2019 AMC 8 Problems/Problem 18

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

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.

Page

Toolbox

Search

2019 AMC 8 Problems/Problem 18

Problem 18

Solution 1

See Also