2016 AMC 10A Problem 12

A product of integers is odd exactly when every factor is odd. Among $1, 2, \dots, 2016$ there are $1008$ odd numbers and $1008$ even numbers.

Pick the three numbers one at a time without replacement. The first is odd with probability $\frac{1008}{2016} = \frac12$. Given that, the second is odd with probability $\frac{1007}{2015}$, and then the third with probability $\frac{1006}{2014}$. So
$$p = \frac12\cdot\frac{1007}{2015}\cdot\frac{1006}{2014}.$$

Now compare each factor to $\frac12$: $2\cdot 1007 = 2014 < 2015$ and $2\cdot 1006 = 2012 < 2014$, so both later factors are strictly less than $\frac12$. Hence
$$p < \frac12\cdot\frac12\cdot\frac12 = \frac18.$$

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

Drawing without replacement makes each success slightly less likely than the previous one when the pool of favorable items shrinks along with the total: $\frac{k-1}{2k-1} < \frac12$ always. The exact value is irrelevant; the answer choices only ask on which side of $\frac18$ the probability lies, so bound each factor instead of computing.

• Treating the picks as independent with replacement and answering exactly 1/8.
• Bounding in the wrong direction (thinking removing an odd number leaves "more than half" odd), which points to choice (C).