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Difference between revisions of "2012 AMC 10A Problems/Problem 25"

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Without loss of generality, assume that <math>0 \le x \le y \le z \le n</math>.
 
Without loss of generality, assume that <math>0 \le x \le y \le z \le n</math>.
  
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Then, the possible choices for <math>x</math>, <math>y</math>, and <math>z</math> is represented by the expression <math>\frac{n^3}{3!}=\frac{n^3}{6}</math>.
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Then, the possible choices for <math>x</math>, <math>y</math>, and <math>z</math> are represented by the expression <math>\frac{n^3}{3!}=\frac{n^3}{6}</math>.
  
 
There are two restrictions: <math>x+1 \le y</math> and <math>y+1 \le z</math>.
 
There are two restrictions: <math>x+1 \le y</math> and <math>y+1 \le z</math>.
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Since <math>0 \le x</math>, then <math>2 \le x'</math>. Thus, <math>2 \le x' \le y' \le z \le n</math>.
 
Since <math>0 \le x</math>, then <math>2 \le x'</math>. Thus, <math>2 \le x' \le y' \le z \le n</math>.
  
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There are <math>n-2</math> ways to choose each number; the successful choices is represented by <math>\frac{(n-2)^3}{6}</math>.
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There are <math>n-2</math> ways to choose each number; the successful choices are represented by <math>\frac{(n-2)^3}{6}</math>.
  
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The probability then, is <math>\text{P}=\frac{\text{successful}}{\text{possible}}=\frac{\frac{n^3}{6}}{\frac{(n-2)^3}{6}}</math> which must be greater than <math>\frac{1}{2}</math>.
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The probability then, is <math>\text{P}=\frac{\text{successful}}{\text{possible}}=\frac{\frac{(n-2)^3}{6}}{\frac{n^3}{6}}</math> which must be greater than <math>\frac{1}{2}</math>.
  
 
Plug in values. Trying <math>(\text{C})</math> gives us <math>\frac{343}{729}</math>, which is less than <math>\frac{1}{2}</math>. Try the next integer, <math>10</math>, which gives us <math>\frac{512}{1000}</math> which is greater than <math>\frac{1}{2}</math>.
 
Plug in values. Trying <math>(\text{C})</math> gives us <math>\frac{343}{729}</math>, which is less than <math>\frac{1}{2}</math>. Try the next integer, <math>10</math>, which gives us <math>\frac{512}{1000}</math> which is greater than <math>\frac{1}{2}</math>.

Revision as of 01:00, 19 February 2012

Problem

Real numbers $x$, $y$, and $z$ are chosen independently and at random from the interval $[0,n]$ for some positive integer $n$. The probability that no two of $x$, $y$, and $z$ are within 1 unit of each other is greater than $\frac {1}{2}$. What is the smallest possible value of $n$?

$\textbf{(A)}\ 7\qquad\textbf{(B)}\ 8\qquad\textbf{(C)}\ 9\qquad\textbf{(D)}\ 10\qquad\textbf{(E)}\ 11$

Solution

Without loss of generality, assume that $0 \le x \le y \le z \le n$.

Then, the possible choices for $x$, $y$, and $z$ are represented by the expression $\frac{n^3}{3!}=\frac{n^3}{6}$.

There are two restrictions: $x+1 \le y$ and $y+1 \le z$.

Now, let $y'=y+1$. Then, $x+2 \le y'$ and $y' \le z$.

Then, let $x'=x+1$. Combining the two inequalities gives us $x' \le y' \le z$.

Since $0 \le x$, then $2 \le x'$. Thus, $2 \le x' \le y' \le z \le n$.

There are $n-2$ ways to choose each number; the successful choices are represented by $\frac{(n-2)^3}{6}$.

The probability then, is $\text{P}=\frac{\text{successful}}{\text{possible}}=\frac{\frac{(n-2)^3}{6}}{\frac{n^3}{6}}$ which must be greater than $\frac{1}{2}$.

Plug in values. Trying $(\text{C})$ gives us $\frac{343}{729}$, which is less than $\frac{1}{2}$. Try the next integer, $10$, which gives us $\frac{512}{1000}$ which is greater than $\frac{1}{2}$.

Thus, our answer is $(\text{D})$.

See Also

2012 AMC 10A (Problems • Answer Key • Resources)
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All AMC 10 Problems and Solutions
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