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"

(→‎Solution 1)
 
(59 intermediate revisions by 25 users not shown)
Line 1: Line 1:
−
==Problem 18==
+
==Problem==
 
The faces of each of two fair dice are numbered <math>1</math>, <math>2</math>, <math>3</math>, <math>5</math>, <math>7</math>, and <math>8</math>. When the two dice are tossed, what is the probability that their sum will be an even number?
 
The faces of each of two fair dice are numbered <math>1</math>, <math>2</math>, <math>3</math>, <math>5</math>, <math>7</math>, and <math>8</math>. When the two dice are tossed, what is the probability that their sum will be an even number?
  
Line 5: Line 5:
  
 
==Solution 1==
 
==Solution 1==
−
The approach to this problem:
+
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.  
−
There are two cases in which the sum can be an even number: both numbers are even and both numbers are odd. This results in only one case where the sum of the numbers are odd (one odd and one even in any order) . We can solve for how many ways the <math>2</math> numbers add up to an odd number and subtract the answer from <math>1</math>.  
 
  
−
How to solve the problem:
+
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.
−
The probability of getting an odd number first is <math>\frac{4}{6}=\frac{2}{3}</math>. In order to make the sum odd, we must select an even number next. The probability of getting an even number is <math>\frac{2}{6}=\frac{1}{3}</math>. Now we multiply the two fractions: <math>\frac{2}{3}\times\frac{1}{3}=2/9</math>. However, this is not the answer because we could pick an even number first then an odd number. The equation is the same except backward and by the Communitive Property of Multiplication, the equations are it does not matter is the equation is backward or not. Thus we do <math>\frac{2}{9}\times2=\frac{4}{9}</math>. This is the probability of getting an odd number. In order to get the probability of getting an even number we do <math>1-\frac{4}{9}=\boxed{\frac{5}{9}}</math> or <math>\boxed{(C)}</math>
 
  
−
- ViratKohli2018 (VK18)
+
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==
+
==Solution 2 (Complementary Counting)==
−
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 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 <math>\frac{1}{3}</math> , and the probability of an odd is <math>\frac{2}{3}</math>. We have to multiply by <math>2!</math> because the even and odd can be in any order. This gets us <math>\frac{4}{9}</math>, so the answer is <math>1 - \frac{4}{9} = \frac{5}{9} = \boxed{(\textbf{C})}</math>.
  
−
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.
+
==Solution 3==
 +
To get an even, you must get either 2 odds or 2 evens. The probability of getting 2 odds is <math>\frac{4}{6} \times \frac{4}{6}</math>. The probability of getting 2 evens is <math>\frac{2}{6} \times \frac{2}{6}</math>. If you add them together, you get <math>\frac{16}{36} + \frac{4}{36}</math> = <math>\boxed{(\textbf{C})  \frac{5}{9}}</math>.
  
−
Ways to roll <math>2</math> evens (Case <math>2</math>): Similarly, we have <math>2*2=4</math> ways to roll <math>2</math> evens.
+
==Video Solution by Math-X (First understand the problem!!!)==
 +
https://youtu.be/IgpayYB48C4?si=UsI0Wu2OeYT813rn&t=5524
  
−
Totally, we have <math>6*6=36</math> ways to roll <math>2</math> dies.
+
~Math-X
  
−
Therefore the answer is <math>\frac{16+4}{36}=\frac{20}{36}=\frac{5}{9}</math>, or <math>\framebox{C}</math>.
 
  
−
~A1337h4x0r
+
https://youtu.be/8fF55uF64mE
  
−
==Solution 3 (Complementary Counting)==
+
- Happytwin
−
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 <math>\frac{1}{3}</math>, and the probability of an odd is <math>\frac{2}{3}</math>. We have to multiply by <math>2!</math> because the even and odd can be in any order. This gets us <math>\frac{4}{9}</math>, so the answer is <math>1 - \frac{4}{9} = \frac{5}{9} = \boxed{(\textbf{C})}</math>. - juliankuang
 
  
−
==Solution 4==
+
https://www.youtube.com/watch?v=_IK58KFUYpk  ~David
−
To get an even, you must get either 2 odds or 2 evens. The probability of getting 2 odds is <math>\frac{4}{6} * \frac{4}{6}</math>. The probability of getting 2 evens is <math>\frac{2}{6} * \frac{2}{6}</math>. If you add them together, you get <math>\frac{16}{36} + \frac{4}{36}</math> = <math>\boxed{(\textbf{C}) \frac{5}{9}}</math>.~heeeeeeeeheeeee
+
 
 +
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(1 to 25 Full Solution)==
 +
https://youtu.be/Xm4ZGND9WoY
 +
 
 +
~Hayabusa1
  
 
==See Also==
 
==See Also==

Latest revision as of 09:32, 9 November 2024

Problem

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 $\frac{4}{6} \times \frac{4}{6}$. The probability of getting 2 evens is $\frac{2}{6} \times \frac{2}{6}$. If you add them together, you get $\frac{16}{36} + \frac{4}{36}$ = $\boxed{(\textbf{C}) \frac{5}{9}}$.

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(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=232590"