Difference between revisions of "2019 AMC 8 Problems/Problem 18"
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==Solution 1== | ==Solution 1== | ||
| − | We have | + | We have <math>2</math> dice with <math>2</math> evens and <math>4</math> odds on each die. For the sum to be even, the 2 rolls must be <math>2</math> odds or <math>2</math> evens. |
Ways to roll <math>2</math> odds (Case <math>1</math>): The total number of ways to obtain <math>2</math> odds on 2 rolls is <math>4*4=16</math>, as there are <math>4</math> possible odds on the first roll and <math>4</math> possible odds on the second roll. | Ways to roll <math>2</math> odds (Case <math>1</math>): The total number of ways to obtain <math>2</math> odds on 2 rolls is <math>4*4=16</math>, as there are <math>4</math> possible odds on the first roll and <math>4</math> possible odds on the second roll. | ||
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==Solution 2 (Complementary Counting)== | ==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 <math>\frac{1}{3}</math>, and the probability of an odd is <math>\frac{2}{3}</math>. We have to multiply by <math>2!</math> because the even and odd can be in any order. This gets us <math>\frac{4}{9}</math>, so the answer is <math>1 - \frac{4}{9} = \frac{5}{9} = \boxed{(\textbf{C})}</math>. | + | 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 <math>\frac{1}{3}</math> , and the probability of an odd is <math>\frac{2}{3}</math>. We have to multiply by <math>2!</math> because the even and odd can be in any order. This gets us <math>\frac{4}{9}</math>, so the answer is <math>1 - \frac{4}{9} = \frac{5}{9} = \boxed{(\textbf{C})}</math>. |
==Solution 3== | ==Solution 3== | ||
| − | To get an even, you must get either 2 odds or 2 evens. The probability of getting 2 odds is <math>\frac{4}{6} | + | To get an even, you must get either 2 odds or 2 evens. The probability of getting 2 odds is <math>\frac{4}{6} \times \frac{4}{6}</math>. The probability of getting 2 evens is <math>\frac{2}{6} \times \frac{2}{6}</math>. If you add them together, you get <math>\frac{16}{36} + \frac{4}{36}</math> = <math>\boxed{(\textbf{C}) \frac{5}{9}}</math>. |
| − | ==Video Solution== | + | ==Video Solution by Math-X (First understand the problem!!!)== |
| − | https://youtu.be/8fF55uF64mE - Happytwin | + | https://youtu.be/IgpayYB48C4?si=UsI0Wu2OeYT813rn&t=5524 |
| + | |||
| + | ~Math-X | ||
| + | |||
| + | |||
| + | https://youtu.be/8fF55uF64mE | ||
| + | |||
| + | - Happytwin | ||
| + | |||
| + | https://www.youtube.com/watch?v=_IK58KFUYpk ~David | ||
| + | https://www.youtube.com/watch?v=EoBZy_WYWEw | ||
| − | Associated video | + | Associated video |
https://www.youtube.com/watch?v=H52AqAl4nt4&t=2s | https://www.youtube.com/watch?v=H52AqAl4nt4&t=2s | ||
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== Video Solution == | == Video Solution == | ||
| − | Solution detailing how to solve the problem: https://www.youtube.com/watch?v=94D1UnH7seo&list=PLbhMrFqoXXwmwbk2CWeYOYPRbGtmdPUhL&index=19 | + | Solution detailing how to solve the problem: |
| + | |||
| + | https://www.youtube.com/watch?v=94D1UnH7seo&list=PLbhMrFqoXXwmwbk2CWeYOYPRbGtmdPUhL&index=19 | ||
==Video Solution== | ==Video Solution== | ||
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~savannahsolver | ~savannahsolver | ||
| + | |||
| + | ==Video Solution (CREATIVE THINKING!!!)== | ||
| + | https://youtu.be/zrmHM_l1UkI | ||
| + | |||
| + | ~Education, the Study of Everything | ||
| + | |||
| + | ==Video Solution by The Power of Logic(Problem 1 to 25 Full Solution)== | ||
| + | https://youtu.be/Xm4ZGND9WoY | ||
| + | |||
| + | ~Hayabusa1 | ||
==See Also== | ==See Also== | ||
Latest revision as of 20:55, 27 October 2024
Contents
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 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. 
