2019 AMC 8 Problems/Problem 18
Contents
- 1 Problem 18
- 2 Solution 1
- 3 Solution 2 (Complementary Counting)
- 4 Solution 3
- 5 Video Solution
- 6 Video Solution by Math-X (First understand the problem!!!)
- 7 Video Solution
- 8 Video Solution
- 9 Video Solution (CREATIVE THINKING!!!)
- 10 Video Solution by The Power of Logic(Problem 1 to 25 Full Solution)
- 11 See Also
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 dice with evens and odds on each die. For the sum to be even, the 2 rolls must be odds or evens.
Ways to roll odds (Case ): The total number of ways to obtain odds on 2 rolls is , as there are possible odds on the first roll and possible odds on the second roll.
Ways to roll evens (Case ): Similarly, we have ways to obtain 2 evens. Probability is , or .
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 .
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 = .
Video Solution
Video Solution by Math-X (First understand the problem!!!)
https://youtu.be/IgpayYB48C4?si=UsI0Wu2OeYT813rn&t=5524
~Math-X
- Happytwin
https://www.youtube.com/watch?v=_IK58KFUYpk ~David
https://www.youtube.com/watch?v=EoBZy_WYWEw
Associated video
https://www.youtube.com/watch?v=H52AqAl4nt4&t=2s
Video Solution
Solution detailing how to solve the problem:
https://www.youtube.com/watch?v=94D1UnH7seo&list=PLbhMrFqoXXwmwbk2CWeYOYPRbGtmdPUhL&index=19
Video Solution
~savannahsolver
Video Solution (CREATIVE THINKING!!!)
~Education, the Study of Everything
Video Solution by The Power of Logic(Problem 1 to 25 Full Solution)
~Hayabusa1
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.