Difference between revisions of "2021 Fall AMC 10A Problems/Problem 9"
Aidensharp (talk | contribs) (Created page with "==Problem== 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...") |
Arcticturn (talk | contribs) (→Solution) |
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| + | ==Solution 2== | ||
| + | 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 <math>\frac{3}{4}\cdot \frac{3}{4} + \frac{1}{4} \cdot \frac{1}{4} = \frac {1}{16} + \frac {9}{16} = \boxed{\textbf{(E)}\ \frac {5}{8}}</math>. | ||
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| + | ~Arcticturn | ||
Revision as of 16:31, 23 November 2021
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
If an even number is times more likely to appear than an odd number, the probability of an even number appearing must be 
. For the sum of two numbers to be even, the numbers must both be even or odd. So, the answer is:
.
-Aidensharp
Solution 2
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 
.
~Arcticturn
