2016 AMC 10A Problems/Problem 12


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 1

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 $\boxed{\textbf{(A) }p<\dfrac{1}{8}}$.

Solution 2

For the product to be odd, all three factors have to be odd. There are a total of $\binom{2016}{3}$ ways to choose 3 numbers at random, and there are $\binom{1008}{3}$ to choose 3 odd numbers. Therefore, the probability of choosing 3 odd numbers is $\frac{\binom{1008}{3}}{\binom{2016}{3}}$. Simplifying this, we obtain $\frac{1008*1007*1006}{2016*2015*2014}$, which is slightly less than $\frac{1}{8}$, so our answer is $\boxed{\textbf{(A) }p<\dfrac{1}{8}}$.

Solution 3

The probability that the product is odd, allowing duplication of the integers, is just $\left( \frac{1}{2} \right) ^3 = \frac{1}{8}$. Since forbidding duplication reduces the probability of all three integers being odd, we see $p<\dfrac{1}{8}$ and our answer is $\boxed{\textbf{(A) }}$.

Video Solution



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

Invalid username
Login to AoPS