Difference between revisions of "2011 AMC 8 Problems/Problem 18"
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==Problem== | ==Problem== | ||
− | A fair 6 | + | A fair <math>6</math> sided die is rolled twice. What is the probability that the first number that comes up is greater than or equal to the second number? |
<math> \textbf{(A) }\dfrac16\qquad\textbf{(B) }\dfrac5{12}\qquad\textbf{(C) }\dfrac12\qquad\textbf{(D) }\dfrac7{12}\qquad\textbf{(E) }\dfrac56 </math> | <math> \textbf{(A) }\dfrac16\qquad\textbf{(B) }\dfrac5{12}\qquad\textbf{(C) }\dfrac12\qquad\textbf{(D) }\dfrac7{12}\qquad\textbf{(E) }\dfrac56 </math> | ||
==Solution== | ==Solution== | ||
− | There are <math>6\cdot6=36</math> ways to roll the two dice, and | + | There are <math>6\cdot6=36</math> ways to roll the two dice, and 6 of them result in two of the same number. Out of the remaining <math>36-6=30</math> ways, the number of rolls where the first dice is greater than the second should be the same as the number of rolls where the second dice is greater than the first. In other words, there are <math>30/2=15</math> ways the first roll can be greater than the second. The probability the first number is greater than or equal to the second number is |
<cmath>\frac{15+6}{36}=\frac{21}{36}=\boxed{\textbf{(D)}\ \frac{7}{12}}</cmath> | <cmath>\frac{15+6}{36}=\frac{21}{36}=\boxed{\textbf{(D)}\ \frac{7}{12}}</cmath> |
Latest revision as of 02:28, 30 October 2020
Problem
A fair sided die is rolled twice. What is the probability that the first number that comes up is greater than or equal to the second number?
Solution
There are ways to roll the two dice, and 6 of them result in two of the same number. Out of the remaining ways, the number of rolls where the first dice is greater than the second should be the same as the number of rolls where the second dice is greater than the first. In other words, there are ways the first roll can be greater than the second. The probability the first number is greater than or equal to the second number is
See Also
2011 AMC 8 (Problems • Answer Key • Resources) | ||
Preceded by Problem 17 |
Followed by Problem 19 | |
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.