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

Revision as of 07:53, 26 March 2023

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 digit. This leads to $5 \cdot 21 \cdot 20 \cdot 10=21,000$ total possible license plates. Therefore, 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 $\dfrac{1}{5}\cdot \dfrac{1}{21} \cdot \dfrac{1}{20}\cdot \dfrac{1}{10}=\dfrac{1}{5\times 21\times 20\times 10}=\dfrac{1}{21\times 100\times 10}=\boxed{\textbf{(B)}~\dfrac{1}{21,000}}$ .

Video Solution

https://youtu.be/fW34t6eFKAU

~savannahsolver

@@ Line 14: / Line 14: @@
 ===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 <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 digit.  This leads to <math>5 \cdot 21 \cdot 20 \cdot 10=21000</math> total possible license plates.  Therefore, the probability of a license plate saying "AMC8" is <math>\boxed{\textbf{(B) } \frac{1}{21,000}}</math>.
+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 digit.  This leads to <math>5 \cdot 21 \cdot 20 \cdot 10=21,000</math> total possible license plates.  Therefore, the probability of a license plate saying "AMC8" is <math>\boxed{\textbf{(B) } \frac{1}{21,000}}</math>.
 ===Solution 2===

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

Page

Toolbox

Search