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

(Solution)
(Added solution)
Line 2: Line 2:
  
 
<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>
 +
 +
==Solution==
 +
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>.
 +
 +
<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>.
 +
 +
==See Also==
 +
{{AMC10 box|year=2016|ab=A|num-b=11|num-a=13}}
 +
{{MAA Notice}}

Revision as of 22:58, 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 (ProblemsAnswer KeyResources)
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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png