Difference between revisions of "2021 Fall AMC 10A Problems/Problem 9"
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==Problem== | ==Problem== | ||
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When a certain unfair die is rolled, an even number is <math>3</math> times as likely to appear as an odd number. The die is rolled twice. What is the probability that the sum of the numbers rolled is even? | When a certain unfair die is rolled, an even number is <math>3</math> times as likely to appear as an odd number. The die is rolled twice. What is the probability that the sum of the numbers rolled is even? | ||
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\frac{9}{16} \qquad\textbf{(E)}\ \frac{5}{8}</math> | \frac{9}{16} \qquad\textbf{(E)}\ \frac{5}{8}</math> | ||
| − | ==Solution== | + | ==Solution 1== |
Since an even number is <math>3</math> times more likely to appear than an odd number, the probability of an even number appearing is <math>\frac{3}{4}</math>. Since the problem states that the sum of the two die must be even, the numbers must both be even or both be odd. We either have EE or OO, so we have <cmath>\frac{3}{4}\cdot \frac{3}{4} + \frac{1}{4} \cdot \frac{1}{4} = \frac {1}{16} + \frac {9}{16} = \frac{10}{16} = \boxed{\textbf{(E)}\ \frac{5}{8}}.</cmath> | Since an even number is <math>3</math> times more likely to appear than an odd number, the probability of an even number appearing is <math>\frac{3}{4}</math>. Since the problem states that the sum of the two die must be even, the numbers must both be even or both be odd. We either have EE or OO, so we have <cmath>\frac{3}{4}\cdot \frac{3}{4} + \frac{1}{4} \cdot \frac{1}{4} = \frac {1}{16} + \frac {9}{16} = \frac{10}{16} = \boxed{\textbf{(E)}\ \frac{5}{8}}.</cmath> | ||
~Arcticturn ~Aidensharp | ~Arcticturn ~Aidensharp | ||
| + | ==Solution 2 (Complementary Counting)== | ||
| + | As explained in the above solution, the probability of an even number appearing is <math>\frac{3}{4}</math>, while the probability of an odd number appearing is <math>\frac{1}{4}</math>. Then the probability of getting an odd and an even (to make an odd number) is <math>\frac{3}{4} \cdot \frac{1}{4} \cdot 2 = \frac{3}{8}.</math> Then the probability of getting an even number is <math>1 - \frac{3}{8} = \boxed{\textbf{(E)}\ \frac{5}{8}}.</math> | ||
| + | ~littlefox_amc | ||
| − | + | ==Solution 3 (Answer Choices)== | |
| + | As explained in the above solutions, the probability of an even number appearing is <math>\frac{3}{4}</math>. Getting two even numbers in a row would result in an even sum, and the probability of this happening is <math>\frac{3}{4} \cdot \frac{3}{4} = \frac{9}{16}.</math> As an even sum can also be a result of the sum of two odd numbers, the probability of an even sum is slightly greater than <math>\frac{9}{16}</math>. The only answer choice with a probability greater than <math>\frac{9}{16}</math> is <math>\boxed{\textbf{(E)}\ \frac{5}{8}}.</math> | ||
| − | ==Video Solution | + | ~TheGoldenRetriever |
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| + | ==Video Solution (HOW TO THINK CREATIVELY!!!)== | ||
| + | https://youtu.be/xZo7pKxrnGA | ||
| + | |||
| + | ~Education, the Study of Everything | ||
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| + | ==Video Solution == | ||
https://youtu.be/ycRZHCOKTVk?t=661 | https://youtu.be/ycRZHCOKTVk?t=661 | ||
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==Video Solution by WhyMath== | ==Video Solution by WhyMath== | ||
Latest revision as of 16:32, 30 October 2024
Contents
Problem
When a certain unfair die is rolled, an even number is 
times as likely to appear as an odd number. The die is rolled twice. What is the probability that the sum of the numbers rolled is even?
Solution 1
Since an even number is times more likely to appear than an odd number, the probability of an even number appearing is 
. Since the problem states that the sum of the two die must be even, the numbers must both be even or both be odd. We either have EE or OO, so we have ![\[\frac{3}{4}\cdot \frac{3}{4} + \frac{1}{4} \cdot \frac{1}{4} = \frac {1}{16} + \frac {9}{16} = \frac{10}{16} = \boxed{\textbf{(E)}\ \frac{5}{8}}.\]](http://latex.artofproblemsolving.com/a/9/d/a9df2f4f8f0bd33c690522e4a4abce345ead2579.png)
~Arcticturn ~Aidensharp
Solution 2 (Complementary Counting)
As explained in the above solution, the probability of an even number appearing is , while the probability of an odd number appearing is 
. Then the probability of getting an odd and an even (to make an odd number) is 
Then the probability of getting an even number is 
~littlefox_amc
Solution 3 (Answer Choices)
As explained in the above solutions, the probability of an even number appearing is . Getting two even numbers in a row would result in an even sum, and the probability of this happening is 
As an even sum can also be a result of the sum of two odd numbers, the probability of an even sum is slightly greater than 
. The only answer choice with a probability greater than is 
~TheGoldenRetriever
Video Solution (HOW TO THINK CREATIVELY!!!)
https://youtu.be/xZo7pKxrnGA
~Education, the Study of Everything
Video Solution
https://youtu.be/ycRZHCOKTVk?t=661
Video Solution by WhyMath
https://youtu.be/oOKx2Wqp_ig ~savannahsolver
See Also
| 2021 Fall AMC 10A (Problems • Answer Key • Resources) | ||
| Preceded by Problem 8 |
Followed by Problem 10 | |
| 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 AMC 10 Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
