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

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

@@ Line 1: / Line 1: @@
 == 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 1 ==
-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>.
+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>.
-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>.
+== Solution 2 ==
+Instead of counting the cases and doing <math>\frac{\text{cases wanted}}{\text{total amount of cases}}</math> 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 <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>
+Case 2: The pair is <math>2-10.</math>
+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>
+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>
+== 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}}
+[[Category:Intermediate Combinatorics Problems]]
 [[Category:Intermediate Probability Problems]]
+{{MAA Notice}}

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