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

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==Problem 18==
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==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?
  
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==Solution 1==
 
==Solution 1==
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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.  
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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.  
  
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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.
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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.
  
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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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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)==
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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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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 4==
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==Solution 3==
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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>.
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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>.
  
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==Solution 5 (Casework)==
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==Video Solution by Math-X (First understand the problem!!!)==
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To get an even number, we must either have two odds or two evens. We will solve this through casework. The probability of rolling a 1 is <math>\frac{1}{6}</math>, and the probability of rolling another odd number after this is 4/6=2/3, so the probability of getting a sum of an even number is (1/6)(2/3)=1/9. The probability of rolling a <math>2</math> is <math>\frac{1}{6}</math>, and the probability of rolling another even number after this is 2/6=1/3, so the probability of rolling a sum of an even number is <math>\frac{1}{6}\cdot\frac{1}{3}=\frac{1}{18}</math>. Now, notice that the probability of getting an even sum with two odd numbers is identical for all odd numbers. This is because the probability of probability of getting an even number is identical for all even numbers, so the probability of getting an even sum with only even numbers is (2)(1/18)=1/9. Adding these two up, we get our desired <math>\boxed{(\textbf{C})  \frac{5}{9}}</math>
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https://youtu.be/IgpayYB48C4?si=UsI0Wu2OeYT813rn&t=5524
  
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==Video Solution==
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~Math-X
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https://youtu.be/8fF55uF64mE - Happytwin
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 +
 
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https://youtu.be/8fF55uF64mE  
 +
 
 +
- Happytwin
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 +
https://www.youtube.com/watch?v=_IK58KFUYpk  ~David
  
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https://www.youtube.com/watch?v=EoBZy_WYWEw
  
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Associated video - https://www.youtube.com/watch?v=EoBZy_WYWEw
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Associated video
  
 
https://www.youtube.com/watch?v=H52AqAl4nt4&t=2s
 
https://www.youtube.com/watch?v=H52AqAl4nt4&t=2s
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== Video Solution ==
 
== Video Solution ==
  
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Solution detailing how to solve the problem: https://www.youtube.com/watch?v=94D1UnH7seo&list=PLbhMrFqoXXwmwbk2CWeYOYPRbGtmdPUhL&index=19
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Solution detailing how to solve the problem:  
 +
 
 +
https://www.youtube.com/watch?v=94D1UnH7seo&list=PLbhMrFqoXXwmwbk2CWeYOYPRbGtmdPUhL&index=19
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 +
==Video Solution==
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https://youtu.be/8gl4rCZMUFI
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~savannahsolver
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==Video Solution (CREATIVE THINKING!!!)==
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https://youtu.be/zrmHM_l1UkI
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 +
~Education, the Study of Everything
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 +
==Video Solution by The Power of Logic(1 to 25 Full Solution)==
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https://youtu.be/Xm4ZGND9WoY
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 +
~Hayabusa1
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==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}}

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

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