Question

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 the second player wins a prize, if the first has won?

Answer 1

Given that, 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 so, the total number of outcomes are n(S) = 1000

If the selected card has a perfect square greater than 500, then player wins a prize.

First, has won i.e., one card is already selected, greater than 500, has a perfect square.

Since repetition is not allowed.

So, one card is removed out of 1000 cards.

So, number of remaining cards is 999.

∴ Total number of remaining outcomes, n(S’) = 999

Let E₂ = Event the second player wins a prize, if the first has won.

= Remaining cards has a perfect square greater than 500 are 8.

∴ n(E₂) = 9 – 1 = 8

So, required probability = `(n(E_2))/(n(S^')) = 8/999`