Difference between revisions of "2021 Fall AMC 10A Problems/Problem 9"
MRENTHUSIASM (talk | contribs) m (Combined solutions since they are incredibly similar. Credit to both authors.) |
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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 == | ||
| + | Denote by <math>p</math> the probability of getting an even number. Hence, <math>p = \frac{3}{4}</math>. | ||
| + | |||
| + | To get the sum of the numbers rolled twice even, these two numbers are either both even or both odd. | ||
| + | |||
| + | Therefore, the probability is | ||
| + | \begin{align*} | ||
| + | p^2 + \left( 1 - p \right)^2 = \frac{5}{8} . | ||
| + | \end{align*} | ||
| + | |||
| + | Therefore, the answer is <math>\boxed{\textbf{(E) }\frac{5}{8}}</math>. | ||
| + | |||
| + | ~Steven Chen (www.professorchenedu.com) | ||
| + | |||
==See Also== | ==See Also== | ||
{{AMC10 box|year=2021 Fall|ab=A|num-b=8|num-a=10}} | {{AMC10 box|year=2021 Fall|ab=A|num-b=8|num-a=10}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 20:03, 25 November 2021
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
Denote by 
the probability of getting an even number. Hence, 
.
To get the sum of the numbers rolled twice even, these two numbers are either both even or both odd.
Therefore, the probability is \begin{align*} p^2 + \left( 1 - p \right)^2 = \frac{5}{8} . \end{align*}
Therefore, the answer is 
.
~Steven Chen (www.professorchenedu.com)
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. 
