Question

Question 42(ii)
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 then 500, has a perfect square. Since, repetition is not allowed. So, one card is removed out of 1000 cards. So, the number of remaining cards is 999.
$\therefore$ Total number remaining outcomes, $n\left(S^{\prime }\right)=999$
Let $E_2$=Event the second player wins a prize, if the first has won.
= Remaining cards has a perfect square greater than 500 are 8.
$\therefore n(E_2=9-1=8$
So, required probability=$\frac{n\left(E_2\right)}{n\left(S^{\prime }\right)}=\frac{8}{999}$