Difference between revisions of "2019 AMC 8 Problems/Problem 18"
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Totally, we have <math>6*6=36</math> ways to roll 2 dies. | Totally, we have <math>6*6=36</math> ways to roll 2 dies. | ||
| − | Therefore the answer is <math>\frac{16+4}{36}=\frac{20}{36}=\frac{5}{9}</math>, or <math>\ | + | Therefore the answer is <math>\frac{16+4}{36}=\frac{20}{36}=\frac{5}{9}</math>, or <math>\framebox{C}</math>. |
==See Also== | ==See Also== | ||
Revision as of 19:18, 20 November 2019
Problem 18
The faces of each of two fair dice are numbered 
, 
, 
, 
, 
, and 
. When the two dice are tossed, what is the probability that their sum will be an even number?
Solution 1
We have a die with evens and 
odds on both dies. For the sum to be even, the rolls must consist of odds or evens.
Ways to roll odds (Case 1): The total number of ways to roll odds is 
, as there are choices for the first odd on the first roll and choices for the second odd on the second roll.
Ways to roll evens (Case 2): Similarly, we have 
ways to roll evens.
Totally, we have 
ways to roll 2 dies.
Therefore the answer is 
, or 
.
See Also
| 2019 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. 
