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Difference between revisions of "2017 AMC 12B Problems/Problem 20"
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==Problem 20==
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==Problem==
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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> ?
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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==
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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>[1/4, 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\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>. Recall that the probability that <math>A</math> or <math>B</math> is the case, where case <math>A</math> and case <math>B</math> are mutually exclusive, is the sum of each individual probability. Symbolically that's <math>P(A \text{ or } B \text{ or } C...) = P(A) + P(B) + P(C)...</math>. Thus, the probability we are looking for is the sum of the probability for each of the cases <math>\lfloor \log_2{x} \rfloor = \lfloor \log_2{y} \rfloor = -1, -2, -3...</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
 
Solution by: vedadehhc
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\\ Edited by: jingwei325
  
 
==See Also==
 
==See Also==
 
{{AMC12 box|year=2017|ab=B|num-b=19|num-a=21}}
 
{{AMC12 box|year=2017|ab=B|num-b=19|num-a=21}}
 
{{MAA Notice}}
 
{{MAA Notice}}
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 +
[[Category:Intermediate Algebra Problems]]
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[[Category:Intermediate Probability Problems]]

Latest revision as of 16:24, 27 September 2022

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 $\frac{1}{4} \cdot \frac{1}{4} = \frac{1}{16}$. Recall that the probability that $A$ or $B$ is the case, where case $A$ and case $B$ are mutually exclusive, is the sum of each individual probability. Symbolically that's $P(A \text{ or } B \text{ or } C...) = P(A) + P(B) + P(C)...$. Thus, the probability we are looking for is the sum of the probability for each of the cases $\lfloor \log_2{x} \rfloor = \lfloor \log_2{y} \rfloor = -1, -2, -3...$. 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 \\ Edited by: jingwei325

See Also

2017 AMC 12B (Problems • Answer Key • Resources)
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

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