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Difference between revisions of "2016 AMC 10A Problems/Problem 12"
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<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>
  
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== Solution ==
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==Solution 1 (Accounts for Order)==
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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>.
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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>.
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==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>.
 +
 
 +
==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>.
  
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===Solution 1===
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==Video Solution (CREATIVE THINKING)==
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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>.
+
https://youtu.be/rioKxSpmlnU
 +
 
 +
~Education, the Study of Everything
 +
 
 +
 
 +
 
 +
 
 +
==Video Solution==
 +
https://youtu.be/dHY8gjoYFXU?t=300
 +
 
 +
~IceMatrix
  
−
<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>.
+
https://youtu.be/8_clljylXwI
  
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===Solution 2===
+
~savannahsolver
−
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}}

Latest revision as of 00:57, 18 October 2024

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 1 (Accounts for Order)

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 $\boxed{\textbf{(A) }p<\dfrac{1}{8}}$.

Solution 2 (Disregards Order)

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 $\boxed{\textbf{(A) }p<\dfrac{1}{8}}$.

Solution 3

The probability that the product is odd, allowing duplication of the integers, is just $\left( \frac{1}{2} \right) ^3 = \frac{1}{8}$. Since forbidding duplication reduces the probability of all three integers being odd, we see $p<\dfrac{1}{8}$ and our answer is $\boxed{\textbf{(A) }}$.

Video Solution (CREATIVE THINKING)

https://youtu.be/rioKxSpmlnU

~Education, the Study of Everything



Video Solution

https://youtu.be/dHY8gjoYFXU?t=300

~IceMatrix

https://youtu.be/8_clljylXwI

~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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