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Difference between revisions of "2000 AIME II Problems/Problem 3"
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== Problem ==
 
== Problem ==
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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 <math>m/n</math> be the probability that two randomly selected cards also form a pair, where <math>m</math> and <math>n</math> are relatively prime positive integers.  Find <math>m + n.</math>
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A deck of forty cards consists of four <math>1</math>'s, four <math>2</math>'s,..., and four <math>10</math>'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 <math>m/n</math> be the [[probability]] that two randomly selected cards also form a pair, where <math>m</math> and <math>n</math> are [[relatively prime]] positive integers.  Find <math>m + n.</math>
  
 
== Solution ==
 
== Solution ==
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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>.
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There are <math>{38 \choose 2} = 703</math> ways we can draw a two cards from the deck. The two cards will form a pair if both are one of the nine numbers that were not removed, which can happen in <math>9{4 \choose 2} = 54</math> ways, or if the two cards are the remaining two cards of the number that was removed, which can happen in <math>1</math> way. Thus, the answer is <math>\frac{54+1}{703} = \frac{55}{703}</math>, and <math>m+n = \boxed{758}</math>.
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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>, and thus the answer is <math>703+55=\boxed{758}</math>.
 
  
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== 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}}
  
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[[Category:Intermediate Combinatorics Problems]]
 
[[Category:Intermediate Probability Problems]]
 
[[Category:Intermediate Probability Problems]]

Revision as of 10:31, 6 September 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

There are ${38 \choose 2} = 703$ ways we can draw a two cards from the deck. The two cards will form a pair if both are one of the nine numbers that were not removed, which can happen in $9{4 \choose 2} = 54$ ways, or if the two cards are the remaining two cards of the number that was removed, which can happen in $1$ way. Thus, the answer is $\frac{54+1}{703} = \frac{55}{703}$, and $m+n = \boxed{758}$.

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
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