At a fete, cards bearing numbers $1$ to $1000$, one number on one card, are put in a box. Each player selects one card at random and that card is not replaced. If the selected card has a perfect square greater than $500$, the player wins a prize. What is the probability that:
$(i)$ the first player wins a prize?
$(ii)$ the second player wins a prize, if the first has won?

(A) Total number of cards in the box is $n(S) = 1000$.
$(i)$ Let $E_1$ be the event that the first player wins a prize. This happens if the player selects a perfect square greater than $500$.
The perfect squares between $1$ and $1000$ are $1^2, 2^2, \dots, 31^2 = 961$.
The perfect squares greater than $500$ are $23^2=529, 24^2=576, 25^2=625, 26^2=676, 27^2=729, 28^2=784, 29^2=841, 30^2=900, 31^2=961$.
There are $9$ such cards.
Thus, $n(E_1) = 9$.
The probability that the first player wins is $P(E_1) = \frac{n(E_1)}{n(S)} = \frac{9}{1000} = 0.009$.
$(ii)$ If the first player has won, one card (which is a perfect square $> 500$) is removed from the box.
The remaining number of cards is $n(S') = 1000 - 1 = 999$.
The number of remaining perfect squares greater than $500$ is $n(E_2) = 9 - 1 = 8$.
The probability that the second player wins given the first has won is $P(E_2|E_1) = \frac{n(E_2)}{n(S')} = \frac{8}{999}$.