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Difference between revisions of "2016 AMC 10A Problems/Problem 12"

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<math>\textbf{(A)}\ p<\dfrac{1}{8}\qquad\textbf{(B)}\ p=\dfrac{1}{8}\qquad\textbf{(C)}\ \dfrac{1}{8}<p<\dfrac{1}{3}\qquad\textbf{(D)}\ p=\dfrac{1}{3}\qquad\textbf{(E)}\ p>\dfrac{1}{3}</math>
 
<math>\textbf{(A)}\ p<\dfrac{1}{8}\qquad\textbf{(B)}\ p=\dfrac{1}{8}\qquad\textbf{(C)}\ \dfrac{1}{8}<p<\dfrac{1}{3}\qquad\textbf{(D)}\ p=\dfrac{1}{3}\qquad\textbf{(E)}\ p>\dfrac{1}{3}</math>
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==Solution==
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There are an equal amount of even and odd numbers from <math>1</math> through <math>2016</math>. This concludes that the probability of choosing an even number is the same of that of choosing an odd number. The only way to get a product of three integers to be odd is for all three integers to be odd. If the probability of choosing an odd number is <math>\frac{1}{2}</math>, then the probability of choosing <math>3</math> odd numbers would be <cmath>\frac{1}{2}\cdot\frac{1}{2}\cdot\frac{1}{2}=\frac{1}{8}.</cmath> Therefore, the answer is <math>\boxed{\text{(B)}}</math>.

Revision as of 19:33, 3 February 2016

Three distinct integers are selected at random between $1$ and $2016$, inclusive. Which of the following is a correct statement about the probability $p$ that the product of the three integers is odd?

$\textbf{(A)}\ p<\dfrac{1}{8}\qquad\textbf{(B)}\ p=\dfrac{1}{8}\qquad\textbf{(C)}\ \dfrac{1}{8}<p<\dfrac{1}{3}\qquad\textbf{(D)}\ p=\dfrac{1}{3}\qquad\textbf{(E)}\ p>\dfrac{1}{3}$

Solution

There are an equal amount of even and odd numbers from $1$ through $2016$. This concludes that the probability of choosing an even number is the same of that of choosing an odd number. The only way to get a product of three integers to be odd is for all three integers to be odd. If the probability of choosing an odd number is $\frac{1}{2}$, then the probability of choosing $3$ odd numbers would be \[\frac{1}{2}\cdot\frac{1}{2}\cdot\frac{1}{2}=\frac{1}{8}.\] Therefore, the answer is $\boxed{\text{(B)}}$.

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