Difference between revisions of "2019 AMC 8 Problems/Problem 18"
(→Solution 3) |
(→Solution 1) |
||
| Line 4: | Line 4: | ||
<math>\textbf{(A) }\frac{4}{9}\qquad\textbf{(B) }\frac{1}{2}\qquad\textbf{(C) }\frac{5}{9}\qquad\textbf{(D) }\frac{3}{5}\qquad\textbf{(E) }\frac{2}{3}</math> | <math>\textbf{(A) }\frac{4}{9}\qquad\textbf{(B) }\frac{1}{2}\qquad\textbf{(C) }\frac{5}{9}\qquad\textbf{(D) }\frac{3}{5}\qquad\textbf{(E) }\frac{2}{3}</math> | ||
| − | ==Solution | + | ==Solution 2== |
We have a <math>2</math> die with <math>2</math> evens and <math>4</math> odds on both dies. For the sum to be even, the rolls must consist of <math>2</math> odds or <math>2</math> evens. | We have a <math>2</math> die with <math>2</math> evens and <math>4</math> odds on both dies. For the sum to be even, the rolls must consist of <math>2</math> odds or <math>2</math> evens. | ||
Revision as of 23:49, 2 January 2020
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 2
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 ): 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 ): Similarly, we have 
ways to roll evens.
Totally, we have 
ways to roll dies.
Therefore the answer is 
, or 
.
~A1337h4x0r
Solution 2 (Complementary Counting)
We count the ways to get an odd. If the sum is odd, then we must have an even and an odd. The probability of an even is 
, and the probability of an odd is 
. We have to multiply by 
because the even and odd can be in any order. This gets us 
, so the answer is 
. - juliankuang
Solution 3
To get an even, you must get either 2 odds or 2 evens. The probability of getting 2 odds is 
. The probability of getting 2 evens is 
. If you add them together, you get 
= 
.~heeeeeeeeheeeee
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. 
