Cards 1–1000 placed in a box. A player wins a prize if the card has a perfect square > 500. Players draw one card each without replacement. What is the probability that (i) the first player wins? (ii) the second player wins, if the first has already won?
(i) \(\dfrac{9}{1000}\) ; (ii) \(\dfrac{8}{999}\)
Step 1: We need to find all perfect squares greater than 500, but less than or equal to 1000.
• The smallest integer whose square is more than 500 is \(23\), because \(22^2 = 484 < 500\) and \(23^2 = 529 > 500\).
• The largest integer whose square is less than or equal to 1000 is \(31\), because \(31^2 = 961 \leq 1000\) and \(32^2 = 1024 > 1000\).
Step 2: So the perfect squares greater than 500 are:
\(23^2 = 529, 24^2 = 576, 25^2 = 625, 26^2 = 676, 27^2 = 729, 28^2 = 784, 29^2 = 841, 30^2 = 900, 31^2 = 961\).
That makes a total of 9 cards.
Step 3 (i): Probability that the first player wins.
Total cards = 1000.
Favorable cards = 9 (the perfect squares above).
So, Probability = \( \tfrac{9}{1000} \).
Step 4 (ii): Probability that the second player wins, given that the first has already won.
If the first player has already drawn one of the 9 winning cards, then:
• Remaining total cards = 999 (since one card is removed).
• Remaining winning cards = 8 (since one winning card is already taken).
So, Probability = \( \tfrac{8}{999} \).
Final Answer: (i) \(\tfrac{9}{1000}\), (ii) \(\tfrac{8}{999}\).