2016 AMC 10A Problems/Problem 12

Problem

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

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 $\boxed{p<\dfrac{1}{8}}$ .

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 $\boxed{p<\dfrac{1}{8}}$ .

2016 AMC 10A (Problems • Answer Key • Resources)
Preceded by Problem 11	Followed by Problem 13
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2016 AMC 10A Problems/Problem 12

Contents

Problem

Solution

Solution 1

Solution 2

See Also