Difference between revisions of "2014 AMC 10A Problems/Problem 17"
(→Solution) |
|||
| Line 6: | Line 6: | ||
==Solution== | ==Solution== | ||
| + | |||
| + | Since there are <math>6</math> possible values for the number on each dice, there are <math>6^3=216</math> total possible rolls. | ||
| + | |||
| + | Note that the possible results of the 3 dice (without respect to order) are <math>(1, 1, 2), (1, 2, 3), (1, 3, 4), (1, 4, 5), (1, 5, 6), (2, 2, 4), (2, 3, 5), (2, 4, 6) (3, 3, 6)</math>. | ||
| + | |||
| + | There are <math>3</math> ways to order the first, sixth, and ninth results, while there are 6 ways to order the other results; therefore, there are a total of 45 ways to roll the dice s.t. 2 of the dice sum to the other, so our answer is <math>\frac{45}{216}=\boxed{\textbf{(D) \frac{5}{24}}}</math> | ||
==See Also== | ==See Also== | ||
Revision as of 17:45, 7 February 2014
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
Since there are 
possible values for the number on each dice, there are 
total possible rolls.
Note that the possible results of the 3 dice (without respect to order) are 
.
There are 
ways to order the first, sixth, and ninth results, while there are 6 ways to order the other results; therefore, there are a total of 45 ways to roll the dice s.t. 2 of the dice sum to the other, so our answer is $\frac{45}{216}=\boxed{\textbf{(D) \frac{5}{24}}}$ (Error compiling LaTeX. Unknown error_msg)
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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
