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

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==Solution==
 
==Solution==
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First let us take the case that <math>\lfloor \log_2{x} \rfloor = \lfloor \log_2{y} \rfloor = -1</math>. In this case, both <math>x</math> and <math>y</math> lie in the interval <math>[1/2, 1)</math>. The probability of this is <math>\frac{1}{2} \cdot \frac{1}{2} = \frac{1}{4}</math>. Similarly, in the case that <math>\lfloor \log_2{x} \rfloor = \lfloor \log_2{y} \rfloor = -2</math>, <math>x</math> and <math>y</math> lie in the interval <math>[\frac{1}{4}, \frac{1}{2})</math>, and the probability is <math>\frac{1}{4} \cdot \frac{1}{4} = \frac{1}{16}</math>. It is easy to see that the probabilities for <math>\lfloor \log_2{x} \rfloor = \lfloor \log_2{y} \rfloor = n</math> for <math>-infty < n < 0</math> are the infinite geometric series that starts at <math>\frac{1}{4}</math> and with common ratio <math>{1}{4}</math>. Using the formula for the sum of an infinite geometric series, we get that the probability is <math>\frac{\frac{1}{4}}{1 - \frac{1}{4}} = \boxed{\textbf{(D)}\frac{1}{3}}</math>.
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First let us take the case that <math>\lfloor \log_2{x} \rfloor = \lfloor \log_2{y} \rfloor = -1</math>. In this case, both <math>x</math> and <math>y</math> lie in the interval <math>[1/2, 1)</math>. The probability of this is <math>\frac{1}{2} \cdot \frac{1}{2} = \frac{1}{4}</math>. Similarly, in the case that <math>\lfloor \log_2{x} \rfloor = \lfloor \log_2{y} \rfloor = -2</math>, <math>x</math> and <math>y</math> lie in the interval <math>[\frac{1}{4}, \frac{1}{2})</math>, and the probability is <math>\frac{1}{4} \cdot \frac{1}{4} = \frac{1}{16}</math>. It is easy to see that the probabilities for <math>\lfloor \log_2{x} \rfloor = \lfloor \log_2{y} \rfloor = n</math> for <math>-\infty < n < 0</math> are the infinite geometric series that starts at <math>\frac{1}{4}</math> and with common ratio <math>{1}{4}</math>. Using the formula for the sum of an infinite geometric series, we get that the probability is <math>\frac{\frac{1}{4}}{1 - \frac{1}{4}} = \boxed{\textbf{(D)}\frac{1}{3}}</math>.

Revision as of 20:55, 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

First let us take the case that $\lfloor \log_2{x} \rfloor = \lfloor \log_2{y} \rfloor = -1$. In this case, both $x$ and $y$ lie in the interval $[1/2, 1)$. The probability of this is $\frac{1}{2} \cdot \frac{1}{2} = \frac{1}{4}$. Similarly, in the case that $\lfloor \log_2{x} \rfloor = \lfloor \log_2{y} \rfloor = -2$, $x$ and $y$ lie in the interval $[\frac{1}{4}, \frac{1}{2})$, and the probability is $\frac{1}{4} \cdot \frac{1}{4} = \frac{1}{16}$. It is easy to see that the probabilities for $\lfloor \log_2{x} \rfloor = \lfloor \log_2{y} \rfloor = n$ for $-\infty < n < 0$ are the infinite geometric series that starts at $\frac{1}{4}$ and with common ratio ${1}{4}$. Using the formula for the sum of an infinite geometric series, we get that the probability is $\frac{\frac{1}{4}}{1 - \frac{1}{4}} = \boxed{\textbf{(D)}\frac{1}{3}}$.

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