Difference between revisions of "2017 AMC 12B Problems/Problem 20"

(Created page with "==Problem 20== Real numbers <math>x</math> and <math>y</math> are chosen independently and uniformly at random from the interval <math>(0,1)</math>. What is the probability th...")
 
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==Solution==
 
==Solution==
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<math>\lfloor\log_2x\rfloor=-1</math> when <math>1/2<x<1</math>. The chance that <math>\lfloor\log_2x\rfloor=\lfloor\log_2y\rfloor=-1</math> is <math>1/2*1/2=1/4</math>.
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<math>\lfloor\log_2x\rfloor=-2</math> when <math>1/4<x<1/2</math>. The chance that <math>\lfloor\log_2x\rfloor=\lfloor\log_2y\rfloor=-2</math> is <math>1/4*1/4=1/16</math>.
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This creates an infinite series, where the common ratio is <math>1/4</math>. The solution is thus <math>1/4 * 1/(1-1/4)=D: 1/3</math>

Revision as of 21:49, 16 February 2017

Problem 20

Real numbers $x$ and $y$ are chosen independently and uniformly at random from the interval $(0,1)$. What is the probability that $\lfloor\log_2x\rfloor=\lfloor\log_2y\rfloor$, where $\lfloor r\rfloor$ denotes the greatest integer less than or equal to the real number $r$ ?

$\textbf{(A)}\ \frac{1}{8}\qquad\textbf{(B)}\ \frac{1}{6}\qquad\textbf{(C)}\ \frac{1}{4}\qquad\textbf{(D)}\ \frac{1}{3}\qquad\textbf{(E)}\ \frac{1}{2}$

Solution

$\lfloor\log_2x\rfloor=-1$ when $1/2<x<1$. The chance that $\lfloor\log_2x\rfloor=\lfloor\log_2y\rfloor=-1$ is $1/2*1/2=1/4$. $\lfloor\log_2x\rfloor=-2$ when $1/4<x<1/2$. The chance that $\lfloor\log_2x\rfloor=\lfloor\log_2y\rfloor=-2$ is $1/4*1/4=1/16$. This creates an infinite series, where the common ratio is $1/4$. The solution is thus $1/4 * 1/(1-1/4)=D: 1/3$

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