Difference between revisions of "2015 AMC 8 Problems/Problem 11"
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===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> | + | 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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==See Also== | ==See Also== | ||
{{AMC8 box|year=2015|num-b=10|num-a=12}} | {{AMC8 box|year=2015|num-b=10|num-a=12}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 17:20, 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"?
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 
choices for the first letter (since it must be a vowel), 
choices for the second letter (since it must be of 21 consonants), 
choices for the third letter (since it must differ from the second letter), and 
choices for the number. This leads to 
total possible license plates. That means the probability of a license plate saying "AMC8" is 
.
Solution 2
The probability of choosing A as the first letter is 
. The probability of choosing 
next is 
. The probability of choosing C as the third letter is 
(since there are 20 other consonants to choos from other then M). The probability of having 
as the last number is 
. We multiply all these to obtain $\dfrac{1}{5}\cdot \dfrac{1}{21} \cdot \dfrac{1}{29}\cdot \dfrac{1}{10}=\dfrac{1}{5}\times 21\times 20\times 10}=\dfrac{1}{21\times 100\times 10}=\boxed{\textbf{(B)}~\dfrac{1}{21000}$ (Error compiling LaTeX. Unknown error_msg)
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. 
