Difference between revisions of "2015 AMC 8 Problems/Problem 11"

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==Problem==
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In the small country of Mathland, all automobile license plates have four symbols. The first must be a vowel (A, E, I, O, or U), the second and third must be two different letters among the 21 non-vowels, and the fourth must be a digit (0 through 9). If the symbols are chosen at random subject to these conditions, what is the probability that the plate will read "AMC8"?
 
In the small country of Mathland, all automobile license plates have four symbols. The first must be a vowel (A, E, I, O, or U), the second and third must be two different letters among the 21 non-vowels, and the fourth must be a digit (0 through 9). If the symbols are chosen at random subject to these conditions, what is the probability that the plate will read "AMC8"?
  
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</math>
 
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==Solution 1==
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==Solutions==
There is one favorable case, which is the license plate says "AMC8".  We must now find how many total cases there are.  There are <math>5</math> choices for the first letter (since it must be a vowel), <math>21</math> choices for the second letter (since it must be of 21 consonants), <math>20</math> choices for the third letter (since it must differ from the second letter), and <math>10</math> choices for the number.  This leads to <math>5 \cdot 21 \cdot 20 \cdot 10=21000</math> total possible license plates.  That means the probability of a license plate saying "AMC8" is <math>\boxed{\textbf{(B) } \frac{1}{21,000}}</math>.
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===Solution 1===
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There is one favorable case, which is the license plate says "AMC8".  We must now find how many total cases there are.  There are <math>5</math> choices for the first letter (since it must be a vowel), <math>21</math> choices for the second letter (since it must be of <math>21</math> consonants), <math>20</math> choices for the third letter (since it must differ from the second letter), and <math>10</math> choices for the number.  This leads to <math>5 \cdot 21 \cdot 20 \cdot 10=21000</math> total possible license plates.  That means the probability of a license plate saying "AMC8" is <math>\boxed{\textbf{(B) } \frac{1}{21,000}}</math>.
  
 
===Solution 2===
 
===Solution 2===
The probability of choosing A as the first letter is <math>\dfrac{1}{5}</math>.  The probability of choosing <math>M</math> next is <math>\dfrac{1}{21}</math>.  The probability of choosing C as the third letter is <math>\dfrac{1}{20}</math> (since there are 20 other consonants to choos from other then M). The probability of having <math>8</math> as the last number is <math>\dfrac{1}{10}</math>.  We multiply all these to obtain <math>\dfrac{1}{5}\cdot \dfrac{1}{21} \cdot \dfrac{1}{29}\cdot \dfrac{1}{10}=\dfrac{1}{5}\times 21\times 20\times 10}=\dfrac{1}{21\times 100\times 10}=\boxed{\textbf{(B)}~\dfrac{1}{21000}</math>
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The probability of choosing A as the first letter is <math>\dfrac{1}{5}</math>.  The probability of choosing <math>M</math> next is <math>\dfrac{1}{21}</math>.  The probability of choosing C as the third letter is <math>\dfrac{1}{20}</math> (since there are <math>20</math> other consonants to choose from other than M). The probability of having <math>8</math> as the last number is <math>\dfrac{1}{10}</math>.  We multiply all these to obtain <math>\dfrac{1}{5}\cdot \dfrac{1}{21} \cdot \dfrac{1}{20}\cdot \dfrac{1}{10}=\dfrac{1}{5\times 21\times 20\times 10}=\dfrac{1}{21\times 100\times 10}=\boxed{\textbf{(B)}~\dfrac{1}{21,000}}</math>
  
 
==See Also==
 
==See Also==

Latest revision as of 16:39, 16 January 2021

Problem

In the small country of Mathland, all automobile license plates have four symbols. The first must be a vowel (A, E, I, O, or U), the second and third must be two different letters among the 21 non-vowels, and the fourth must be a digit (0 through 9). If the symbols are chosen at random subject to these conditions, what is the probability that the plate will read "AMC8"?

$\textbf{(A) } \frac{1}{22,050} \qquad \textbf{(B) } \frac{1}{21,000}\qquad \textbf{(C) } \frac{1}{10,500}\qquad \textbf{(D) } \frac{1}{2,100} \qquad \textbf{(E) } \frac{1}{1,050}$

Solutions

Solution 1

There is one favorable case, which is the license plate says "AMC8". We must now find how many total cases there are. There are $5$ choices for the first letter (since it must be a vowel), $21$ choices for the second letter (since it must be of $21$ consonants), $20$ choices for the third letter (since it must differ from the second letter), and $10$ choices for the number. This leads to $5 \cdot 21 \cdot 20 \cdot 10=21000$ total possible license plates. That means the probability of a license plate saying "AMC8" is $\boxed{\textbf{(B) } \frac{1}{21,000}}$.

Solution 2

The probability of choosing A as the first letter is $\dfrac{1}{5}$. The probability of choosing $M$ next is $\dfrac{1}{21}$. The probability of choosing C as the third letter is $\dfrac{1}{20}$ (since there are $20$ other consonants to choose from other than M). The probability of having $8$ as the last number is $\dfrac{1}{10}$. We multiply all these to obtain $\dfrac{1}{5}\cdot \dfrac{1}{21} \cdot \dfrac{1}{20}\cdot \dfrac{1}{10}=\dfrac{1}{5\times 21\times 20\times 10}=\dfrac{1}{21\times 100\times 10}=\boxed{\textbf{(B)}~\dfrac{1}{21,000}}$

See Also

2015 AMC 8 (ProblemsAnswer KeyResources)
Preceded by
Problem 10
Followed by
Problem 12
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All AJHSME/AMC 8 Problems and Solutions

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