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

m (Solution)
 
(9 intermediate revisions by 5 users not shown)
Line 1: Line 1:
==Problem 20==
+
==Problem==
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 that <math>\lfloor\log_2x\rfloor=\lfloor\log_2y\rfloor</math>, where <math>\lfloor r\rfloor</math> denotes the greatest integer less than or equal to the real number <math>r</math> ?
+
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 that <math>\lfloor\log_2x\rfloor=\lfloor\log_2y\rfloor</math>?
  
 
<math>\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}</math>
 
<math>\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}</math>
  
 
==Solution==
 
==Solution==
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>.
+
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\over2}, 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>[{1\over4}, {1\over2})</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>\frac{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>.
 +
 
 +
Solution by: vedadehhc
 +
 
 +
==See Also==
 +
{{AMC12 box|year=2017|ab=B|num-b=19|num-a=21}}
 +
{{MAA Notice}}
 +
 
 +
[[Category:Intermediate Algebra Problems]]
 +
[[Category:Intermediate Probability Problems]]

Latest revision as of 13:56, 15 February 2021

Problem

Real numbers $x$ and $y$ are chosen independently and uniformly at random from the interval $(0,1)$. What is the probability that $\lfloor\log_2x\rfloor=\lfloor\log_2y\rfloor$?

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

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 $[{1\over2}, 1)$. 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 $[{1\over4}, {1\over2})$, 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 $\frac{1}{4}$. Using the formula for the sum of an infinite geometric series, we get that the probability is $\frac{\frac{1}{4}}{1 - \frac{1}{4}} = \boxed{\textbf{(D)}\frac{1}{3}}$.

Solution by: vedadehhc

See Also

2017 AMC 12B (ProblemsAnswer KeyResources)
Preceded by
Problem 19
Followed by
Problem 21
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 12 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Invalid username
Login to AoPS