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?
$\textbf{(A)}\ p\dfrac{1}{3}$
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 $\boxed{\textbf{(A) }p<\dfrac{1}{8}}$.