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

If an even number is $3$ times more likely to appear than an odd number, the probability of an even number appearing must be $\frac{3}{4}$ . For the sum of two numbers to be even, the numbers must both be even or odd. So, the answer is:

$\frac{3}{4}\cdot \frac{3}{4} + \frac{1}{4}\cdot \frac{1}{4} = \frac{10}{16} = \boxed{\textbf{(E)}\ \frac{5}{8}}$ .

-Aidensharp

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2021 Fall AMC 10A Problems/Problem 9

Problem

Solution