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Q.

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?

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a

\frac{8}{999}

b

\frac{15}{999}

c

\frac{19}{999}

d

\frac{17}{339}

answer is A.

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Detailed Solution

Given that at a fete, cards bearing numbers 1 to 1000, one number on one card, are put in a box.
Suppose that the first has won, that means, one card is already selected, which is a perfect square greater than 500. As repetition is not allowed, one card is taken out of 1000 cards. Therefore, the total number of remaining cards is

1000 − 1 = 999

. Also, as one perfect square is selected, the remaining favorable outcomes are

9 − 1 = 8

.
The probability is,

P A = Number of favourable outcomes Total number of outcomes P A = 8999

The probability that the second player wins a prize, if the first has won is

8999

.
Therefore, 1 is the correct option.

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