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

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

Video Solution by TheBeautyofMath

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

~IceMatrix

Video Solution by WhyMath

https://youtu.be/oOKx2Wqp_ig ~savannahsolver

See Also

2021 Fall AMC 10A (ProblemsAnswer KeyResources)
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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