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

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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.  
 
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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Ways to roll <math>2</math> odds (Case 1): 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 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> evens (Case 2): Similarly, we have <math>2*2=4</math> ways to roll <math>2</math> evens.
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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>2</math> evens.
  
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Totally, we have <math>6*6=36</math> ways to roll 2 dies.
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Totally, we have <math>6*6=36</math> ways to roll <math>2</math> dies.
  
 
Therefore the answer is <math>\frac{16+4}{36}=\frac{20}{36}=\frac{5}{9}</math>, or <math>\framebox{C}</math>.
 
Therefore the answer is <math>\frac{16+4}{36}=\frac{20}{36}=\frac{5}{9}</math>, or <math>\framebox{C}</math>.

Revision as of 19:20, 20 November 2019

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 $2$ evens.

Totally, we have $6*6=36$ ways to roll $2$ dies.

Therefore the answer is $\frac{16+4}{36}=\frac{20}{36}=\frac{5}{9}$, or $\framebox{C}$.

~A1337h4x0r

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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