Difference between revisions of "2016 AMC 10A Problems/Problem 12"
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<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>. | <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 (Disregards Order)== |
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{\textbf{(A) }p<\dfrac{1}{8}}</math>. | 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{\textbf{(A) }p<\dfrac{1}{8}}</math>. | ||
+ | |||
==Solution 3== | ==Solution 3== | ||
The probability that the product is odd, allowing duplication of the integers, is just <math>\left( \frac{1}{2} \right) ^3 = \frac{1}{8}</math>. Since forbidding duplication reduces the probability of all three integers being odd, we see <math>p<\dfrac{1}{8}</math> and our answer is <math>\boxed{\textbf{(A) }}</math>. | The probability that the product is odd, allowing duplication of the integers, is just <math>\left( \frac{1}{2} \right) ^3 = \frac{1}{8}</math>. Since forbidding duplication reduces the probability of all three integers being odd, we see <math>p<\dfrac{1}{8}</math> and our answer is <math>\boxed{\textbf{(A) }}</math>. |
Revision as of 00:57, 18 October 2024
Contents
[hide]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 (Account for Order)
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 (Disregards Order)
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 .
Solution 3
The probability that the product is odd, allowing duplication of the integers, is just . Since forbidding duplication reduces the probability of all three integers being odd, we see and our answer is .
Video Solution (CREATIVE THINKING)
~Education, the Study of Everything
Video Solution
https://youtu.be/dHY8gjoYFXU?t=300
~IceMatrix
~savannahsolver
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 |
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