 # At a fete, cards bearing numbers 1 to 1000,

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

(i)  the first player wins a prize?

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

Solution:

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.

(i) Let $E_{1}=$ Event first player wins a prize $=$ Player select a card which is a perfect square greater than 500

$=\{529,576,625,676,729,784,841,900,961\}$

$=\left\{(23)^{2},(24)^{2},(25)^{2},(26)^{2},(27)^{2},(28)^{2},(29)^{2},(30)^{2},(31)^{2}\right\}$

$\therefore \quad n\left(E_{1}\right)=9$

So, required probability $=\frac{n\left(E_{1}\right)}{n(S)}=\frac{9}{1000}=0.009$

(ii) First, has won i.e., one card is already selected, greater than 500 , has a perfect square. Since, repeatition is not allowed. So, one card is removed out of 1000 cards. So, number of remaining cards is 999 .

$\therefore$ Total number of remaining outcomes, $n\left(S^{\prime}\right)=999$

Let $E_{2}=$ Event the second player wins a . rize, if the first has won.

$=$ Remaining cari is has a perfect square greater than 500 are 8 .

$\therefore \quad n\left(E_{2}\right)=9-1=8$

So, required probability $=\frac{n\left(E_{2}\right)}{n\left(S^{\prime}\right)}=\frac{8}{999}$