Difference between revisions of "2016 AMC 10A Problems/Problem 12"
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For the product to be odd, all three factors have to be odd. The probability of this is <math>\frac{1008}{2016} \cdot \frac{1007}{2015} \cdot \frac{1006}{2014}</math>. | For the product to be odd, all three factors have to be odd. The probability of this is <math>\frac{1008}{2016} \cdot \frac{1007}{2015} \cdot \frac{1006}{2014}</math>. | ||
| − | <math>\frac{1008}{2016} = \frac{1}{2}</math>, but <math>\frac{1007}{2015}</math> and <math>\frac{1006}{2014}</math> are slightly less than <math>\frac{1}{2}</math>. Thus, the whole product is slightly less than <math>\frac{1}{2} \cdot \frac{1}{2} \cdot \frac{1}{2} = \frac{1}{8}</math>, so <math>\boxed{p<\dfrac{1}{8}}</math>. | + | <math>\frac{1008}{2016} = \frac{1}{2}</math>, but <math>\frac{1007}{2015}</math> and <math>\frac{1006}{2014}</math> are slightly less than <math>\frac{1}{2}</math>. Thus, the whole product is slightly less than <math>\frac{1}{2} \cdot \frac{1}{2} \cdot \frac{1}{2} = \frac{1}{8}</math>, so <math>\boxed{\textbf{(A) }p<\dfrac{1}{8}}</math>. |
==Solution 2== | ==Solution 2== | ||
Revision as of 16:47, 13 October 2018
Contents
Problem
Three distinct integers are selected at random between 
and 
, inclusive. Which of the following is a correct statement about the probability 
that the product of the three integers is odd?
Solution 1
For the product to be odd, all three factors have to be odd. The probability of this is 
.
, but 
and 
are slightly less than 
. Thus, the whole product is slightly less than 
, so 
.
Solution 2
For the product to be odd, all three factors have to be odd. There are a total of 
ways to choose 3 numbers at random, and there are 
to choose 3 odd numbers. Therefore, the probability of choosing 3 odd numbers is 
. Simplifying this, we obtain 
, which is slightly less than 
, so our answer is 
.
See Also
| 2016 AMC 10A (Problems • Answer Key • Resources) | ||
| Preceded by Problem 11 |
Followed by Problem 13 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
| All AMC 10 Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
