Difference between revisions of "2000 AIME II Problems/Problem 3"

m (See also)
Line 7: Line 7:
 
<math>703+55=758</math>
 
<math>703+55=758</math>
  
== 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 20:30, 18 March 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 $m/n$ 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 (ProblemsAnswer KeyResources)
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