2021 Fall AMC 10A Problems/Problem 9

Problem

When a certain unfair die is rolled, an even number is $3$ times as likely to appear as an odd number. The die is rolled twice. What is the probability that the sum of the numbers rolled is even?

$\textbf{(A)}\ \frac{3}{8} \qquad\textbf{(B)}\ \frac{4}{9} \qquad\textbf{(C)}\ \frac{5}{9} \qquad\textbf{(D)}\ \frac{9}{16} \qquad\textbf{(E)}\ \frac{5}{8}$

Solution 1

Since an even number is $3$ times more likely to appear than an odd number, the probability of an even number appearing is $\frac{3}{4}$ . Since the problem states that the sum of the two die must be even, the numbers must both be even or both be odd. We either have EE or OO, so we have $\frac{3}{4}\cdot \frac{3}{4} + \frac{1}{4} \cdot \frac{1}{4} = \frac {1}{16} + \frac {9}{16} = \frac{10}{16} = \boxed{\textbf{(E)}\ \frac{5}{8}}.$

~Arcticturn ~Aidensharp

Solution 2 (Complementary Counting)

As explained in the above solution, the probability of an even number appearing is $\frac{3}{4}$ , while the probability of an odd number appearing is $\frac{1}{4}$ . Then the probability of getting an odd and an even (to make an odd number) is $\frac{3}{4} \cdot \frac{1}{4} \cdot 2 = \frac{3}{8}.$ Then the probability of getting an even number is $1 - \frac{3}{8} = \boxed{\textbf{(E)}\ \frac{5}{8}}.$

~littlefox_amc

Solution 3 (Answer Choices)

As explained in the above solutions, the probability of an even number appearing is $\frac{3}{4}$ . Getting two even numbers in a row would result in an even sum, and the probability of this happening is $\frac{3}{4} \cdot \frac{3}{4} = \frac{9}{16}.$ As an even sum can also be a result of the sum of two odd numbers, the probability of an even sum is slightly greater than $\frac{9}{16}$ . The only answer choice with a probability greater than $\frac{9}{16}$ is $\boxed{\textbf{(E)}\ \frac{5}{8}}.$

~TheGoldenRetriever

Video Solution (HOW TO THINK CREATIVELY!!!)

https://youtu.be/xZo7pKxrnGA

~Education, the Study of Everything

Video Solution

https://youtu.be/ycRZHCOKTVk?t=661

Video Solution by WhyMath

https://youtu.be/oOKx2Wqp_ig ~savannahsolver

2021 Fall AMC 10A (Problems • Answer Key • Resources)
Preceded by Problem 8	Followed by Problem 10
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 AMC 10 Problems and Solutions

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