Difference between revisions of "2012 AMC 10A Problems/Problem 9"
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Assume we roll the die with only evens first. For whatever value rolled, there are exactly 2 faces on the odd die that makes the sum 7. The odd die has 6 faces, so our probability is <math>\boxed{\textbf{(D)}\ \frac13}</math>. | Assume we roll the die with only evens first. For whatever value rolled, there are exactly 2 faces on the odd die that makes the sum 7. The odd die has 6 faces, so our probability is <math>\boxed{\textbf{(D)}\ \frac13}</math>. | ||
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| + | ==Video Solution (CREATIVE THINKING)== | ||
| + | https://youtu.be/0ELm7O84mBY | ||
| + | |||
| + | ~Education, the Study of Everything | ||
== See Also == | == See Also == | ||
Latest revision as of 12:54, 1 July 2023
Problem
A pair of six-sided dice are labeled so that one die has only even numbers (two each of 2, 4, and 6), and the other die has only odd numbers (two of each 1, 3, and 5). The pair of dice is rolled. What is the probability that the sum of the numbers on the tops of the two dice is 7?
Solution
The total number of combinations when rolling two dice is 
.
There are three ways that a sum of 7 can be rolled. 
, 
, and 
. There are two 2's on one die and two 5's on the other, so there are a total of 4 ways to roll the combination of 2 and 5. There are two 4's on one die and two 3's on the other, so there are a total of 4 ways to roll the combination of 4 and 3. There are two 6's on one die and two 1's on the other, so there are a total of 4 ways to roll the combination of 6 and 1. Add 
.
Thus, our probability is 
. The answer is 
.
Solution 2
Assume we roll the die with only evens first. For whatever value rolled, there are exactly 2 faces on the odd die that makes the sum 7. The odd die has 6 faces, so our probability is .
Video Solution (CREATIVE THINKING)
~Education, the Study of Everything
See Also
| 2012 AMC 10A (Problems • Answer Key • Resources) | ||
| Preceded by Problem 8 |
Followed by Problem 10 | |
| 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 AMC 10 Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
