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Difference between revisions of "2002 AIME I Problems/Problem 1"

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=== Solution 2 ===
 
=== Solution 2 ===
 
Using complementary counting, we count all of the license plates that do not have the desired property.  In order to not be a palindrome, the first and third characters of each string must be different.  Therefore, there are <math>10\cdot 10\cdot 9</math> three digit non-palindromes, and there are <math>26\cdot 26\cdot 25</math> three letter non palindromes.  As there are <math>10^3\cdot 26^3</math> total three-letter three-digit arrangements, the probability that a license plate does not have the desired property is <math>\frac{10\cdot 10\cdot 9\cdot 26\cdot 26\cdot 25}{10^3\cdot 26^3}=\frac{45}{52}</math>.  We subtract this from 1 to get <math>1-\frac{45}{52}=\frac{7}{52}</math> as our probability.  Therefore, our answer is <math>7+52=\boxed{059}</math>.
 
Using complementary counting, we count all of the license plates that do not have the desired property.  In order to not be a palindrome, the first and third characters of each string must be different.  Therefore, there are <math>10\cdot 10\cdot 9</math> three digit non-palindromes, and there are <math>26\cdot 26\cdot 25</math> three letter non palindromes.  As there are <math>10^3\cdot 26^3</math> total three-letter three-digit arrangements, the probability that a license plate does not have the desired property is <math>\frac{10\cdot 10\cdot 9\cdot 26\cdot 26\cdot 25}{10^3\cdot 26^3}=\frac{45}{52}</math>.  We subtract this from 1 to get <math>1-\frac{45}{52}=\frac{7}{52}</math> as our probability.  Therefore, our answer is <math>7+52=\boxed{059}</math>.
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== Video Solution by OmegaLearn ==
 +
https://youtu.be/jRZQUv4hY_k?t=98
 +
 +
~ pi_is_3.14
 +
 
== See also ==
 
== See also ==
 
{{AIME box|year=2002|n=I|before=First Question|num-a=2}}
 
{{AIME box|year=2002|n=I|before=First Question|num-a=2}}

Revision as of 04:24, 4 November 2022

Problem

Many states use a sequence of three letters followed by a sequence of three digits as their standard license-plate pattern. Given that each three-letter three-digit arrangement is equally likely, the probability that such a license plate will contain at least one palindrome (a three-letter arrangement or a three-digit arrangement that reads the same left-to-right as it does right-to-left) is $\dfrac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. Find $m+n.$

Solution

Solution 1

Consider the three-digit arrangement, $\overline{aba}$. There are $10$ choices for $a$ and $10$ choices for $b$ (since it is possible for $a=b$), and so the probability of picking the palindrome is $\frac{10 \times 10}{10^3} = \frac 1{10}$. Similarly, there is a $\frac 1{26}$ probability of picking the three-letter palindrome.

By the Principle of Inclusion-Exclusion, the total probability is

$\frac{1}{26}+\frac{1}{10}-\frac{1}{260}=\frac{35}{260}=\frac{7}{52}\quad\Longrightarrow\quad7+52=\boxed{059}$

Solution 2

Using complementary counting, we count all of the license plates that do not have the desired property. In order to not be a palindrome, the first and third characters of each string must be different. Therefore, there are $10\cdot 10\cdot 9$ three digit non-palindromes, and there are $26\cdot 26\cdot 25$ three letter non palindromes. As there are $10^3\cdot 26^3$ total three-letter three-digit arrangements, the probability that a license plate does not have the desired property is $\frac{10\cdot 10\cdot 9\cdot 26\cdot 26\cdot 25}{10^3\cdot 26^3}=\frac{45}{52}$. We subtract this from 1 to get $1-\frac{45}{52}=\frac{7}{52}$ as our probability. Therefore, our answer is $7+52=\boxed{059}$.


Video Solution by OmegaLearn

https://youtu.be/jRZQUv4hY_k?t=98

~ pi_is_3.14

See also

2002 AIME I (Problems • Answer Key • Resources)
Preceded by
First Question 		Followed by
Problem 2
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions

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