Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2021 Fall AMC 10A Problems/Problem 9"

(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...")
 
m
 
(15 intermediate revisions by 9 users not shown)
Line 5: Line 5:
 
  \frac{9}{16} \qquad\textbf{(E)}\ \frac{5}{8}</math>
 
  \frac{9}{16} \qquad\textbf{(E)}\ \frac{5}{8}</math>
  
−
==Solution==
+
==Solution 1==
−
If an even number is <math>3</math> times more likely to appear than an odd number, the probability of an even number appearing must be <math>\frac{3}{4}</math>. For the sum of two numbers to be even, the numbers must both be even or odd. So, the answer is:
+
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>
  
−
<math>\frac{3}{4}\cdot \frac{3}{4} + \frac{1}{4}\cdot \frac{1}{4} = \frac{10}{16} = \boxed{\textbf{(E)}\ \frac{5}{8}}</math>.
+
~Arcticturn ~Aidensharp
  
−
-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
 +
 
 +
==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==
 +
{{AMC10 box|year=2021 Fall|ab=A|num-b=8|num-a=10}}
 +
{{MAA Notice}}

Latest revision as of 18:33, 2 November 2023

Problem

When a certain unfair die is rolled, an even number is $3$ 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?

$\textbf{(A)}\ \frac{3}{8} \qquad\textbf{(B)}\ \frac{4}{9} \qquad\textbf{(C)}\ \frac{5}{9} \qquad\textbf{(D)}\ \frac{9}{16} \qquad\textbf{(E)}\ \frac{5}{8}$

Solution 1

Since an even number is $3$ times more likely to appear than an odd number, the probability of an even number appearing is $\frac{3}{4}$. 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 \[\frac{3}{4}\cdot \frac{3}{4} + \frac{1}{4} \cdot \frac{1}{4} = \frac {1}{16} + \frac {9}{16} = \frac{10}{16} = \boxed{\textbf{(E)}\ \frac{5}{8}}.\]

~Arcticturn ~Aidensharp

Solution 2 (Complementary Counting)

As explained in the above solution, the probability of an even number appearing is $\frac{3}{4}$, while the probability of an odd number appearing is $\frac{1}{4}$. Then the probability of getting an odd and an even (to make an odd number) is $\frac{3}{4} \cdot \frac{1}{4} \cdot 2 = \frac{3}{8}.$ Then the probability of getting an even number is $1 - \frac{3}{8} = \boxed{\textbf{(E)}\ \frac{5}{8}}.$

~littlefox_amc

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. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2021_Fall_AMC_10A_Problems/Problem_9&oldid=200638"