Difference between revisions of "2014 AMC 10A Problems/Problem 17"
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Therefore, there are a total of <math>3\times3+6\times6=45</math> ways to roll the dice such that 2 of the dice sum to the other die, so our answer is <cmath>\frac{45}{216}=\boxed{\textbf{(D)} \ \frac{5}{24}}</cmath> | Therefore, there are a total of <math>3\times3+6\times6=45</math> ways to roll the dice such that 2 of the dice sum to the other die, so our answer is <cmath>\frac{45}{216}=\boxed{\textbf{(D)} \ \frac{5}{24}}</cmath> | ||
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==See Also== | ==See Also== |
Revision as of 14:39, 26 January 2015
Problem
Three fair six-sided dice are rolled. What is the probability that the values shown on two of the dice sum to the value shown on the remaining die?
Solution 1 (Clean Counting)
First, we note that there are and ways to get sums of respectively--this is not too hard to see. With any specific sum, there is exactly one way to attain it on the other die. This means that the probability that two specific dice have the same sum as the other is Since there are ways to choose which die will be the one with the sum of the other two, our answer is .
Solution 2 (Casework)
Since there are possible values for the number on each dice, there are total possible rolls.
The possible results of the 3 dice such that the sum of the values of two of the die is equal to the value of the third die are, without considering the order of the die,
There are ways to order the first, sixth, and ninth results, and there are ways to order the other results.
Therefore, there are a total of ways to roll the dice such that 2 of the dice sum to the other die, so our answer is
See Also
2014 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 16 |
Followed by Problem 18 | |
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 |
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