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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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For the product to be odd, all three factors have to be odd. The probability of this is <math>\frac{1008}{2016} \cdot \frac{1007}{2015} \cdot \frac{1006}{2014}</math>.
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<math>\frac{1008}{2016} = \frac{1}{2}</math>, but <math>\frac{1007}{2015}</math> and <math>\frac{1006}{2014}</math> are slightly less than <math>\frac{1}{2}</math>. Thus, the whole product is slightly less than <math>\frac{1}{2} \cdot \frac{1}{2} \cdot \frac{1}{2} = \frac{1}{8}</math>, so <math>\boxed{p<\dfrac{1}{8}}</math>.
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==See Also==
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{{AMC10 box|year=2016|ab=A|num-b=11|num-a=13}}
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{{MAA Notice}}

Revision as of 21:58, 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

For the product to be odd, all three factors have to be odd. The probability of this is $\frac{1008}{2016} \cdot \frac{1007}{2015} \cdot \frac{1006}{2014}$.

$\frac{1008}{2016} = \frac{1}{2}$, but $\frac{1007}{2015}$ and $\frac{1006}{2014}$ are slightly less than $\frac{1}{2}$. Thus, the whole product is slightly less than $\frac{1}{2} \cdot \frac{1}{2} \cdot \frac{1}{2} = \frac{1}{8}$, so $\boxed{p<\dfrac{1}{8}}$.

See Also

2016 AMC 10A (Problems • Answer Key • Resources)
Preceded by
Problem 11 		Followed by
Problem 13
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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