2019 AMC 8 Problems/Problem 18

Revision as of 19:18, 20 November 2019 by A1337h4x0r (talk | contribs) (Solution 1)

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 (ProblemsAnswer KeyResources)
Preceded by
Problem 17
Followed by
Problem 19
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All AJHSME/AMC 8 Problems and Solutions

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