Difference between revisions of "2019 AMC 8 Problems/Problem 18"
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==Solution 1== | ==Solution 1== | ||
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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. | ||
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Ways to roll <math>2</math> evens (Case <math>2</math>): Similarly, we have <math>2*2=4</math> ways to roll <math>{36}=\frac{20}{36}=\frac{5}{9}</math>, or <math>\framebox{C}</math>. | Ways to roll <math>2</math> evens (Case <math>2</math>): Similarly, we have <math>2*2=4</math> ways to roll <math>{36}=\frac{20}{36}=\frac{5}{9}</math>, or <math>\framebox{C}</math>. | ||
− | ==Solution | + | ==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 | + | ==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} * \frac{4}{6}</math>. The probability of getting 2 evens is <math>\frac{2}{6} * \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>. | To get an even, you must get either 2 odds or 2 evens. The probability of getting 2 odds is <math>\frac{4}{6} * \frac{4}{6}</math>. The probability of getting 2 evens is <math>\frac{2}{6} * \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>. | ||
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==Video Solution== | ==Video Solution== | ||
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https://www.youtube.com/watch?v=H52AqAl4nt4&t=2s | https://www.youtube.com/watch?v=H52AqAl4nt4&t=2s | ||
+ | == Video Solution == | ||
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+ | Solution detailing how to solve the problem: https://www.youtube.com/watch?v=94D1UnH7seo&list=PLbhMrFqoXXwmwbk2CWeYOYPRbGtmdPUhL&index=19 | ||
==See Also== | ==See Also== | ||
{{AMC8 box|year=2019|num-b=17|num-a=19}} | {{AMC8 box|year=2019|num-b=17|num-a=19}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 19:27, 27 April 2021
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 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 , 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
https://youtu.be/8fF55uF64mE - Happytwin
Associated video - https://www.youtube.com/watch?v=EoBZy_WYWEw
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
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 |
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