Difference between revisions of "2012 AMC 10A Problems/Problem 9"
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| − | == Problem | + | == 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? | 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? | ||
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There are three ways that a sum of 7 can be rolled. <math>2+5</math>, <math>4+3</math>, and <math>6+1</math>. 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 <math>4 + 4 + 4 = 12</math>. | There are three ways that a sum of 7 can be rolled. <math>2+5</math>, <math>4+3</math>, and <math>6+1</math>. 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 <math>4 + 4 + 4 = 12</math>. | ||
| − | Thus, our probability is <math>\frac{12}{36} = \frac{1}{3}</math>. The answer is <math>\ | + | Thus, our probability is <math>\frac{12}{36} = \frac{1}{3}</math>. The answer is <math>\boxed{\textbf{(D)}\ \frac13}</math>. |
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| + | == See Also == | ||
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| + | {{AMC10 box|year=2012|ab=A|num-b=8|num-a=10}} | ||
Revision as of 22:53, 8 February 2012
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
To solve this, we need to find the number of ways that we can roll a sum of 7 divided by the total possible rolls.
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 
.
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 | ||
