose U.
Then we can select TU in 10 x 9 = 90 ways
Favorable outcome ={90}
P(2 digits NOT ending in 5) = 90/100 = 0.9