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

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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>
 
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>
  
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== Solution ==
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== Solution 1 ==
 
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>.
 
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>.
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== Solution 2 ==
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Instead of counting the cases and doing <math>\frac{\text{cases wanted}}{\text{total amount of cases}}</math> we can use probability directly.
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For sake of simplicity, WLOG, assume that a pair of ones were removed from the deck of forty cards. We can split it into two cases:
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Case 1: The pair is ones.
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The probability that a one is chosen is <math>\frac{2}{38}.</math> The probability that a second one is chosen is <math>\frac{1}{37}</math> because one card was removed. Therefore, the probability that the pair is ones is <math>\frac{2}{38} \cdot \frac{1}{37}.</math>
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Case 2: The pair is <math>2-10.</math>
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The probability that any other number is chosen is <math>\frac{36}{38}.</math> The probability that a number that is equal to this number of chosen (for example, if two was chosen originally then another two being chosen) is <math>\frac{3}{37}.</math> Therefore, the probability that the pair is <math>2-10</math> is <math>\frac{36}{38} \cdot \frac{3}{37}.</math>
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Adding these two probabilities gives <math>\frac{2}{38} \cdot \frac{1}{37} + \frac{36}{38} \cdot \frac{3}{37} = \frac{110}{38 \cdot 37} = \frac{55}{703}</math>, and <math>m+n = \boxed{758}.</math>
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== Video Solution by OmegaLearn ==
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https://youtu.be/mIJ8VMuuVvA?t=59
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~ pi_is_3.14
  
 
== See also ==
 
== See also ==
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[[Category:Intermediate Combinatorics Problems]]
 
[[Category:Intermediate Combinatorics Problems]]
 
[[Category:Intermediate Probability Problems]]
 
[[Category:Intermediate Probability Problems]]
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{{MAA Notice}}

Latest revision as of 01:46, 31 July 2024

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 1

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 $m+n = \boxed{758}$.

Solution 2

Instead of counting the cases and doing $\frac{\text{cases wanted}}{\text{total amount of cases}}$ we can use probability directly.


For sake of simplicity, WLOG, assume that a pair of ones were removed from the deck of forty cards. We can split it into two cases:


Case 1: The pair is ones.

The probability that a one is chosen is $\frac{2}{38}.$ The probability that a second one is chosen is $\frac{1}{37}$ because one card was removed. Therefore, the probability that the pair is ones is $\frac{2}{38} \cdot \frac{1}{37}.$

Case 2: The pair is $2-10.$

The probability that any other number is chosen is $\frac{36}{38}.$ The probability that a number that is equal to this number of chosen (for example, if two was chosen originally then another two being chosen) is $\frac{3}{37}.$ Therefore, the probability that the pair is $2-10$ is $\frac{36}{38} \cdot \frac{3}{37}.$

Adding these two probabilities gives $\frac{2}{38} \cdot \frac{1}{37} + \frac{36}{38} \cdot \frac{3}{37} = \frac{110}{38 \cdot 37} = \frac{55}{703}$, and $m+n = \boxed{758}.$

Video Solution by OmegaLearn

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

~ pi_is_3.14

See also

2000 AIME II (Problems • Answer Key • Resources)
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