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> | ||
| − | == Solution == | + | == 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>. | ||
| + | == 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> | ||
| + | |||
| + | Thus, 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 == | == Video Solution by OmegaLearn == | ||
Revision as of 02:45, 31 July 2024
Problem
A deck of forty cards consists of four 
's, four 
's,..., and four 
'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 
be the probability that two randomly selected cards also form a pair, where 
and 
are relatively prime positive integers. Find 
Solution 1
There are 
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 
ways, or if the two cards are the remaining two cards of the number that was removed, which can happen in way. Thus, the answer is 
, and 
.
Solution 2
Instead of counting the cases and doing $\frac{\text{cases wanted}{\text{total amount of cases}}$ (Error compiling LaTeX. Unknown error_msg) 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 
The probability that a second one is chosen is 
because one card was removed. Therefore, the probability that the pair is ones is 
Case 2: The pair is 
The probability that any other number is chosen is 
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 
Therefore, the probability that the pair is 
is 
Thus, adding these two probabilities gives 
, and 
Video Solution by OmegaLearn
https://youtu.be/mIJ8VMuuVvA?t=59
~ pi_is_3.14
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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
