2000 AIME II Problems/Problem 3

Problem

A deck of forty cards consists of four 1's, four 2's,..., and four 10's. A matching pair (two cards with the same number) is removed from the deck. Given that these cards are not returned to the deck, let $m/n$ be the probability that two randomly selected cards also form a pair, where $m$ and $n$ are relatively prime positive integers. Find $m + n.$

Solution

WLOG, assume that the pair removed is a pair of 1s. Therefore, the probability that a pair of 1's is drawn is $\frac{2}{38}*\frac{1}{37}=\frac{1}{19*37}$ . The probability that it is a pair of 2, 3, ..., or 10 is $9*\frac{4}{38}*\frac{3}{37}=\frac{54}{19*37}$ . $\frac{1}{19*37}+\frac{54}{19*37}=\frac{55}{703}$

$703+55=758$

2000 AIME II (Problems • Answer Key • Resources)
Preceded by Problem 2	Followed by Problem 4
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2000 AIME II Problems/Problem 3

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