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2021 Fall AMC 10A Problems/Problem 9

Revision as of 15:31, 23 November 2021 by Arcticturn (talk | contribs) (→‎Solution)

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

Solution 2

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 $\frac{3}{4}\cdot \frac{3}{4} + \frac{1}{4} \cdot \frac{1}{4} = \frac {1}{16} + \frac {9}{16} = \boxed{\textbf{(E)}\ \frac {5}{8}}$.

~Arcticturn

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