2016年AMC10竞赛第12题5分
從$$1$$到$$2016$$的整數中隨意地選出三個不同的整數，若這三個數的乘積爲奇數的機率爲$$p$$，則下列何者正確？  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?
For the product to be odd, all three factors have to be odd. The probability of this is $$\frac {1008}{2016}\cdot \frac {1007}{2015}\cdot \frac {1006}{2014}$$.  $$\frac {1008}{2016}=\frac 12$$, but $$\frac {1007}{2015}$$ and $$\frac {1006}{2014}$$ are slightly less than $$\frac 12$$. Thus, the whole product is slightly less than $$\frac 12\cdot \frac 12\cdot \frac 12=\frac 18$$, so $$p\textless\frac 18$$.  For the product to be odd, all three factors have to be odd. There are a total of $$\left(\frac {2016}3\right)$$ ways to choose $$3$$ numbers at random, and there are $$\left(\frac {1008}3\right)$$ to choose $$3$$ odd numbers. Therefore, the probability of choosing $$3$$ odd numbers is $$\frac {(\frac {1008}3)}{(\frac {2016}3)}$$. Simplifying this, we obtain $$\frac {1008\cdot 1007\cdot 1006}{2016\cdot 2015\cdot 2014}$$, which is slightly less than $$\frac 18$$, so our answer is $$p\textless\frac 18$$. 