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Difference between revisions of "2019 AMC 8 Problems/Problem 18"

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==Problem 18==
 
==Problem 18==
 
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?
 +
 
<math>\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}</math>
 
<math>\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}</math>
  
 
==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.
 +
 +
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> 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>.
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 +
==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 <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>.
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==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} * \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>.
 +
 +
==Video Solution==
 +
https://youtu.be/8fF55uF64mE - Happytwin
 +
 +
 +
Associated video - https://www.youtube.com/watch?v=EoBZy_WYWEw
 +
 +
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
 
==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 19:27, 27 April 2021

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 a $2$ die with $2$ evens and $4$ odds on both dies. For the sum to be even, the rolls must consist of $2$ odds or $2$ evens.

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

Ways to roll $2$ evens (Case $2$): Similarly, we have $2*2=4$ ways to roll ${36}=\frac{20}{36}=\frac{5}{9}$, 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

https://youtu.be/8fF55uF64mE - Happytwin


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

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

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

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