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

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===Solution 2===
 
===Solution 2===
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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 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}{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}{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 20 other consonants to choose 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}{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}{21000}}</math>
  
 
==See Also==
 
==See Also==

Revision as of 18:22, 25 November 2015

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"?

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

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 then M). The probability of having $8$ as the last number is $\dfrac{1}{10}$. We multiply all these to obtain $\dfrac{1}{5}\cdot \dfrac{1}{21} \cdot \dfrac{1}{20}\cdot \dfrac{1}{10}=\dfrac{1}{5\times 21\times 20\times 10}=\dfrac{1}{21\times 100\times 10}=\boxed{\textbf{(B)}~\dfrac{1}{21000}}$

See Also

2015 AMC 8 (Problems • Answer Key • Resources)
Preceded by
Problem 10 		Followed by
Problem 12
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AJHSME/AMC 8 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

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