2016 AMC 10A Problem 12

1. A product is odd only when every number in it is odd. One even number makes the whole product even. So we need all three picks to be odd.
2. From $1$ to $2016$ there are $1008$ odd and $1008$ even numbers.
3. Pick the numbers one at a time.
- First pick: $1008$ odd out of $2016$, a chance of exactly $\tfrac12$.
- Second pick: one odd number is used up, so $1007$ odd out of $2015$ left.
- Third pick: $1006$ odd out of $2014$ left.
4. So
$$p = \frac12 \cdot \frac{1007}{2015} \cdot \frac{1006}{2014}.$$
5. We do not need the exact value, only how it compares with $\tfrac18 = \tfrac12 \cdot \tfrac12 \cdot \tfrac12$. Compare each fraction with $\tfrac12$:
- Half of $2015$ is $1007.5$, and $1007$ is less, so $\tfrac{1007}{2015} < \tfrac12$.
- Half of $2014$ is $1007$, and $1006$ is less, so $\tfrac{1006}{2014} < \tfrac12$.
6. Two of the three factors are below $\tfrac12$, so $p < \tfrac18$.

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

When you pick without putting numbers back, each odd pick removes one odd number but the total shrinks by only one too, so the odd numbers become slightly less than half of what is left. The answer choices only ask which side of $\tfrac18$ the probability is on, so compare each factor with $\tfrac12$ instead of multiplying big numbers.

Test the idea on a tiny case. From $1, 2, 3, 4$ pick three different numbers: there are $4$ ways, and none of them is all odd (there are only two odd numbers). So the chance is $0$, far below $\tfrac18$. With more numbers the effect shrinks, but it always pushes the probability below $\tfrac18$.