Difference between revisions of "2016 AMC 10A Problems/Problem 12"
(Added solution) |
(added solution 2) |
||
| Line 3: | Line 3: | ||
<math>\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}</math> | <math>\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}</math> | ||
| − | ==Solution== | + | ==Solution 1== |
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{p<\dfrac{1}{8}}</math>. | ||
| + | |||
| + | ==Solution 2== | ||
| + | Again, for the product to be odd, all three factors have to be odd. There are a total of <math>\binom{2016}{3}</math> ways to choose 3 numbers at random, and there are <math>\binom{1008}{3}</math> to choose 3 odd numbers. Therefore, the probability of choosing 3 odd numbers is <math>\frac{\binom{1008}{3}}{\binom{2016}{3}}</math>. Simplifying this, we obtain <math>\frac{1008*1007*1006}{2016*2015*2014}</math>, which is slightly less than <math>\frac{1}{8}</math>, so our answer is <math>\boxed{p<\dfrac{1}{8}}</math>. | ||
==See Also== | ==See Also== | ||
{{AMC10 box|year=2016|ab=A|num-b=11|num-a=13}} | {{AMC10 box|year=2016|ab=A|num-b=11|num-a=13}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 22:45, 3 February 2016
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
Again, 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. 
