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Difference between revisions of "2021 Fall AMC 10A Problems/Problem 9"
m (Combined solutions since they are incredibly similar. Credit to both authors.)
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==Solution==
 
==Solution==
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If an even number is <math>3</math> times more likely to appear than an odd number, the probability of an even number appearing must be <math>\frac{3}{4}</math>. For the sum of two numbers to be even, the numbers must both be even or odd. So, the answer is:
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Since an even number is <math>3</math> times more likely to appear than an odd number, the probability of an even number appearing is <math>\frac{3}{4}</math>. 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 <cmath>\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}}.</cmath>
  
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<math>\frac{3}{4}\cdot \frac{3}{4} + \frac{1}{4}\cdot \frac{1}{4} = \frac{10}{16} = \boxed{\textbf{(E)}\ \frac{5}{8}}</math>.
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~Arcticturn ~Aidensharp
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-Aidensharp
 
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==Solution 2==
 
−
Since an even number is <math>3</math> times more likely to appear than an odd number, the probability of an even number appearing is <math>\frac{3}{4}</math>. 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 <math>\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}}</math>.
 
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~Arcticturn
 
  
 
==See Also==
 
==See Also==
 
{{AMC10 box|year=2021 Fall|ab=A|num-b=8|num-a=10}}
 
{{AMC10 box|year=2021 Fall|ab=A|num-b=8|num-a=10}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 18:52, 24 November 2021

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

See Also

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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