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 $2$ 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 $\frac{4}{6} * \frac{4}{6}$. 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

Video Solution by Math-X (First understand the problem!!!)




- Happytwin

https://www.youtube.com/watch?v=_IK58KFUYpk ~David


Associated video


Video Solution

Solution detailing how to solve the problem:


Video Solution



Video Solution (CREATIVE THINKING!!!)


~Education, the Study of Everything

Video Solution by The Power of Logic(Problem 1 to 25 Full Solution)



See Also

2019 AMC 8 (ProblemsAnswer KeyResources)
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. AMC logo.png