Difference between revisions of "2000 AIME II Problems/Problem 3"
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== Solution == | == Solution == | ||
| − | {{ | + | WLOG, assume that the pair removed is a pair of 1s. Therefore, the probability that a pair of 1's is drawn is <math>\frac{2}{38}*\frac{1}{37}=\frac{1}{19*37}</math>. The probability that it is a pair of 2, 3, ..., or 10 is <math>9*\frac{4}{38}*\frac{3}{37}=\frac{54}{19*37}</math>. <math>\frac{1}{19*37}+\frac{54}{19*37}=\frac{55}{703}</math> |
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| + | <math>703+55=758</math> | ||
== See also == | == See also == | ||
{{AIME box|year=2000|n=II|num-b=2|num-a=4}} | {{AIME box|year=2000|n=II|num-b=2|num-a=4}} | ||
Revision as of 20:57, 3 January 2008
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 
be the probability that two randomly selected cards also form a pair, where 
and 
are relatively prime positive integers. Find 
Solution
WLOG, assume that the pair removed is a pair of 1s. Therefore, the probability that a pair of 1's is drawn is 
. The probability that it is a pair of 2, 3, ..., or 10 is 
. 
See also
| 2000 AIME II (Problems • Answer Key • Resources) | ||
| Preceded by Problem 2 |
Followed by Problem 4 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
| All AIME Problems and Solutions | ||
