$A$ box contains $90$ discs which are numbered from $1$ to $90$. If one disc is drawn at random from the box, find the probability that it bears:
$(i)$ a two-digit number
$(ii)$ a perfect square number
$(iii)$ a number divisible by $5$.

(N/A) Total number of discs $= 90$.
$(i)$ The two-digit numbers from $1$ to $90$ are $10, 11, \dots, 90$. The total count is $90 - 9 = 81$.
$P(\text{two-digit number}) = \frac{81}{90} = \frac{9}{10}$.
$(ii)$ The perfect square numbers between $1$ and $90$ are $1^2=1, 2^2=4, 3^2=9, 4^2=16, 5^2=25, 6^2=36, 7^2=49, 8^2=64, 9^2=81$. Total count is $9$.
$P(\text{perfect square}) = \frac{9}{90} = \frac{1}{10}$.
$(iii)$ The numbers between $1$ and $90$ divisible by $5$ are $5, 10, 15, 20, 25, 30, 35, 40, 45, 50, 55, 60, 65, 70, 75, 80, 85, 90$. Total count is $18$.
$P(\text{divisible by } 5) = \frac{18}{90} = \frac{1}{5}$.