Difference between revisions of "2021 Fall AMC 10A Problems/Problem 9"
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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> | ||
| + | |||
| + | ~Eric X | ||
==Video Solution by TheBeautyofMath== | ==Video Solution by TheBeautyofMath== | ||
Revision as of 16:09, 24 October 2022
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 
~Eric X
Video Solution by TheBeautyofMath
https://youtu.be/ycRZHCOKTVk?t=661
~IceMatrix
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. 
