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 "2019 AMC 8 Problems/Problem 18"

m (→‎Solution 4 (Casework))
(20 intermediate revisions by 6 users not shown)
Line 5: Line 5:
  
 
==Solution 1==
 
==Solution 1==
−
We have a <math>2</math> die with <math>2</math> evens and <math>4</math> odds on both dies. For the sum to be even, the rolls must consist of <math>2</math> odds or <math>2</math> evens.  
+
We have <math>2</math> dice with <math>2</math> evens and <math>4</math> odds on each die. For the sum to be even, the 2 rolls must be <math>2</math> odds or <math>2</math> evens.  
  
−
Ways to roll <math>2</math> odds (Case <math>1</math>): The total number of ways to roll <math>2</math> odds is <math>4*4=16</math>, as there are <math>4</math> choices for the first odd on the first roll and <math>4</math> choices for the second odd on the second roll.
+
Ways to roll <math>2</math> odds (Case <math>1</math>): The total number of ways to obtain <math>2</math> odds on 2 rolls is <math>4*4=16</math>, as there are <math>4</math> possible odds on the first roll and <math>4</math> possible odds on the second roll.
  
−
Ways to roll <math>2</math> evens (Case <math>2</math>): Similarly, we have <math>2*2=4</math> ways to roll <math>{36}=\frac{20}{36}=\frac{5}{9}</math>, or <math>\framebox{C}</math>.
+
Ways to roll <math>2</math> evens (Case <math>2</math>): Similarly, we have <math>2*2=4</math> ways to obtain 2 evens. Probability is <math>\frac{20}{36}=\frac{5}{9}</math>, or <math>\framebox{C}</math>.
  
 
==Solution 2 (Complementary Counting)==
 
==Solution 2 (Complementary Counting)==
Line 18: Line 18:
  
 
==Video Solution==
 
==Video Solution==
−
https://youtu.be/8fF55uF64mE - Happytwin
 
  
 +
==Video Solution by Math-X (First understand the problem!!!)==
 +
https://youtu.be/IgpayYB48C4?si=UsI0Wu2OeYT813rn&t=5524
  
−
Associated video - https://www.youtube.com/watch?v=EoBZy_WYWEw
+
~Math-X
 +
 
 +
 
 +
https://youtu.be/8fF55uF64mE
 +
 
 +
- Happytwin
 +
 
 +
https://www.youtube.com/watch?v=_IK58KFUYpk  ~David
 +
 
 +
https://www.youtube.com/watch?v=EoBZy_WYWEw
 +
 
 +
Associated video
  
 
https://www.youtube.com/watch?v=H52AqAl4nt4&t=2s
 
https://www.youtube.com/watch?v=H52AqAl4nt4&t=2s
Line 27: Line 39:
 
== Video Solution ==
 
== Video Solution ==
  
−
Solution detailing how to solve the problem: https://www.youtube.com/watch?v=94D1UnH7seo&list=PLbhMrFqoXXwmwbk2CWeYOYPRbGtmdPUhL&index=19
+
Solution detailing how to solve the problem:  
 +
 
 +
https://www.youtube.com/watch?v=94D1UnH7seo&list=PLbhMrFqoXXwmwbk2CWeYOYPRbGtmdPUhL&index=19
 +
 
 +
==Video Solution==
 +
https://youtu.be/8gl4rCZMUFI
 +
 
 +
~savannahsolver
 +
 
 +
==Video Solution (CREATIVE THINKING!!!)==
 +
https://youtu.be/zrmHM_l1UkI
 +
 
 +
~Education, the Study of Everything
 +
 
 +
==Video Solution by The Power of Logic(Problem 1 to 25 Full Solution)==
 +
https://youtu.be/Xm4ZGND9WoY
 +
 
 +
~Hayabusa1
 +
 
 
==See Also==
 
==See Also==
 
{{AMC8 box|year=2019|num-b=17|num-a=19}}
 
{{AMC8 box|year=2019|num-b=17|num-a=19}}
  
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 15:05, 30 December 2023

Problem 18

The faces of each of two fair dice are numbered $1$, $2$, $3$, $5$, $7$, and $8$. When the two dice are tossed, what is the probability that their sum will be an even number?

$\textbf{(A) }\frac{4}{9}\qquad\textbf{(B) }\frac{1}{2}\qquad\textbf{(C) }\frac{5}{9}\qquad\textbf{(D) }\frac{3}{5}\qquad\textbf{(E) }\frac{2}{3}$

Solution 1

We have $2$ dice with $2$ evens and $4$ odds on each die. For the sum to be even, the 2 rolls must be $2$ odds or $2$ evens.

Ways to roll $2$ odds (Case $1$): The total number of ways to obtain $2$ odds on 2 rolls is $4*4=16$, as there are $4$ possible odds on the first roll and $4$ possible odds on the second roll.

Ways to roll $2$ evens (Case $2$): Similarly, we have $2*2=4$ ways to obtain 2 evens. Probability is $\frac{20}{36}=\frac{5}{9}$, or $\framebox{C}$.

Solution 2 (Complementary Counting)

We count the ways to get an odd. If the sum is odd, then we must have an even and an odd. The probability of an even is $\frac{1}{3}$, and the probability of an odd is $\frac{2}{3}$. We have to multiply by $2!$ because the even and odd can be in any order. This gets us $\frac{4}{9}$, so the answer is $1 - \frac{4}{9} = \frac{5}{9} = \boxed{(\textbf{C})}$.

Solution 3

To get an even, you must get either 2 odds or 2 evens. The probability of getting 2 odds is $\frac{4}{6} * \frac{4}{6}$. The probability of getting 2 evens is $\frac{2}{6} * \frac{2}{6}$. If you add them together, you get $\frac{16}{36} + \frac{4}{36}$ = $\boxed{(\textbf{C}) \frac{5}{9}}$.

Video Solution

Video Solution by Math-X (First understand the problem!!!)

https://youtu.be/IgpayYB48C4?si=UsI0Wu2OeYT813rn&t=5524

~Math-X


https://youtu.be/8fF55uF64mE

- Happytwin

https://www.youtube.com/watch?v=_IK58KFUYpk ~David

https://www.youtube.com/watch?v=EoBZy_WYWEw

Associated video

https://www.youtube.com/watch?v=H52AqAl4nt4&t=2s

Video Solution

Solution detailing how to solve the problem:

https://www.youtube.com/watch?v=94D1UnH7seo&list=PLbhMrFqoXXwmwbk2CWeYOYPRbGtmdPUhL&index=19

Video Solution

https://youtu.be/8gl4rCZMUFI

~savannahsolver

Video Solution (CREATIVE THINKING!!!)

https://youtu.be/zrmHM_l1UkI

~Education, the Study of Everything

Video Solution by The Power of Logic(Problem 1 to 25 Full Solution)

https://youtu.be/Xm4ZGND9WoY

~Hayabusa1

See Also

2019 AMC 8 (Problems • Answer Key • Resources)
Preceded by
Problem 17 		Followed by
Problem 19
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 AJHSME/AMC 8 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=2019_AMC_8_Problems/Problem_18&oldid=208961"