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

m (See also)
(Solution)
 
(4 intermediate revisions by 4 users not shown)
Line 1: Line 1:
 
== Problem ==
 
== 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 <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>
+
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 ==
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>
+
There are <math>{38 \choose 2} = 703</math> ways we can draw two cards from the reduced 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>.
  
<math>703+55=758</math>
 
  
 +
== Video Solution by OmegaLearn ==
 +
https://youtu.be/mIJ8VMuuVvA?t=59
 +
 +
~ pi_is_3.14
 +
 +
== 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}}
 +
 +
[[Category:Intermediate Combinatorics Problems]]
 +
[[Category:Intermediate Probability Problems]]
 +
{{MAA Notice}}

Latest revision as of 04:27, 4 November 2022

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 two cards from the reduced 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 $m+n = \boxed{758}$.


Video Solution by OmegaLearn

https://youtu.be/mIJ8VMuuVvA?t=59

~ pi_is_3.14

See also

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png